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Page 1: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

OverviewOverview

61 Discrete Random Variables

62 Binomial Probability Distribution

63 Continuous Random Variables and the Normal Probability Distribution

64 Standard Normal Distribution

65 Applications of the Normal Distribution

61 Discrete Random Variables61 Discrete Random Variables

ObjectivesBy the end of this section I will beable tohellip

1) Identify random variables2)Explain what a discrete probability

distribution is and construct probability distribution tables and graphs

3)Calculate the mean variance and standard deviation of a discrete random variable

Random VariablesRandom Variables

A variable whose values are determined by chance

Chance in the definition of a random variable is crucial

Example 62 - Notation Example 62 - Notation for random variablesfor random variables

Suppose our experiment is to toss a single fair die and we are interested in the numberrolled We define our random variable X to be the outcome of a single die rolla Why is the variable X a random variableb What are the possible values that the random variable X can takec What is the notation used for rolling a 5d Use random variable notation to express the

probability of rolling a 5

Example 62 continuedExample 62 continuedSolution

a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment

b) Possible values for X 1 2 3 4 5 and 6

c) When a 5 is rolled then X equals the outcome 5 or X = 5

d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16

Main types of random variablesMain types of random variables

Discrete random variable - a finite or a countable number of values

Continuous random variable can take infinitely many values

Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables

For the following random variables (i) determine

whether they are discrete or continuousand (ii) indicate the possible values they can

takea The number of automobiles owned by a

familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs

Example 63 continuedExample 63 continued

Solution

aSince the possible number of automobiles

owned by a family is finite and may be written as a list of numbers it represents a discrete random variable

The possible values are 0 1 2 3 4

Example 63 continuedExample 63 continued

Solution

bWidth is something that must be measured not counted Width can take infinitely many different

possible values with these values forming an interval on the number line

Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft

Example 63 continuedExample 63 continued

Solution

c The number of games played in the next

World Series can be counted and thus represents a discrete random variable

The possible values are 4 5 6 7

Example 63 continuedExample 63 continued

Solution

d The weight of model year 2007 SUVs must

be measured not counted and so represents a continuous random variable

Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 2: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

61 Discrete Random Variables61 Discrete Random Variables

ObjectivesBy the end of this section I will beable tohellip

1) Identify random variables2)Explain what a discrete probability

distribution is and construct probability distribution tables and graphs

3)Calculate the mean variance and standard deviation of a discrete random variable

Random VariablesRandom Variables

A variable whose values are determined by chance

Chance in the definition of a random variable is crucial

Example 62 - Notation Example 62 - Notation for random variablesfor random variables

Suppose our experiment is to toss a single fair die and we are interested in the numberrolled We define our random variable X to be the outcome of a single die rolla Why is the variable X a random variableb What are the possible values that the random variable X can takec What is the notation used for rolling a 5d Use random variable notation to express the

probability of rolling a 5

Example 62 continuedExample 62 continuedSolution

a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment

b) Possible values for X 1 2 3 4 5 and 6

c) When a 5 is rolled then X equals the outcome 5 or X = 5

d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16

Main types of random variablesMain types of random variables

Discrete random variable - a finite or a countable number of values

Continuous random variable can take infinitely many values

Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables

For the following random variables (i) determine

whether they are discrete or continuousand (ii) indicate the possible values they can

takea The number of automobiles owned by a

familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs

Example 63 continuedExample 63 continued

Solution

aSince the possible number of automobiles

owned by a family is finite and may be written as a list of numbers it represents a discrete random variable

The possible values are 0 1 2 3 4

Example 63 continuedExample 63 continued

Solution

bWidth is something that must be measured not counted Width can take infinitely many different

possible values with these values forming an interval on the number line

Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft

Example 63 continuedExample 63 continued

Solution

c The number of games played in the next

World Series can be counted and thus represents a discrete random variable

The possible values are 4 5 6 7

Example 63 continuedExample 63 continued

Solution

d The weight of model year 2007 SUVs must

be measured not counted and so represents a continuous random variable

Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 3: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Random VariablesRandom Variables

A variable whose values are determined by chance

Chance in the definition of a random variable is crucial

Example 62 - Notation Example 62 - Notation for random variablesfor random variables

Suppose our experiment is to toss a single fair die and we are interested in the numberrolled We define our random variable X to be the outcome of a single die rolla Why is the variable X a random variableb What are the possible values that the random variable X can takec What is the notation used for rolling a 5d Use random variable notation to express the

probability of rolling a 5

Example 62 continuedExample 62 continuedSolution

a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment

b) Possible values for X 1 2 3 4 5 and 6

c) When a 5 is rolled then X equals the outcome 5 or X = 5

d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16

Main types of random variablesMain types of random variables

Discrete random variable - a finite or a countable number of values

Continuous random variable can take infinitely many values

Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables

For the following random variables (i) determine

whether they are discrete or continuousand (ii) indicate the possible values they can

takea The number of automobiles owned by a

familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs

Example 63 continuedExample 63 continued

Solution

aSince the possible number of automobiles

owned by a family is finite and may be written as a list of numbers it represents a discrete random variable

The possible values are 0 1 2 3 4

Example 63 continuedExample 63 continued

Solution

bWidth is something that must be measured not counted Width can take infinitely many different

possible values with these values forming an interval on the number line

Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft

Example 63 continuedExample 63 continued

Solution

c The number of games played in the next

World Series can be counted and thus represents a discrete random variable

The possible values are 4 5 6 7

Example 63 continuedExample 63 continued

Solution

d The weight of model year 2007 SUVs must

be measured not counted and so represents a continuous random variable

Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 4: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 62 - Notation Example 62 - Notation for random variablesfor random variables

Suppose our experiment is to toss a single fair die and we are interested in the numberrolled We define our random variable X to be the outcome of a single die rolla Why is the variable X a random variableb What are the possible values that the random variable X can takec What is the notation used for rolling a 5d Use random variable notation to express the

probability of rolling a 5

Example 62 continuedExample 62 continuedSolution

a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment

b) Possible values for X 1 2 3 4 5 and 6

c) When a 5 is rolled then X equals the outcome 5 or X = 5

d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16

Main types of random variablesMain types of random variables

Discrete random variable - a finite or a countable number of values

Continuous random variable can take infinitely many values

Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables

For the following random variables (i) determine

whether they are discrete or continuousand (ii) indicate the possible values they can

takea The number of automobiles owned by a

familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs

Example 63 continuedExample 63 continued

Solution

aSince the possible number of automobiles

owned by a family is finite and may be written as a list of numbers it represents a discrete random variable

The possible values are 0 1 2 3 4

Example 63 continuedExample 63 continued

Solution

bWidth is something that must be measured not counted Width can take infinitely many different

possible values with these values forming an interval on the number line

Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft

Example 63 continuedExample 63 continued

Solution

c The number of games played in the next

World Series can be counted and thus represents a discrete random variable

The possible values are 4 5 6 7

Example 63 continuedExample 63 continued

Solution

d The weight of model year 2007 SUVs must

be measured not counted and so represents a continuous random variable

Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 5: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 62 continuedExample 62 continuedSolution

a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment

b) Possible values for X 1 2 3 4 5 and 6

c) When a 5 is rolled then X equals the outcome 5 or X = 5

d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16

Main types of random variablesMain types of random variables

Discrete random variable - a finite or a countable number of values

Continuous random variable can take infinitely many values

Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables

For the following random variables (i) determine

whether they are discrete or continuousand (ii) indicate the possible values they can

takea The number of automobiles owned by a

familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs

Example 63 continuedExample 63 continued

Solution

aSince the possible number of automobiles

owned by a family is finite and may be written as a list of numbers it represents a discrete random variable

The possible values are 0 1 2 3 4

Example 63 continuedExample 63 continued

Solution

bWidth is something that must be measured not counted Width can take infinitely many different

possible values with these values forming an interval on the number line

Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft

Example 63 continuedExample 63 continued

Solution

c The number of games played in the next

World Series can be counted and thus represents a discrete random variable

The possible values are 4 5 6 7

Example 63 continuedExample 63 continued

Solution

d The weight of model year 2007 SUVs must

be measured not counted and so represents a continuous random variable

Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 6: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Main types of random variablesMain types of random variables

Discrete random variable - a finite or a countable number of values

Continuous random variable can take infinitely many values

Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables

For the following random variables (i) determine

whether they are discrete or continuousand (ii) indicate the possible values they can

takea The number of automobiles owned by a

familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs

Example 63 continuedExample 63 continued

Solution

aSince the possible number of automobiles

owned by a family is finite and may be written as a list of numbers it represents a discrete random variable

The possible values are 0 1 2 3 4

Example 63 continuedExample 63 continued

Solution

bWidth is something that must be measured not counted Width can take infinitely many different

possible values with these values forming an interval on the number line

Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft

Example 63 continuedExample 63 continued

Solution

c The number of games played in the next

World Series can be counted and thus represents a discrete random variable

The possible values are 4 5 6 7

Example 63 continuedExample 63 continued

Solution

d The weight of model year 2007 SUVs must

be measured not counted and so represents a continuous random variable

Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 7: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables

For the following random variables (i) determine

whether they are discrete or continuousand (ii) indicate the possible values they can

takea The number of automobiles owned by a

familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs

Example 63 continuedExample 63 continued

Solution

aSince the possible number of automobiles

owned by a family is finite and may be written as a list of numbers it represents a discrete random variable

The possible values are 0 1 2 3 4

Example 63 continuedExample 63 continued

Solution

bWidth is something that must be measured not counted Width can take infinitely many different

possible values with these values forming an interval on the number line

Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft

Example 63 continuedExample 63 continued

Solution

c The number of games played in the next

World Series can be counted and thus represents a discrete random variable

The possible values are 4 5 6 7

Example 63 continuedExample 63 continued

Solution

d The weight of model year 2007 SUVs must

be measured not counted and so represents a continuous random variable

Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 8: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 63 continuedExample 63 continued

Solution

aSince the possible number of automobiles

owned by a family is finite and may be written as a list of numbers it represents a discrete random variable

The possible values are 0 1 2 3 4

Example 63 continuedExample 63 continued

Solution

bWidth is something that must be measured not counted Width can take infinitely many different

possible values with these values forming an interval on the number line

Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft

Example 63 continuedExample 63 continued

Solution

c The number of games played in the next

World Series can be counted and thus represents a discrete random variable

The possible values are 4 5 6 7

Example 63 continuedExample 63 continued

Solution

d The weight of model year 2007 SUVs must

be measured not counted and so represents a continuous random variable

Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 9: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 63 continuedExample 63 continued

Solution

bWidth is something that must be measured not counted Width can take infinitely many different

possible values with these values forming an interval on the number line

Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft

Example 63 continuedExample 63 continued

Solution

c The number of games played in the next

World Series can be counted and thus represents a discrete random variable

The possible values are 4 5 6 7

Example 63 continuedExample 63 continued

Solution

d The weight of model year 2007 SUVs must

be measured not counted and so represents a continuous random variable

Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 10: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 63 continuedExample 63 continued

Solution

c The number of games played in the next

World Series can be counted and thus represents a discrete random variable

The possible values are 4 5 6 7

Example 63 continuedExample 63 continued

Solution

d The weight of model year 2007 SUVs must

be measured not counted and so represents a continuous random variable

Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 11: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 63 continuedExample 63 continued

Solution

d The weight of model year 2007 SUVs must

be measured not counted and so represents a continuous random variable

Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 12: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Discrete Probability DistributionsDiscrete Probability Distributions

Provides all the possible values that the random variable can assume

Together with the probability associated with each value

Can take the form of a table graph or formula

Describe populations not samples

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 13: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 65 - Probability Example 65 - Probability distribution tabledistribution table

Construct the probability distribution table of the number of heads observed whentossing a fair coin twice

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 14: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 65 continuedExample 65 continued

Solution

The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261

Table 62 Probability distribution of number of heads on two fair coin tosses

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 15: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

RequirementsRequirements

Probability Distribution of a Discrete Random Variable

The sum of the probabilities of all the possible values of a discrete random variable must equal 1

That is ΣP(X) = 1

The probability of each value of X must be between 0 and 1 inclusive

That is 0 le P(X ) le 1

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 16: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in

probability distribution tables

Identify patterns more quickly

FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 17: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Mean of a Discrete Random Mean of a Discrete Random VariableVariable

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times

Also called the expected value or expectation of the random variable X

Denoted as E(X )

Holds for discrete and continuous random variables

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 18: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable

Multiply each possible value of X by its probability

Add the resulting products

X P X

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 19: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable

Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 20: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 68 continuedExample 68 continued

Solution

Multiply each possible outcome (value of X) by its probability P(X)

Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on

Add these four products to find the mean

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 21: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 68 continuedExample 68 continued

Solution

μ=15(007) + 16(017) + 17(029) +18(047) = 1716

The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 22: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable

22

2

Definition Formulas

X P X

X P X

2 2 2

2 2

Computational Formulas

X P X

X P X

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 23: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

SummarySummary

Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text

Random variables are variables whose value is determined at least partly by chance

Discrete random variables take values that are either finite or countable and may be put in a list

Continuous random variables take an infinite number of possible values represented by an interval on the number line

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 24: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

SummarySummary

Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable

Such a distribution can take the form of a table graph or formula

Probability distributions describe populations not samples

We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 25: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Explain what constitutes a binomial experiment

2)Compute probabilities using the binomial probability formula

3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance

and standard deviation of the binomial random variable

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 26: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Binomial ExperimentBinomial Experiment

Four Requirements

1)Each trial of the experiment has only two possible outcomes (success or failure)

2)Fixed number of trials

3)Experimental outcomes are independent of each other

4)Probability of observing a success remains the same from trial to trial

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 27: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

NotationNotation

Table 66 Notation for binomial experiments and the binomial distribution

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 28: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Formula for the Number of Formula for the Number of CombinationsCombinations

The number of combinations of X items chosen from n different items is given by

where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1

n X

nC

X n X

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 29: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league

Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 30: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 613 continuedExample 613 continued

Solution

Number of combinations of n = 5

Volleyball teams taken X = 2 at a time

Ten games will be held

5 2

5 5 4 3 2 1 12010

2 5 2 2 1 3 2 1 2 6C

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 31: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Binomial Probability Binomial Probability Distribution FormulaDistribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

P(X) = (nCX) pX (1 - p)n-X

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 32: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Binomial Distribution TablesBinomial Distribution Tables

n is the number of trialsX is the number of successesp is the probability of observing a success

FIGURE 67 Excerpt from the binomial tables

See Example 616 on page 278 for more information

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 33: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation

Mean (or expected value) μ = n p

Variance σ2 = n p (1 - p)

Standard deviation 1n p p

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 34: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students

Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 35: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 618 continuedExample 618 continued

Solution

The binomial random variable here is X = the number of left-handed students

a Here n = 200 and p = 010

So the expected number of left-handed students in a sample of 200 is

E(X) = μ = n p = (200)(010) = 20

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 36: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

SummarySummary

The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials

The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 37: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

SummarySummary

Binomial probabilities can also be found using the binomial tables or using technology

There are formulas for finding the mean variance and standard deviation of a binomial random variable

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 38: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution

ObjectivesBy the end of this section I will beable tohellip

1) Identify a continuous probability distribution

2)Explain the properties of the normal probability distribution

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 39: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

(a) Relatively small sample (n = 100)with large classwidths (05 lb)

FIGURE 615FIGURE 615

(b) Large sample(n = 200) with smaller class widths (02 lb)

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 40: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Figure 615 continuedFigure 615 continued

(c) Very large sample (n = 400) with very small class widths (01 lb)

(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 41: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Continuous Probability DistributionsContinuous Probability Distributions

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take

Density curve is drawn above the horizontal axis

Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 42: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

RequirementsRequirements

1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)

2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 43: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

ProbabilityProbability

Probability for Continuous Distributions is represented by area under the curve above an interval

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 44: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the

world

Population is said to be normally distributed the data values follow a normal probability distribution

Specific population mean μ

Specific population standard deviation σ

μ and σ are parameters of the normal distribution

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 45: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

FIGURE 619FIGURE 619

The normal distribution is symmetric about its mean μ

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 46: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ

2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution

3) It has inflection points at μ-σ and μ+σ

4) The total area under the curve equals 1

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 47: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under

the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)

6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 48: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems

1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X

Insert the mean μ in the center of the number line

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 49: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems

2) Mark on the number line the value of X indicated in the problem

Shade in the desired area under the normal curve to the right or left of this value

3) Proceed to find the desired area or probability

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 50: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

SummarySummary

Continuous random variables assume infinitely many possible values with no gap between the values

Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 51: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

SummarySummary

The normal distribution is the most important continuous probability distribution

It is symmetric about its mean μ and has standard deviation σ

One should always sketch a picture of a normal probability problem to help solve it

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 52: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

64 Standard Normal 64 Standard Normal DistributionDistribution

ObjectivesBy the end of this section I will beable tohellip

1)Find areas under the standard normal curve given a Z-value

2)Find the standard normal Z-value given an area

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 53: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution

A normal distribution with

mean μ = 0 and

standard deviation σ = 1

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 54: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Steps for finding areas under the standard normal curve

Table 67Table 67

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 55: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities

Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 56: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 629 continuedExample 629 continuedSolution

a

The probability that Z is greater than -125 is 08944

That is P(Z gt 125) = 08944

bThe probability that Z is between -1 and 1 is

06826

That is P(1 lt Z lt 1) = 06826

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 57: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 630 Example 630

Find the value of Z with area 090 to the left

Find the standard normal Z-value that has area 090 to the left of it

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 58: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 630 continuedExample 630 continuedSolution

Step 1Draw the standard normal curve Label the Z-value Z1

Step 2Shade the area to the left of Z1 Remember that we are given an area and

are looking for a value of Z Label the area to the left of Z1 with the given

area (090) as shown in Figure 635

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 59: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 630 continuedExample 630 continued

Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see

Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 60: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 630 continuedExample 630 continued

FIGURE 635

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 61: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

SummarySummary

The standard normal distribution has mean μ=0 and standard deviation σ=1

This distribution is often called the Z distribution

The Z table and technology can be used to find areas under the standard normal curve

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 62: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

SummarySummary

In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z

The Z table and technology can also be used to find a value of Z given a probability or an area under the curve

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 63: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip

1)Compute probabilities for a given value of any normal random variable

2)Find the appropriate value of any normal random variable given an area or probability

3)Calculate binomial probabilities using the normal approximation to the binomial distribution

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 64: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Standardizing a Normal Standardizing a Normal Random VariableRandom Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

xZ

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 65: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean

μ and the standard deviation σ

Draw the normal curve for X and shade the desired area

Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 66: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade

the area corresponding to the shaded area in the graph of X

Step4Find the area under the standard normal

curve using either (a) the Z table or (b) technology

This area is equal to the area under the normal curve for X drawn in Step 1

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 67: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 1

Determine X μ and σ and draw the normal curve for X

Shade the desired area

Mark the position of X1 the unknown value of X

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 68: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2

Find the Z-value corresponding to the desired area

Look up the area you identified in Step 1 on the inside of the Z table

If you do not find the exact value of your area by convention choose the area that is closest

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 69: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities

Step 3

Transform this value of Z into a value of X which is the solution

Use the formula X1 = Zσ + μ

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 70: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues

Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 71: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 638 continuedExample 638 continued

Solution

Step 1 Determine X μ and σ and draw the normal

curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1

and X2 as shown in Figure 649

FIGURE 649

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 72: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 638 continuedExample 638 continued

Solution

Step 2

Find the Z-values corresponding to the desired area

The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975

Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196

Looking up area 0975 on the inside of the Z table gives us Z2 = 196

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 73: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ

X1 = Z1σ + μ

=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ

=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of

2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 74: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution

For the binomial random variable X with probability of success p and number of trials n

if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean

μX = np

and standard deviation σX = 1 np p

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 75: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

SummarySummary

Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64

Methods for finding probabilities for a given value of the normal random variable X were discussed

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5

Page 76: Overview 6.1 Discrete Random Variables 6.2 Binomial Probability Distribution 6.3 Continuous Random Variables and the Normal Probability Distribution 6.4

SummarySummary

For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ

The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5