chapter 8 random variables

Copyright © 2013, 2009, and 2007, Pearson Education, Inc.

Chapter 8Random Variables

• Probability Distributions

• Discrete Random Variables

• Continuous Random Variables

• Binomial Probability Model

Copyright © 2013, 2009, and 2007, Pearson Education, Inc.2

A random variable is a numerical measurement of the outcome of a random phenomenon. It is a quantity that can take on different values.

Often, the randomness results from selecting a random sample for a population

or performing a randomized experiment

Randomness


Use letters near the end of the alphabet, such as x, to symbolize: variables a particular value of the random variable

Use a capital letter, such as X, to refer to the random variable itself.

Example: Flip a coin three times X= number of heads in the 3 flips; defines the random

variable x=2; represents a possible value of the random variable

Random Variable


The probability distribution of a random variable specifies its possible values and their probabilities.

Note: It is the randomness of the variable that allows us to specify probabilities for the outcomes.

Probability Distribution


A discrete random variable X takes a set of separate values (such as 0,1,2,…) as its possible outcomes.

Its probability distribution assigns a probability P(x) to each possible value x:

For each x, the probability P(x) falls between 0 and 1.

The sum of the probabilities for all the possible x

values equals 1.

Probability Distribution of a Discrete Random Variable


Let

X = the discrete random variable.k = a number the discrete r.v. could assume.P(X = k) is the probability that X equals k.

Probability distribution function (pdf) of X is a table or rule that assigns probabilities to possible values of X.

Probability Distribution of a Discrete Random Variable


• 35% of students taking four courses,

• 45% taking five,

• and remaining 20% are taking six courses.

• Let X = number of courses a randomly selected student is taking

• The probability distribution function of X can be given by:

Example: Number of Courses


Example: Number of Courses


Family has 3 children. Probability of a girl is ½.What are the probabilities of having 0, 1, 2, or 3 girls?

Sample Space: For each birth, write either B or G. There are eight possible arrangements of B and G for three births. These are the simple events.

Sample Space and Probabilities: The eight simple events are equally likely.

Random Variable X: number of girls in three births. For each simple event, the value of X is the number of G’s listed.

Example: Number of Girls


Probability Distribution Table



Probability DistributionFunction for Number of Girls X: Graph of the pdf of X:



What is the estimated probability of at least three home runs?

Example: Number of Home Runs in a Game

Table 8.1 Probability Distribution of Number of Home Runs in a Game for San Francisco Giants


The probability of at least three home runs in a game is

P(3)+P(4)+P(5 or more)= 0.0556 + 0.0185 + 0 = 0.0741




Cumulative distribution function (cdf) for a random variable X is a rule or table that provides the probabilities P(X ≤ k) for any real number k.

Cumulative probability = probability that X is less than or equal to a particular value.

Cumulative Distribution Function for a Discrete Random Variable


Cumulative Distribution Function for Number of Girls


What is the probability that a family with 3 children will have at least one child of each sex?

If X = Number of Girls then either family has one girl and two boys (X = 1) or two girls and one boy (X = 2).

P(X = 1 or X = 2) = P(X = 1) + P(X = 2)

= 3/8 + 3/8 = 6/8 = 3/4

Probability distribution for number of girls:

Example: A Mixture of Children


The mean of a probability distribution for a discrete random variable is:

where the sum is taken over all possible values of x.

The mean of a probability distribution is denoted by the parameter, .

The mean is a weighted average; values of x that are more likely receive greater weight P(x).

Mean of a Discrete Probability Distribution

)(xpx


The mean of a probability distribution of a random variable X is also called the expected value of X.

The expected value reflects not what we’ll observe in a single observation, but rather what we expect for the average in a long run of observations.

It is not unusual for the expected value of a random variable to equal a number that is NOT a possible outcome.

Expected Value of X


Find the mean of this probability distribution.




The mean:

= 0(0.3889) + 1(0.3148) + 2(0.2222) + 3(0.0556) + 4(0.0185)

= 0 * P(0) + 1 * P(1) + 2 * P(2) + 3 * P(3) + 4 * P(4)

= 1

)(xpx



The standard deviation of a probability distribution, denoted by the parameter, , measures variability from the mean.

Larger values of correspond to greater spread.

Roughly, describes how far the random variable falls, on the average, from the mean of its distribution.

The Standard Deviation of a Discrete Random Variable


The standard deviation of a random variable is roughly the average distance the random variable falls from its mean, or expected value, over the long run.

If X is a random variable with possible values x1, x2, x3, . . . , occurring with probabilities p1, p2, p3, . . . , and expected value E(X) = , then

Standard Deviation of a Discrete Random Variable

ii

ii

pxX

pxXVX

2

22

ofDeviation Standard

of Variance


Two plans for investing $100 – which would you choose?

Example: Stability or Excitement?


Expected Value For Each Plan:

Plan 1:E(X ) = $5,000(.001) + $1,000(.005) + $0(.994) = $10.00

Plan 2: E(Y ) = $20(.3) + $10(.2) + $4(.5) = $10.00



Variability for Each Plan:



Plan 1: V(X ) = $29,900.00 and = $172.92

Plan 2: V(X ) = $48.00 and = $6.93

The possible outcomes for Plan 1 are much more variable. If you wanted to invest cautiously, you would choose Plan 2, but if you wanted to have the chance to gain a large amount of money, you would choose Plan 1.



More Examples From Midterm II Practice Sheet


A continuous random variable has an infinite continuum of possible values in an interval.

Examples are: time, age and size measures such as height and weight.

Continuous variables are usually measured in a discrete manner because of rounding.

Continuous Random Variable


A continuous random variable has possible values that form an interval. Its probability distribution is specified by a curve.

Each interval has probability between 0 and 1.

The interval containing all possible values has probability equal to 1.

Probability Distribution of a Continuous Random Variable


Figure 8.2 Probability Distribution of Commuting Time. The area under the curve for values higher than 45 is 0.15. Question: Identify the area under the curve represented by the probability that commuting time is less than 15 minutes, which equals 0.29.

Smooth curve approximation

Probability Distribution of a Continuous Random Variable


Useful Probability Relationships


Probabilities for Bell – Shaped Distributions or Continuous

Random Variables

Bell-Shaped Distributions


The normal distribution is symmetric, bell-shaped and characterized by its mean and standard deviation .

The normal distribution is the most important distribution in statistics.

Many distributions have an approximately normal distribution.

The normal distribution also can approximate many discrete distributions well when there are a large number of possible outcomes.

Many statistical methods use it even when the data are not bell shaped.

Normal Distribution


Normal distributions are Bell shaped Symmetric around the mean

The mean ( ) and the standard deviation ( ) completely describe the density curve.

Increasing/decreasing moves the curve along the horizontal axis.

Increasing/decreasing controls the spread of the curve.

Normal Distribution


Within what interval do almost all of the men’s heights fall? Women’s height?

Figure 8.4 Normal Distributions for Women’s Height and Men’s Height. For each different combination of and values, there is a normal distribution with mean and standard deviation . Question: Given that = 70 and = 4, within what interval do almost all of the men’s heights fall?

Normal Distribution


≈ 68% of the observations fall within one standard deviation of the mean.

≈ 95% of the observations fall within two standard deviations of the mean.

≈ 99.7% of the observations fall within three standard deviations of the mean.

Empirical Rule or 68-95-99.7 Rule for Any Normal Curve

Figure 8.5 The Normal Distribution. The probability equals approximately 0.68 within 1 standard deviation of the mean, approximately 0.95 within 2 standard deviations, and approximately 0.997 within 3 standard deviations. Question: How do these probabilities relate to the empirical rule?


Empirical Rule or 68 – 95 – 99.7% Rule


Heights of adult women can be approximated by a normal

distribution, inches; inches

68-95-99.7 Rule for women’s heights:

68% are between 61.5 and 68.5 inches

95% are between 58 and 72 inches

99.7% are between 54.5 and 75.5 inches

Example : 68-95-99.7% Rule

65 3.5

[ 65 3.5]

[ 2 65 2(3.5) 65 7]

[ 3 65 3(3.5) 65 10.5]


The z-score for a value x of a random variable is the number of standard deviations that x falls from the mean.

A negative (positive) z-score indicates that the value is below (above) the mean.

Z-scores can be used to calculate the probabilities of a normal random variable using the normal tables in the back of the book.

Z-Scores and the Standard Normal Distribution

zx


The formula for converting any value x to a z-score is

.

A z-score measures the number of standard deviations that a value falls from the mean

Standard Score (z – score)

x

zdeviation Standard

MeanValue


Shifting data: Adding (or subtracting) a constant to every data

value adds (or subtracts) the same constant to measures of position.

Adding (or subtracting) a constant to each value will increase (or decrease) measures of position: center, percentiles, max or min by the same constant.

Its shape and spread - range, IQR, standard deviation - remain unchanged.

Shifting and Rescaling Data


The following histograms show a shift from men’s actual weights to kilograms above recommended weight:



Rescaling data: When we multiply (or divide) all the data values

by any constant, all measures of position (such as the mean, median, and percentiles) and measures of spread (such as the range, the IQR, and the standard deviation) are multiplied (or divided) by that same constant.



The men’s weight data set measured weights in kilograms. If we want to think about these weights in pounds, we would rescale the data:



Examples: Midterm Exam II Practice Sheet


A standard normal distribution has mean and standard deviation .

When a random variable has a normal distribution and its values are converted to z-scores by subtracting the mean and dividing by the standard deviation, the z-scores follow the standard normal distribution.

Z-Scores and the Standard Normal Distribution

0 1


Table Z enables us to find normal probabilities. It tabulates the normal cumulative probabilities

falling below the point .

To use the table: Find the corresponding z-score. Look up the closest standardized score (z) in the

table. First column gives z to the first decimal place. First row gives the second decimal place of z.

The corresponding probability found in the body of the table gives the probability of falling below the z-score.

Table Z: Standard Normal Probabilities

z


Finding Probabilities Using The Standard Normal Table (Table Z)

The figure shows us how to find the area to the left when we have a z-score of 1.80:


Find the probability that a normal random variable takes a value less than 1.43 standard deviations above ;

Example: Using Table Z

( 1.43) 0.9236P z


Find the probability that a normal random variable assumes a value within 1.43 standard deviations of .

Probability below

Probability below


1.43 0.9236

1.43 0.0764

( 1.43 1.43) 0.9236 0.0764 0.8472P z


Figure 8.7 The Normal Cumulative Probability, Less than z Standard Deviations above the Mean. Table Z lists a cumulative probability of 0.9236 for , so 0.9236 is the probability less than 1.43 standard deviations above the mean of any normal distribution (that is, below ). The complement probability of 0.0764 is the probability above in the right tail.


1.43z

1.43 1.43


Sometimes we start with areas and need to find the corresponding z-score or even the original data value.

Example: What z-score represents the first quartile in a Normal model?

From Percentiles to z - Scores


Look in Table Z for an area of 0.2500.

The exact area is not there, but 0.2514 is pretty close.

This figure is associated with z = -0.67, so the first quartile is 0.67 standard deviations below the mean.

From Percentiles to z - Scores


To solve some of our problems, we will need to find the value of z that corresponds to a certain normal cumulative probability.

To do so, we use Table A in reverse.

Rather than finding z using the first column (value of z up to one decimal) and the first row (second decimal of z). Find the probability in the body of the table. The z-score is given by the corresponding

values in the first column and row.

How Can We Find the Value of z for a Certain Cumulative Probability?


Example: Find the value of z for a cumulative probability of 0.025.

Look up the cumulative probability of 0.025 in the body of Table A.

A cumulative probability of 0.025

corresponds to .

Thus, the probability that a normal

random variable falls at least

1.96 standard deviations

below the mean is 0.025.

How Can We Find the Value of z for a Certain Cumulative Probability?

1.96z


When you actually have your own data, you must check to see whether a Normal model is reasonable.Looking at a histogram of the data is a good way to check that the underlying distribution is roughly unimodal and symmetric. A more specialized graphical display that can help you decide whether a Normal model is appropriate is the Normal probability plot.If the distribution of the data is roughly Normal, the Normal probability plot approximates a diagonal straight line. Deviations from a straight line indicate that the distribution is not Normal.

Are You Normal? Normal Probability Plots


Nearly Normal data have a histogram and a Normal probability plot that look somewhat like this example:



A skewed distribution might have a histogram and Normal probability plot like below. In such cases it is unwise to use the Normal Model.



Z-scores can be used to compare observations from different normal distributions.

Picture the Scenario: There are two primary standardized tests used by

college admissions, the SAT and the ACT.

You score 650 on the SAT which has and and 30 on the ACT which has and .

How can we compare these scores to tell which score is relatively higher?

Example: Comparing Test Scores That Use Different Scales

500 100 21.0 4.7


Compare z-scores:

SAT:

ACT:

Since your z-score is greater for the ACT, you performed relatively better on this exam.

Using Z-scores to Compare Distributions

z650 500

1001.5

z30 21

4.71.91


If we’re given a value x and need to find a probability, convert x to a z-score using , use a table of normal probabilities (or software, or a calculator) to get a cumulative probability and then convert it to the probability of interest

If we’re given a probability and need to find the value of x , convert the probability to the related cumulative probability, find the z-score using a normal table (or software, or a calculator), and then evaluate .

SUMMARY: Using Z-Scores to Find Normal Probabilities or Random Variable x Values

( ) /z x

x z


Examples: Midterm Exam II Practice Sheet


Probabilities When Each Observation Has

Two Possible Outcomes

Binomial Probability Model


We use the binomial distribution when each observation is binary: it has one of two possible outcomes.

Examples: Accept or decline an offer from a bank for a

credit card. Have or do not have health insurance. Vote yes or no on a referendum.

The Binomial Distribution: Probabilitiesfor Counts with Binary Data


Each of n trials has two possible outcomes: “success” or “failure”.

Each trial has the same probability of success, denoted by p, so the probability of a failure is denoted by .

The n trials are independent, That is, the result for one trial does not depend on the results of other trials.

The binomial random variable X is the number of successes in the n trials.

Conditions for the Binomial Distribution

1 p


Probabilities for a Binomial Distribution

P(x) n!

x!(n - x)!px (1 p)n x, x 0,1,2,...,n

When the number of trials n is large, it’s tedious to write out all the possible outcomes in the sample space. There is a formula you can use to find binomial probabilities for any n.

Denote the probability of success on a trial by p. For n independent trials, the probability of x successes equals:


Rules for factorials:

For example,

Factorials

! *( 1)*( 2)...2*1n n n n

1! 1

0! 1

4! 4*3*2*1 24


Example: Midterm II Exam Practice Sheet


John Doe claims to possess extrasensory perception (ESP).

An experiment is conducted: A person in one room picks one of the integers 1, 2,

3, 4, 5 at random. In another room, John Doe identifies the number he

believes was picked. Three trials are performed for the experiment. John Doe got the correct answer twice.

Example: An ESP Experiment


If John Doe does not actually have ESP and is actually guessing the number, what is the probability that he’d make a correct guess on two of the three trials? The three ways John Doe could make two correct

guesses in three trials are: SSF, SFS, and FSS.

Each of these has probability: .

The total probability of two correct guesses is

.


2(0.2) (0.8) 0.032

3(0.032) 0.096


The probability of exactly 2 correct guesses is the binomial probability with n = 3 trials, x = 2 correct guesses and p = 0.2 probability of a correct guess.

P(2) 3!

2!1!(0.2)2(0.8)1 3(0.04)(0.8) 0.096



Before using the binomial distribution, check that its three conditions apply:

Binary data (success or failure)

The same probability of success for each trial (denoted by p)

Independent trials

Warning!!!!!!!!!!!!!!!!!!


Example: Midterm Exam II Practice Sheet


Binomial Mean and Standard Deviation:

The binomial probability distribution for n trials with probability p of success on each trial has mean and standard deviation given by:

Mean and Standard Deviation of the Binomial Distribution

np, np(1-p)


Example: Midterm II Exam Practice Sheet


In 2006, the New York City Police Department (NYPD) confronted approximately 500,000 pedestrians for suspected criminal violations.

88.9% were non-white. Meanwhile, according to the 2006 American

Community Survey conducted by the U.S. Census Bureau, of the more than 8 million individuals living in New York City, 44.6% were white.

Are the data presented above evidence of racial profiling in police officers’ decisions to confront particular individuals?

Example: Checking for Racial Profiling


Assume: 500,000 confrontations as n = 500,000 trials P(driver is non-white) is p = 0.554

Calculate the mean and standard deviation of this binomial distribution:

351)554)(0.446500,000(0.

277,000 554)500,000(0.



Recall: Empirical Rule When a distribution is bell-shaped, close to 100% of the

observations fall within 3 standard deviations of the mean:

If no racial profiling is taking place, we would not be surprised if between about 275,947 and 278,053 of the 500,000 people stopped were non-white. However, 88.9% of all stops, or 500,000(0.889) = 444,500 involved non-whites. This suggests that the number of non-whites stopped is much higher than we would expect if the probability of confronting a pedestrian were the same for each resident, regardless of their race.

278,0533(351)277,000 3

275,947 3(351) - 000,2773 -u



Extra – Credit Work Follows


A linear combination of random variables, X, Y, . . . is a combination of the form:

L = aX + bY + …

where a, b, etc. are numbers – positive or negative.

Most common:Sum = X + YDifference = X – Y

Section 8.8 Sums, Differences of Discrete RVs


L = aX + bY + …

The mean of L is:

Mean(L) = a Mean(X) + b Mean(Y) + …

Most common:

Mean( X + Y) = Mean(X) + Mean(Y)

Mean(X – Y) = Mean(X) – Mean(Y)

Section 8.8 Means of Linear Combination


If X, Y, . . . are independent random variables, then

Variance(L) = a2 Variance(X) + b2 Variance(Y) + …

Most common:

Variance( X + Y) = Variance(X) + Variance(Y)

Variance(X – Y) = Variance(X) + Variance(Y)

Section 8.8 Variances of Linear Combination


Extra – Credit Exercises

chapter 8 random variables

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