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8. Probability Distributions and Statistics and 10.Probability and Calculus(Combined Chapters)

Calculus with Business Applications II Math 1690

Spring 2011

Calculus with Business Applications II Math 1690 8. Probability Distributions and Statistics and 10. Probability and Calculus(Combined Chapters)

8.1 and 10.1 Probability Distributions and RandomVariables

Random VariableA random variable is a rule that assigns a number to each outcomeof a chance experiment.There are two kinds of random variables: Discrete andContinuous:A random variable X that assumes only discrete values(e.gintegers) is called a discrete random variable.

For Example, suppose an experiment consists of throwing a die andobserving the face that lands up. If we let X denote the outcome ofthe experiment, then X assumes one of the values 1, 2, 3, 4, 5, 6.

A random variable X that can assume any value in an interval iscalled a continuous random variable.

Examples of a continuous random variables are the life span of alight bulb, the length of a telephone call, the length of an infant atbirth.


8.1 and 10.1 Probability Distributions and RandomVariables(Contd.)

Example 1: A pair of dice is rolled. Let X be the random variablethat gives the sum of the faces that fall uppermost.

a. List the outcomes of the experiment; that is, find the domain ofthe function X.b. Find the value assigned to each outcome of the experiment bythe random variable X.c. Find the event that gives X=7.

Probability Distribution of a random Variable X is a table thatgives the distinct(numerical) values of the random variable X andthe probabilities associated with these values.

Example 2: Find the probability distribution of the randomvariable X given in Example 1.

HistogramsA probability distribution of a random variable may be exhibitedgraphically by means of a histogram.



Example 3: Draw the histogram for the probability distribution inExample 2.

Note: In histogram, the probability associated with more than onevalue of the random variable X is given by the sum of the areas ofthe rectangles associated with those values of X. For Example, inExample 1, the probability of the event of getting sum 4 or 5(thatis, X=4 or X=5) is P(X=4)+P(X=5).

Example 4: The probability distribution of the random variable Xis shown in the accompanying table:

x -5 -3 -2 0 2 3

P(X=x) 0.17 0.13 0.33 0.16 0.11 0.10

Finda. P(X ≤ 0) b. P(X ≤ −3) c. P(−2 ≤ X ≤ 2)



Discrete Probability FunctionA discrete probability function P with domain {x1, x2, . . . , xn}satisfies these conditions:

1. 0 ≤ P(xi ) ≤ 1, for 1 ≤ i ≤ n2. P(x1) + P(x2) + . . .P(xn) = 1

Probability Density Function(Continuous Case)A probability density function of a random variable X in aninterval I, where I may be bounded or unbounded, is a nonnegativefunction f having the property that the total area of the regionunder the graph of f is equal to 1.

The probability that an observed value of the random variable Xlies in the interval [a, b] is given by

P(a ≤ X ≤ b) =

∫ b

af (x)dx



Example 5: a. Determine the value of the constant k such thatthe function f (x) = k(5− x) is a probability density function onthe interval [0, 5].b. If X is a continuous random variable with the probability densityfunction given in part a, compute the probability that X willassume a value between 2 and 3.c. Find the probability that X will assume a value 2.

Joint Probability Density FunctionA joint probability density function of the random variables X andY on a region D is a nonnegative function f (x , y) having theproperty ∫∫

D

f (x , y)dA = 1

Thus, the volume of the solid under the graph of f is equal to 1.Calculus with Business Applications II Math 1690 8. Probability Distributions and Statistics and 10. Probability and Calculus(Combined Chapters)


The probability that the observed values of the random variables Xand Y lie in a region R ∈ D is given by

P[(X ,Y ) in R] =

∫∫R

f (x , y)dA

Example 6: Let f (x , y) = xy be the joint probability densityfunction for the random variables X and Y onD = {(x , y)|0 ≤ x ≤ 1; 0 ≤ y ≤ 2}. Find

a. P(0 ≤ X ≤ 1; 0 ≤ Y ≤ 2)b. P(X + 2Y ≤ 1)

Home Work: Section 8.1 on Page 423 problems 5, 13, 15, 19.Section 10.1 on Page 654 problems 17, 21, 31, 41, 43.


8.2, 8.3 and 10.2 Expected Value and Standard Deviation

Mean, Median and ModeThe average, or mean , of the n numbers x1, x2, . . . , xn is x̄ (read”x bar”), where

x̄ =x1 + x2 + · · ·+ xn

n

The median of a group of numbers arranged in increasing ordecreasing order is (a) the middle number if there is an oddnumber of entries or (b) the mean of the two middle numbers ifthere is an even number of entries.

The mode of a group of numbers is the number in the group thatoccurs most frequently.

Example 1: The weights, in ounces, of ten packages of potatochips are16.1 16 15.8 16 15.9 16.1 15.9 1616.2Find the mean, median and mode weights.


8.2, 8.3 and 10.2 Expected Value and StandardDeviation(Contd.)

Expected Value of a Discrete Random variable

Let X denote a random variable that assumes the valuesx1, x2, . . . , xn with associated probabilities p1, p2, . . . , pn,respectively. Then the expected value of X, denoted by E (X ), isgiven by

E (X ) = x1p1 + x2p2 + · · ·+ xnpn

Expected Value of a Continuous Random variable

Suppose the function f defined on the interval [a, b] is theprobability density function associated with a continuous randomvariable X. Then, the expected value of X is

E (X ) =

∫ b

axf (x)dx



Example 2: If a sample of three batteries is selected from a lot often, of which two are defective, what is the expected number ofdefective batteries?(Hint: First find the probability distribution for the randomvariable associated with the number of defective batteries)

Example 3: Find the expected value of a random variable Xassociated with the probability density function f (x) = 3

125x2 overthe interval [0, 5]

OddsIf P(E ) is the probability of an event E occurring, then

1. The odds in favor of E occurring are

P(E )

1− P(E )=

P(E )

P(E c)P(E ) 6= 1



2. The odds against E occurring are

1− P(E )

P(E )=

P(E c)

P(E )P(E ) 6= 0

Probability of an Event(Given the Odds)

If the odds in favor of an event E occurring are a to b, then theprobability of E occurring is

P(E ) =a

a + b

Example 4: Suppose the probability that it will rain tomorrow is0.3.

a. What are the odds that it will rain tomorrow?b. What are the odds that it will not rain tomorrow?



Example 5: The odds in favor of an event E occurring are 9 to 7.What is the probability of E occurring?

Variance and Standard Deviation of a Discrete RandomVariable X

Suppose a random variable has the probability distributionx x1 x2 x3 . . . xn

P(X=x) p1 p2 p3 . . . pn

and expected valueE (X ) = µ

Then the variance of the random variable X is

Var(X ) = p1(x1 − µ)2 + p2(x2 − µ)2 + · · ·+ pn(xn − µ)2

and the standard deviation of X is

σ =√

Var(X )Calculus with Business Applications II Math 1690 8. Probability Distributions and Statistics and 10. Probability and Calculus(Combined Chapters)


Variance and Standard Deviation of a Continuous RandomVariable X

Let X be a continuous random variable with probability densityfunction f (x) on [a, b]. Then the variance of X is

Var(X ) =

∫ b

a(x − µ)2f (x)dx

and the standard deviation of X is

σ =√

Var(X )

Example 6: Find the variance and standard deviation for theExamples 2 and 3.



Chebychev’s Inequality

Let X be a random variable with expected value µ and standarddeviation σ. Then the probability that a randomly chosen outcomeof the experiment lies between µ− kσ and µ + kσ is at least1− 1

k2 ; that is,

P(µ− kσ ≤ X ≤ µ + kσ) ≥ 1− 1

k2

Example 7: Suppose X is a random variable with mean µ andstandard deviation σ. If a large number of trials is observed, atleast what percentage of these values is expected to lie betweenµ− 2σ and µ + 2σ?

Home Work: Odd numbers in Sections 8.2, 8.3 and 10.2.


8.4 The Binomial Distribution

Bernoulli TrialsIn general, experiments with two outcomes are called Bernoullitrials, or binomial trials. We usually label one of the outcomes asuccess and the other a failure.A sequence of Bernoulli(binomial) trials is called a binomialexperiment. A binomial experiment has the following properties:1. The number of trials in the experiment is fixed.2. There are two outcomes of each trial: ”success” and ”failure.”3. The probability of success in each trial is the same.4. The trials are independent of each other.

Note that in a binomial experiment it is customary to denote thenumber of trials by n, the probability of success by p, and theprobability of failure by q.

Computation of Probabilities in Bernoulli TrialsIn a binomial experiment in which the probability of success in anytrial is p, the probability of exactly x successes in n independent


8.4 The Binomial Distribution(Contd.)

trials is given byC (n, x)pxqn−x

If we let X be the random variable that gives the number ofsuccesses in a binomial experiment, then the probability of exactlyx successes in n independent trials may be written

P(X = x) = C (n, x)pxqn−x (x = 0, 1, 2, . . . , n)

The random variable X is called a binomial random variable, andthe probability distribution of X is called a binomial distribution.

Example 1: A fair dice is rolled four times. If a 6 lands uppermostin a trial, then the throw is considered a success. Otherwise, thethrow is considered a failure.a. Find the probability of obtaining exactly 0, 1, 2, 3, and 4successes, respectively, in this experiment.b. Construct the binomial distribution for this experiment and draw



a histogram associated with it.c. What is the probability of getting 0 or 1 success in theexperiment?

Example 2: If the probability that a certain tennis player will servean ace is 1

4 , what is the probability that he will serve exactly twoaces out of five serves? (Assume that the five serves areindependent.)

Mean, Variance, and Standard Deviation of a RandomVariable XIf X is a binomial random variable associated with a binomialexperiment consisting of n trials with probability of success p andprobability of failure q, then the mean(expected value), Variance,and Standard Deviation of X are

µ = E (X ) = np



Var(X ) = npq

σX =√

npq

Example 3: Find the mean, variance, and standard deviation forthe experiment in Example 2.

Home Work: Section 8.4 on page 459 problems 11, 19, 23, 33.


8.5 and 10.3 The Normal Distribution

Normal Distributions are important type of continous probabilitydistributions. The graph of a normal distribution, which is bellshaped, is called a normal curve.The normal curve(and therefore the corresponding normaldistribution) is completely determined by its mean µ and standarddeviation σ. In fact, the normal curve has the followingcharacteristics:1. The curve has a peak at x = µ.2. The curve is symmetric with respect to the vertical line x = µ.3. The curve always lies above the x-axis but approaches the x-axisas x extends indefinitely in either direction.4. The area under the curve is 1.5. For any normal curve, 68% of the area under the curve lieswithin 1 standard deviation of the mean(that is, between µ− σand µ + σ), 95% of the area lies within 2 standard deviations ofthe mean, and 99.7% of the area lies within 3 standard deviationsof the mean.


8.5 and 10.3 The Normal Distribution(Contd.)

The following is the normal curve with mean µ and standarddeviation σ:



The mean µ of a normal distribution determines where the centerof the curve is located, whereas the standard deviation σ of anormal distribution determines the peakedness(or flatness) of thecurve.

The normal curve with mean µ = 0 and standard deviation σ = 1is called the standard normal curve. The correspondingdistribution is called the standard normal distribution. Therandom variable itself is called the standard normal randomvariable and is commonly denoted by Z.

Note that the general normal probability density function withmean µ and standard deviation σ is defined to be

f (x) =1

σ√

2πe−(x−µ)2

2σ2 (−∞ < x < ∞)



Computations of Probabilities Associated with NormalDistributionsThe table for Standard Normal Distribution is found on AppendixD on page 577. We will use the table to solve the followingexamples:

Example 1: Let Z be the standard normal variable. Make a sketchof the appropriate region under the standard normal curve, andthen find the values of

a. P(Z < 1.45) b. P(Z > 1.11)c. P(−1.32 < Z < 1.74)

Example 2: Let Z be the standard normal random variable. Findthe value of z if z satisfies

a. P(Z < z) = 0.8907 b. P(Z > z) = 0.9678c. P(−z < Z < z) = 0.8354



If X is a normal random variable with mean µ and standarddeviation σ, then it can be transformed into the standard normalrandom variable Z by means of the substitution

Z =X − µ

σ

The area of the region under the normal curve(with randomvariable X) between x = a and x = b is equal to the area of theregion under the standard normal curve between z = a−µ

σ and

z = b−µσ . In terms of probabilities associated with these

distributions, we have

P(a < X < b) = P(a− µ

σ< Z <

b − µ

σ)

P(X < b) = P(Z <b − µ

σ)

andCalculus with Business Applications II Math 1690 8. Probability Distributions and Statistics and 10. Probability and Calculus(Combined Chapters)


P(X > a) = P(Z >a− µ

σ)

Example 3: Suppose X is a normal random variable with meanµ = 500 and σ = 75. Find the value of

a. P(X < 750) b. P(X > 350)c. P(400 < X < 600)

Home Work: Section 8.5 on page 469 problems 7, 11, 13, 16, 17,19.


8.6 Applications of the Normal Distribution

Applications Involving Normal Random VariablesExample 1: The medical records of infants delivered at KaiserMemorial Hospital show that the infants’ lengths at birth(ininches) are normally distributed with a mean of 20 and a standarddeviation of 2.6. Find the probability that an infant selected atrandom from among those delivered at the hospital measures

a. More than 22 in.b. Less than 18 in.c. Between 19 and 21 in.(Hint: Let X be the random variable associated with the infants’lengths at birth)

Example 2: The scores on a sociology examination are normallydistributed with a mean of 70 and a standard deviation of 10. Ifthe instructor assigns A’s to 15%, B’s to 25%, C’s to 40%, D’s to15%, and F’s to 5% of the class, find the cutoff points for gradesA-D.



Approximating Binomial DistributionsRecall that a binomial distribution is a probability distribution ofthe form

P(X = x) = C (n, x)pxqn−x x = 0, 1, 2, . . . , n

Theorem 1: Suppose we are given a binomial distributionassociated with a binomial experiment involving n trials, each witha probability of success p and a probability of failure q. Then, if nis large and p is not close to 0 or 1, the binomial distribution maybe approximated by a normal distribution with

µ = np and σ =√

npq

Applications Involving Binomial Random Variables

Example 3: A coin is weighted so that the probability of obtaininga head in a single toss is 0.4. If the coin is



tossed 25 times, what is the probability of obtaining

a. Fewer than 10 heads?b. Between 10 and 12 heads, inclusive?c. More than 15 heads?

Example 4: Colorado Mining and Mineral has 800 employeesengaged in its mining operations. It has been estimated that theprobability of a worker meeting with an accident during a 1-yrperiod is 0.1. What is the probability that more than 70 workerswill meet with an accident during the 1-yr period?

Home Work: Section 8.6 on page 477 problems 3, 5, 10, 14, 15


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