Chapter 5 Discrete Probability Distributions.pdf

USING STATISTICS @ Saxon HomeImprovement

5.1 The Probability Distributionfor a Discrete RandomVariableExpected Value of a Discrete

Random VariableVariance and Standard

Deviation of a DiscreteRandom Variable

5.2 Covariance and ItsApplication in FinanceCovarianceExpected Value, Variance, and

Standard Deviation of theSum of Two RandomVariables

Portfolio Expected Returnand Portfolio Risk

5.3 Binomial Distribution

5.4 Poisson Distribution

5.5 HypergeometricDistribution

5.6 Online Topic Using thePoisson Distribution toApproximate the BinomialDistribution

USING STATISTICS @ Saxon HomeImprovement Revisited

CHAPTER 5 EXCEL GUIDE

CHAPTER 5 MINITABGUIDE

Discrete ProbabilityDistributions5

Learning ObjectivesIn this chapter, you learn:

The properties of a probability distribution

To compute the expected value and variance of a probability distribution

To calculate the covariance and understand its use in finance

To compute probabilities from the binomial, Poisson, and hypergeometricdistributions

How the binomial, Poisson, and hypergeometric distributions can be used to solve business problems

181

You are an accountant for the Saxon Home Improvement Company, which uses a state-of-the-art accounting information system to manage its accounting andfinancial operations.Accounting information systems collect, process, store, transform, and distribute financial information to decision makers both internal and external to a business organization (see reference 7). These systems continuously audit accounting information,looking for errors or incomplete or improbable information. For example, when customers of theSaxon Home Improvement Company submit online orders, the companys accounting informationsystem reviews the order forms for possible mistakes. Any questionable invoices are tagged and

included in a daily exceptions report. Recent data collected by the company show that the likelihood is 0.10 that an order form will be tagged. Saxon would like to

determine the likelihood of finding a certain number of taggedforms in a sample of a specific size. For example, what would

be the likelihood that none of the order forms are taggedin a sample of four forms? That one

of the order forms is tagged?

U S I N G S TAT I S T I C S

@ Saxon Home Improvement

182 CHAPTER 5 Discrete Probability Distributions

H ow could the Saxon Home Improvement Company determine the solution to thistype of probability problem? One way is to use a model, or small-scale representa-tion, that approximates the process. By using such an approximation, Saxon man-agers could make inferences about the actual order process. In this case, the Saxon managerscan use probability distributions, mathematical models suited for solving the type of probabil-ity problems the managers are facing.

This chapter introduces you to the concept and characteristics of probability distributions.You will learn how the knowledge about a probability distribution can help you choose be-tween alternative investment strategies. You will also learn how the binomial, Poisson, and hy-pergeometric distributions can be applied to help solve business problems.

5.1 The Probability Distribution for a Discrete Random VariableIn Section 1.4, a numerical variable was defined as a variable that yields numerical responses,such as the number of magazines you subscribe to or your height. Numerical variables are ei-ther discrete or continuous. Continuous numerical variables produce outcomes that come froma measuring process (e.g., your height). Discrete numerical variables produce outcomes thatcome from a counting process (e.g., the number of magazines you subscribe to). This chapterdeals with probability distributions that represent discrete numerical variables.

T A B L E 5 . 1

Probability Distributionof the Number ofInterruptions per Day

Interruptions per Day Probability

0 0.351 0.252 0.203 0.104 0.055 0.05

0 2 3 4 5 X

P (X)

.3

.2

Interruptions per Days

.1

.4

1

F I G U R E 5 . 1Probability distribution of the number of interruptions per day

Expected Value of a Discrete Random VariableThe mean, of a probability distribution is the expected value of its random variable. Tocalculate the expected value, you multiply each possible outcome, x, by its correspondingprobability, and then sum these products.P1X = xi2,

m,

PROBABILITY DISTRIBUTION FOR A DISCRETE RANDOM VARIABLEA probability distribution for a discrete random variable is a mutually exclusive listof all the possible numerical outcomes along with the probability of occurrence of eachoutcome.

For example, Table 5.1 gives the distribution of the number of interruptions per day in a largecomputer network. The list in Table 5.1 is collectively exhaustive because all possible outcomesare included. Thus, the probabilities sum to 1. Figure 5.1 is a graphical representation of Table 5.1.

5.1 The Probability Distribution for a Discrete Random Variable 183

T A B L E 5 . 2

Computing the Expected Value of theNumber of Interruptionsper Day

Interruptions per Day 1xi2 P1X = xi2 xiP1X = xi2

0 0.35 10210.352 = 0.001 0.25 11210.252 = 0.252 0.20 12210.202 = 0.403 0.10 13210.102 = 0.304 0.05 14210.052 = 0.205 0.05 15210.052 = 0.25

1.00 m = E1X2 = 1.40

EXPECTED VALUE, OF A DISCRETE RANDOM VARIABLE

(5.1)

where

outcome of the discrete random variable X

probability of occurrence of the ith outcome of XP1X = xi2 =xi = the ith

m = E1X2 = aN

i=1xi P1X = xi2

m,

For the probability distribution of the number of interruptions per day in a large computernetwork (Table 5.1), the expected value is computed as follows, using Equation (5.1), and isalso shown in Table 5.2:

= 1.40

= 0 + 0.25 + 0.40 + 0.30 + 0.20 + 0.25

= 10210.352 + 11210.252 + 12210.202 + 13210.102 + 14210.052 + 15210.052m = E1X2 = a

N

i=1xi P1X = xi2

The expected value is 1.40. The expected value of 1.4 for the number of interruptions perday is not a possible outcome because the actual number of interruptions in a given day must bean integer value. The expected value represents the mean number of interruptions in a given day.

Variance and Standard Deviation of a DiscreteRandom VariableYou compute the variance of a probability distribution by multiplying each possible squareddifference by its corresponding probability, and then summing theresulting products. Equation (5.2) defines the variance of a discrete random variable.

P1X = xi2,[xi - E1X2]2

VARIANCE OF A DISCRETE RANDOM VARIABLE

(5.2)

where

the ith outcome of the discrete random variable X

of occurrence of the ith outcome of XP1X = xi2 = probabilityxi =

s2 = aN

i=1[xi - E1X2]2 P1X = xi2


Equation (5.3) defines the standard deviation of a discrete random variable.

LEARNING THE BASICS5.1 Given the following probability distributions:

T A B L E 5 . 3

Computing the Varianceand Standard Deviationof the Number ofInterruptions per Day

Interruptions per Day 1xi2 P1X = xi2 xiP1X = xi2 [xi - E1X2]2 P1X = xi2

0 0.35 10210.352 = 0.00 10 - 1.42210.352 = 0.6861 0.25 11210.252 = 0.25 11 - 1.42210.252 = 0.0402 0.20 12210.202 = 0.40 12 - 1.42210.202 = 0.0723 0.10 13210.102 = 0.30 13 - 1.42210.102 = 0.2564 0.05 14210.052 = 0.20 14 - 1.42210.052 = 0.3385 0.05 15210.052 = 0.25 15 - 1.42210.052 = 0.648

1.00 m = E1X2 = 1.40 s2 = 2.04

Number of Accidents Daily (X) P1X = xi20 0.101 0.202 0.453 0.154 0.055 0.05

STANDARD DEVIATION OF A DISCRETE RANDOM VARIABLE

(5.3)s = 2s2 = AaNi=1 [xi - E1X 2]2 P1X = xi2The variance and the standard deviation of the number of interruptions per day are com-

puted as follows and in Table 5.3, using Equations (5.2) and (5.3):

= 2.04

= 0.686 + 0.040 + 0.072 + 0.256 + 0.338 + 0.648

+ 14 - 1.42210.052 + 15 - 1.42210.052= 10 - 1.42210.352 + 11 - 1.42210.252 + 12 - 1.42210.202 + 13 - 1.42210.102

s2 = aN

i=1[xi - E1X2]2 P1X = xi2

and

Thus, the mean number of interruptions per day is 1.4, the variance is 2.04, and the standarddeviation is approximately 1.43 interruptions per day.

s = 2s2 = 22.04 = 1.4283

Distribution A Distribution B

X P1X = xi2 X P1X = xi20 0.50 0 0.051 0.20 1 0.102 0.15 2 0.153 0.10 3 0.204 0.05 4 0.50

APPLYING THE CONCEPTS

5.2 The following table contains the probabilitydistribution for the number of traffic accidents

daily in a small city:

SELFTest

Problems for Section 5.1

a. Compute the expected value for each distribution.b. Compute the standard deviation for each distribution.c. Compare the results of distributions A and B.

a. Compute the mean number of accidents per day.b. Compute the standard deviation.

5.2 Covariance and Its Application in Finance 185

5.3 Recently, a regional automobile dealership sent outfliers to perspective customers, indicating that they hadalready won one of three different prizes: a Kia Optima val-ued at $15,000, a $500 gas card, or a $5 Wal-Mart shoppingcard. To claim his or her prize, a prospective customerneeded to present the flier at the dealerships showroom. Thefine print on the back of the flier listed the probabilities ofwinning. The chance of winning the car was 1 out of 31,478,the chance of winning the gas card was 1 out of 31,478, andthe chance of winning the shopping card was 31,476 out31,478.a. How many fliers do you think the automobile dealership

sent out?b. Using your answer to (a) and the probabilities listed on

the flier, what is the expected value of the prize won by aprospective customer receiving a flier?

c. Using your answer to (a) and the probabilities listed onthe flier, what is the standard deviation of the value of theprize won by a prospective customer receiving a flier?

d. Do you think this is an effective promotion? Why or whynot?

5.4 In the carnival game Under-or-Over-Seven, a pair of fairdice is rolled once, and the resulting sum determines whetherthe player wins or loses his or her bet. For example, the playercan bet $1 that the sum will be under 7that is, 2, 3, 4, 5, or6. For this bet, the player wins $1 if the result is under 7 andloses $1 if the outcome equals or is greater than 7. Similarly,the player can bet $1 that the sum will be over 7that is, 8, 9,10, 11, or 12. Here, the player wins $1 if the result is over 7but loses $1 if the result is 7 or under. A third method of playis to bet $1 on the outcome 7. For this bet, the player wins $4if the result of the roll is 7 and loses $1 otherwise.a. Construct the probability distribution representing the

different outcomes that are possible for a $1 bet onunder 7.

b. Construct the probability distribution representing the dif-ferent outcomes that are possible for a $1 bet on over 7.

c. Construct the probability distribution representing thedifferent outcomes that are possible for a $1 bet on 7.

d. Show that the expected long-run profit (or loss) to theplayer is the same, no matter which method of play is used.

Arrivals Frequency

0 141 312 473 414 295 216 107 58 2

5.5 The number of arrivals per minute at a bank located inthe central business district of a large city was recordedover a period of 200 minutes, with the following results:

a. Compute the expected number of mortgages approvedper week.

b. Compute the standard deviation.

a. Compute the expected number of arrivals per minute.b. Compute the standard deviation.

5.6 The manager of the commercial mortgage departmentof a large bank has collected data during the past two yearsconcerning the number of commercial mortgages approvedper week. The results from these two years (104 weeks)indicated the following:

Number of CommercialMortgages Approved Frequeny

0 131 252 323 174 95 66 17 1

5.2 Covariance and Its Application in FinanceIn Section 5.1, the expected value, variance, and standard deviation of a discrete random variableof a probability distribution are discussed. In this section, the covariance between two variablesis introduced and applied to portfolio management, a topic of great interest to financial analysts.

CovarianceThe covariance, , measures the strength of the relationship between two numerical randomvariables, X and Y. A positive covariance indicates a positive relationship. A negative covari-ance indicates a negative relationship. A covariance of 0 indicates that the two variables are in-dependent. Equation (5.4) defines the covariance for a discrete probability distribution.

sXY


To illustrate the covariance, suppose that you are deciding between two different invest-ments for the coming year. The first investment is a mutual fund that consists of the stocks thatcomprise the Dow Jones Industrial Average. The second investment is a mutual fund that is ex-pected to perform best when economic conditions are weak. Table 5.4 summarizes your esti-mate of the returns (per $1,000 investment) under three economic conditions, each with agiven probability of occurrence.

T A B L E 5 . 4

Estimated Returns forEach Investment UnderThree EconomicConditions

Investment

P1xi yi2 Economic Condition Dow Jones Fund Weak-Economy Fund0.2 Recession -$300 +$2000.5 Stable economy +100 +500.3 Expanding economy +250 -100

COVARIANCE

(5.4)

where

random variable X

outcome of X

random variable Y

outcome of Y

of occurrence of the ith outcome of X and the ith outcome of Y

for X and Yi = 1, 2, , N

P1xi yi2 = probabilityyi = ith

Y = discrete

xi = ith

X = discrete

sXY = aN

i=1[xi - E1X2][yi - E1Y2] P1xi yi2

The expected value and standard deviation for each investment and the covariance of thetwo investments are computed as follows:

= -19,275

= -12,045 + 262.5 - 7,492.5

+ 1250 - 6521-100 - 35210.32sXY = 1-300 - 6521200 - 35210.22 + 1100 - 652150 - 35210.52sY = $105.00

= 11,025

Var1Y2 = s2Y = 1200 - 352210.22 + 150 - 352210.52 + 1-100 - 352210.32sX = $193.71

= 37,525

Var1X2 = s2X = 1-300 - 652210.22 + 1100 - 652210.52 + 1250 - 652210.32E1Y2 = mY = 1+200210.22 + 150210.52 + 1-100210.32 = $35E1X2 = mX = 1-300210.22 + 1100210.52 + 1250210.32 = $65Let X = Dow Jones fund and Y = weak-economy fund

5.2 Covariance and Its Application in Finance 187

Thus, the Dow Jones fund has a higher expected value (i.e., larger expected return) than theweak-economy fund but also has a higher standard deviation (i.e., more risk). The covarianceof between the two investments indicates a negative relationship in which the two in-vestments are varying in the opposite direction. Therefore, when the return on one investmentis high, typically, the return on the other is low.

Expected Value, Variance, and Standard Deviation of the Sum of Two Random VariablesEquations (5.1) through (5.3) define the expected value, variance, and standard deviation of aprobability distribution, and Equation (5.4) defines the covariance between two variables, Xand Y. The expected value of the sum of two random variables is equal to the sum of the ex-pected values. The variance of the sum of two random variables is equal to the sum of thevariances plus twice the covariance. The standard deviation of the sum of two random vari-ables is the square root of the variance of the sum of two random variables.

-19,275

EXPECTED VALUE OF THE SUM OF TWO RANDOM VARIABLES

(5.5)

VARIANCE OF THE SUM OF TWO RANDOM VARIABLES

(5.6)

STANDARD DEVIATION OF THE SUM OF TWO RANDOM VARIABLES

(5.7)sX+Y = 2s2X+YVar1X + Y2 = s2X+Y = s2X + s2Y + 2sXY

E1X + Y2 = E1X2 + E1Y2

To illustrate the expected value, variance, and standard deviation of the sum of two ran-dom variables, consider the two investments previously discussed. If fund and

fund, using Equations (5.5), (5.6), and (5.7),

The expected value of the sum of the Dow Jones fund and the weak-economy fund is $100,with a standard deviation of $100. The standard deviation of the sum of the two investments isless than the standard deviation of either single investment because there is a large negativecovariance between the investments.

Portfolio Expected Return and Portfolio RiskNow that the covariance and the expected value and standard deviation of the sum of tworandom variables have been defined, these concepts can be applied to the study of agroup of assets referred to as a portfolio. Investors combine assets into portfolios to re-duce their risk (see references 1 and 2). Often, the objective is to maximize the returnwhile minimizing the risk. For such portfolios, rather than study the sum of two randomvariables, the investor weights each investment by the proportion of assets assigned tothat investment. Equations (5.8) and (5.9) def ine the portfolio expected return andportfolio risk.

sX+Y = $100

= 10,000

= 37,525 + 11,025 + 1221-19,2752s2X+Y = s2X + s2Y + 2sXY

E1X + Y2 = E1X2 + E1Y2 = 65 + 35 = $100Y = weak-economy

X = Dow Jones


Problems for Section 5.2LEARNING THE BASICS5.7 Given the following probability distributions for vari-ables X and Y:

Computea. andb. andc.d. E1X + Y2.sXY.

sY.sX

E1Y2.E1X2

P(XiYi) X Y

0.2 -100 500.4 50 300.3 200 200.1 300 20

P(XiYi) X Y

0.4 100 2000.6 200 100

PORTFOLIO EXPECTED RETURNThe portfolio expected return for a two-asset investment is equal to the weight assigned toasset X multiplied by the expected return of asset X plus the weight assigned to asset Ymultiplied by the expected return of asset Y.

(5.8)

where

portfolio expected return

portion of the portfolio value assigned to asset X

portion of the portfolio value assigned to asset Y

expected return of asset X

expected return of asset Y

PORTFOLIO RISK

(5.9)sp = 2w2s2X + 11 - w22s2Y + 2w11 - w2sXYE1Y2 =E1X2 =

11 - w2 =w =

E1P2 =

E1P2 = wE1X2 + 11 - w2E1Y2

In the previous section, you evaluated the expected return and risk of two differentinvestments, a Dow Jones fund and a weak-economy fund. You also computed the covarianceof the two investments. Now, suppose that you want to form a portfolio of these two invest-ments that consists of an equal investment in each of these two funds. To compute the portfo-lio expected return and the portfolio risk, using Equations (5.8) and (5.9), with

and

Thus, the portfolio has an expected return of $50 for each $1,000 invested (a return of 5%) and hasa portfolio risk of $50. The portfolio risk here is smaller than the standard deviation of either invest-ment because there is a large negative covariance between the two investments. The fact that eachinvestment performs best under different circumstances reduces the overall risk of the portfolio.

= 22,500 = $50sp = 210.522137,5252 + 11 - 0.522111,0252 + 210.5211 - 0.521-19,2752

E1P2 = 10.521652 + 11 - 0.521352 = $50sXY = -19,275,E1X2 = $65, E1Y2 = $35,s2X = 37,525,s2Y = 11,025,

w = 0.50,

5.8 Given the following probability distributions forvariables X and Y:

Computea. andb. andc.d. E1X + Y2.sXY.

sY.sX

E1Y2.E1X2

Problems for Section 5.2 189

5.13 Suppose that in Problem 5.12 you wanted to create aportfolio that consists of stock X and stock Y. Compute theportfolio expected return and portfolio risk for each of thefollowing percentages invested in stock X:a. 30%b. 50%c. 70%d. On the basis of the results of (a) through (c), which port-

folio would you recommend? Explain.

5.14 You are trying to develop a strategy for investing in twodifferent stocks. The anticipated annual return for a $1,000investment in each stock under four different economic con-ditions has the following probability distribution:

5.9 Two investments, X and Y, have the following charac-teristics:

If the weight of portfolio assets assigned to investment X is0.4, compute thea. portfolio expected return.b. portfolio risk.

APPLYING THE CONCEPTS

5.10 The process of being served at a bank consists of twoindependent partsthe time waiting in line and the time ittakes to be served by the teller. Suppose that the time waitingin line has an expected value of 4 minutes, with a standarddeviation of 1.2 minutes, and the time it takes to be served bythe teller has an expected value of 5.5 minutes, with a stan-dard deviation of 1.5 minutes. Compute thea. expected value of the total time it takes to be served at the

bank.b. standard deviation of the total time it takes to be served

at the bank.

5.11 In the portfolio example in this section (seepage 186), half the portfolio assets are invested in the DowJones fund and half in a weak-economy fund. Recalculatethe portfolio expected return and the portfolio risk ifa. 30% of the portfolio assets are invested in the Dow Jones

fund and 70% in a weak-economy fund.b. 70% of the portfolio assets are invested in the Dow Jones

fund and 30% in a weak-economy fund.c. Which of the three investment strategies (30%, 50%, or

70% in the Dow Jones fund) would you recommend?Why?

5.12 You are trying to develop a strategy forinvesting in two different stocks. The anticipated

annual return for a $1,000 investment in each stock underfour different economic conditions has the followingprobability distribution:

SELFTest

s2Y = 15,000, andsXY = 7,500.

E1X2 = $50, E1Y2 = $100,s2X = 9,000,

Returns

Probability Economic Condition Stock X Stock Y

0.1 Recession -100 500.3 Slow growth 0 1500.3 Moderate growth 80 -200.3 Fast growth 150 -100

Compute thea. expected return for stock X and for stock Y.b. standard deviation for stock X and for stock Y.c. covariance of stock X and stock Y.d. Would you invest in stock X or stock Y? Explain.

5.15 Suppose that in Problem 5.14 you wanted to create aportfolio that consists of stock X and stock Y. Compute theportfolio expected return and portfolio risk for each of thefollowing percentages invested in stock X:a. 30%b. 50%c. 70%d. On the basis of the results of (a) through (c), which port-


5.16 You plan to invest $1,000 in a corporate bond fund or ina common stock fund. The following information about theannual return (per $1,000) of each of these investments underdifferent economic conditions is available, along with theprobability that each of these economic conditions will occur:

Compute thea. expected return for stock X and for stock Y.b. standard deviation for stock X and for stock Y.c. covariance of stock X and stock Y.d. Would you invest in stock X or stock Y? Explain.

Returns

Probability Economic Condition Stock X Stock Y

0.1 Recession -50 -1000.3 Slow growth 20 500.4 Moderate growth 100 1300.2 Fast growth 150 200

ProbabilityEconomicCondition

CorporateBond Fund

CommonStock Fund

0.01 Extreme recession 200 9990.09 Recession 70 3000.15 Stagnation 30 1000.35 Slow growth 80 1000.30 Moderate growth 100 1500.10 High growth 120 350

Compute thea. expected return for the corporate bond fund and for the

common stock fund.


5.3 Binomial DistributionThe next three sections use mathematical models to solve business problems.

MATHEMATICAL MODELA mathematical model is a mathematical expression that represents a variable of interest.

First Order Second Order Third Order Fourth Order

Tagged Tagged Not tagged Tagged

What is the probability of having three tagged order forms in a sample of four orders inthis particular sequence? Because the historical probability of a tagged order is 0.10, the prob-ability that each order occurs in the sequence is

First Order Second Order Third Order Fourth Order

p = 0.10 p = 0.10 1 - p = 0.90 p = 0.10

b. standard deviation for the corporate bond fund and forthe common stock fund.

c. covariance of the corporate bond fund and the commonstock fund.

d. Would you invest in the corporate bond fund or the com-mon stock fund? Explain.

e. If you chose to invest in the common stock fund in (d),what do you think about the possibility of losing $999 ofevery $1,000 invested if there is an extreme recession?

5.17 Suppose that in Problem 5.16 you wanted to createa portfolio that consists of the corporate bond fund and

the common stock fund. Compute the portfolio expectedreturn and portfolio risk for each of the followingsituations:a. $300 in the corporate bond fund and $700 in the common

stock fund.b. $500 in each fund.c. $700 in the corporate bond fund and $300 in the common

stock fund.d. On the basis of the results of (a) through (c), which port-


When a mathematical expression is available, you can compute the exact probability ofoccurrence of any particular outcome of the variable.

The binomial distribution is one of the most useful mathematical models. You use thebinomial distribution when the discrete random variable is the number of events of interest ina sample of n observations. The binomial distribution has four basic properties:

The sample consists of a fixed number of observations, n. Each observation is classified into one of two mutually exclusive and collectively

exhaustive categories. The probability of an observation being classified as the event of interest, is constant

from observation to observation. Thus, the probability of an observation beingclassified as not being the event of interest, is constant over all observations.

The outcome of any observation is independent of the outcome of any other observation.

Returning to the Saxon Home Improvement scenario presented on page 181 concerningthe accounting information system, suppose the event of interest is defined as a tagged orderform. You are interested in the number of tagged order forms in a given sample of orders.

What results can occur? If the sample contains four orders, there could be none, one, two,three, or four tagged order forms. No other value can occur because the number of taggedorder forms cannot be more than the sample size, n, and cannot be less than zero. Therefore,the range of the binomial random variable is from 0 to n.

Suppose that you observe the following result in a sample of four orders:

1 - p,

p,

5.3 Binomial Distribution 191

Each outcome is independent of the others because the order forms were selected from anextremely large or practically infinite population and each order form could only be selectedonce. Therefore, the probability of having this particular sequence is

This result indicates only the probability of three tagged order forms (events of interest)from a sample of four order forms in a specific sequence. To find the number of ways ofselecting x objects from n objects, irrespective of sequence, you use the rule of combinationsgiven in Equation (5.10) and previously defined in Equation (4.14) on page 170.

= 0.0009

= 10.10210.10210.10210.902= 10.102310.9021

pp11 - p2p = p311 - p21

COMBINATIONSThe number of combinations of selecting x objects1 out of n objects is given by

(5.10)

where

is called n factorial. By definition, 0! = 1.n! = 1n21n - 12 112

nCx =n!

x!1n - x2!

1On many scientific calculators,there is a button labeled that al-lows you to compute the number ofcombinations. On these calculators,the symbol r is used instead of x.

nCr

With and there are

such sequences. The four possible sequences are

Sequence with probability




Therefore, the probability of three tagged order forms is equal to

You can make a similar, intuitive derivation for the other possible outcomes of the randomvariablezero, one, two, and four tagged order forms. However, as n, the sample size, gets large,the computations involved in using this intuitive approach become time-consuming. Equation

= 142 * 10.00092 = 0.00361Number of possible sequences2 * 1Probability of a particular sequence2

11 - p2ppp = p311 - p21 = 0.00094 = not tagged, tagged, tagged, tagged,

p11 - p2pp = p311 - p21 = 0.00093 = tagged, not tagged, tagged, tagged,

pp11 - p2p = p311 - p21 = 0.00092 = tagged, tagged, not tagged, tagged,

ppp11 - p2 = p311 - p21 = 0.00091 = tagged, tagged, tagged, not tagged,

nCx =n!

x!1n - x2! =4!

3!14 - 32! =4 * 3 * 2 * 113 * 2 * 12112 = 4

x = 3,n = 4

If the likelihood of a tagged order form is 0.1, what is the probability that there are threetagged order forms in the sample of four?

SOLUTION Using Equation (5.11), the probability of three tagged orders from a sample of four is

= 410.1210.1210.1210.92 = 0.0036=

4!

3!112!10.12310.921P1X = 3|n = 4,p = 0.12 = 4!

3!14 - 32!10.12311 - 0.124-3

EXAMPLE 5.1Determining

Givenand p = 0.1n = 4

P(X = 3),


BINOMIAL DISTRIBUTION

(5.11)

where

that events of interest, given n and

of observations

of an event of interest

of not having an event of interest

of events of interest in the sample

the number of combinations of x events of interest out of nobservations

n!

x!1n - x2! =1X = 0, 1, 2, , n2x = number

1 - p = probability

p = probability

n = number

pX = xP1X = x|n,p2 = probability

P1X = x|n,p2 = n!x!1n - x2!px11 - p2n-x

Equation (5.11) restates what was intuitively derived previously. The binomial variable Xcan have any integer value x from 0 through n. In Equation (5.11), the product

represents the probability of exactly x events of interest from n observations in a particular se-quence.

The term

is the number of combinations of the x events of interest from the n observations possible.Hence, given the number of observations, n, and the probability of an event of interest, theprobability of x events of interest is

Example 5.1 illustrates the use of Equation (5.11).

=n!

x!1n - x2! px11 - p2n-xP1X = x|n,p2 = 1Number of combinations2*1Probability of a particular combination2

p,

n!

x!1n - x2!

px11 - p2n-x

(5.11) is the mathematical model that provides a general formula for computing any probabilityfrom the binomial distribution with the number of events of interest, x, given n and p.


Examples 5.2 and 5.3 show the computations for other values of X.

If the likelihood of a tagged order form is 0.1, what is the probability that there are three ormore (i.e., at least three) tagged order forms in the sample of four?

SOLUTION In Example 5.1, you found that the probability of exactly three tagged orderforms from a sample of four is 0.0036. To compute the probability of at least three tagged or-der forms, you need to add the probability of three tagged order forms to the probability offour tagged order forms. The probability of four tagged order forms is

Thus, the probability of at least three tagged order forms is

There is a 0.37% chance that there will be at least three tagged order forms in a sample of four.

= 0.0037

= 0.0036 + 0.0001

P1X 32 = P1X = 32 + P1X = 42

= 110.1210.1210.1210.12(1) = 0.0001=

4!

4!102! 10.12410.920P1X = 4|n = 4,p = 0.12 = 4!

4!14 - 42!10.12411 - 0.124-4



P(X 3),

If the likelihood of a tagged order form is 0.1, what is the probability that there are fewer thanthree tagged order forms in the sample of four?

SOLUTION The probability that there are fewer than three tagged order forms is

Using Equation (5.11) on page 192, these probabilities are

Therefore, could also becalculated from its complement, as follows:

= 1 - 0.0037 = 0.9963

P(X 6 3) = 1 - P(X 3)

P1X 32,P1X 6 32 = 0.6561 + 0.2916 + 0.0486 = 0.9963. P(X 6 3)

P1X = 2|n = 4,p = 0.12 = 4!2!14 - 22! 10.12211 - 0.124-2 = 0.0486

P1X = 1|n = 4,p = 0.12 = 4!1!14 - 12! 10.12111 - 0.124-1 = 0.2916

P1X = 0|n = 4,p = 0.12 = 4!0!14 - 02! 10.12011 - 0.124-0 = 0.6561

P1X 6 32 = P1X = 02 + P1X = 12 + P1X = 22



P(X63),

Computing binomial probabilities become tedious as n gets large. Figure 5.2 shows howbinomial probabilities can be computed by Excel (left) and Minitab (right). Binomial proba-bilities can also be looked up in a table of probabilities, as discussed in the Binomial onlinetopic available on this books companion website. (See Appendix C to learn how to access theonline topic files.)


F I G U R E 5 . 2Excel worksheet and Minitab results for computing binomial probabilities with and p = 0.1n = 4

F I G U R E 5 . 3Histogram of thebinomial probabilitydistribution with and p = 0.1

n = 4

The shape of a binomial probability distribution depends on the values of n andWhenever the binomial distribution is symmetrical, regardless of how large orsmall the value of n. When the distribution is skewed. The closer is to 0.5 andthe larger the number of observations, n, the less skewed the distribution becomes. For ex-ample, the distribution of the number of tagged order forms is highly right skewed because

and (see Figure 5.3).n = 4p = 0.1

pp Z 0.5,p = 0.5,

p.

Observe from Figure 5.3 that unlike the histogram for continuous variables in Section 2.6,the bars for the values are very thin, and there is a large gap between each pair of values. Thatis because the histogram represents a discrete variable. (Theoretically, the bars should have nowidth. They should be vertical lines.)

The mean (or expected value) of the binomial distribution is equal to the product of n andInstead of using Equation (5.1) on page 183 to compute the mean of the probability distri-

bution, you can also use Equation (5.12) to compute the mean for variables that follow the bi-nomial distribution.

p.


On the average, over the long run, you theoretically expecttagged order form in a sample of four orders.

The standard deviation of the binomial distribution can be calculated using Equation (5.13).14210.12 = 0.4 m = E1X2 = np =

MEAN OF THE BINOMIAL DISTRIBUTIONThe mean, of the binomial distribution is equal to the sample size, n, multiplied by theprobability of an event of interest,

(5.12)m = E1X2 = npp.

m,

STANDARD DEVIATION OF THE BINOMIAL DISTRIBUTION

(5.13)s = 2s2 = 2Var1X2 = 2np11 - p2The standard deviation of the number of tagged order forms is

You get the same result if you use Equation (5.3) on page 184.Example 5.4 applies the binomial distribution to service at a fast-food restaurant.

s = 2410.1210.92 = 0.60

EXAMPLE 5.4ComputingBinomialProbabilities

Accuracy in taking orders at a drive-through window is important for fast-food chains.Periodically, QSR Magazine (data extracted from http://www.qsrmagazine.com/reports/drive-thru_time_study/2009/2009_charts/whats_your_preferred_way_to_order_fast_food.html) publishes the results of its surveys. Accuracy is measured as the percentage of or-ders that are filled correctly. Recently, the percentage of orders filled correctly at Wendyswas approximately 89%. Suppose that you go to the drive-through window at Wendys andplace an order. Two friends of yours independently place orders at the drive-through windowat the same Wendys. What are the probabilities that all three, that none of the three, and thatat least two of the three orders will be filled correctly? What are the mean and standard devi-ation of the binomial distribution for the number of orders filled correctly?

SOLUTION Because there are three orders and the probability of a correct order isand Using Equations (5.12) and (5.13),

Using Equation (5.11) on page 192,

= 110.89210.89210.892112 = 0.7050=

3!

3!13 - 32! 10.892310.1120P1X = 3|n = 3,p = 0.892 = 3!

3!13 - 32! 10.892311 - 0.8923-3

= 20.2937 = 0.5419= 2310.89210.112s = 2s2 = 2Var1X2 = 2np11 - p2m = E1X2 = np = 310.892 = 2.67

p = 0.890.89, n = 3

http://www.qsrmagazine.com/reports/drive-thru_time_study/2009/2009_charts/whats_your_preferred_way_to_order_fast_food.htmlhttp://www.qsrmagazine.com/reports/drive-thru_time_study/2009/2009_charts/whats_your_preferred_way_to_order_fast_food.htmlhttp://www.qsrmagazine.com/reports/drive-thru_time_study/2009/2009_charts/whats_your_preferred_way_to_order_fast_food.html


In this section, you have been introduced to the binomial distribution. The binomial distri-bution is an important mathematical model in many business situations.

Problems for Section 5.3LEARNING THE BASICS5.18 If and what is the probability thata.b.c.d.

5.19 Determine the following:a. For and what isb. For and what isc. For and what isd. For and what is

5.20 Determine the mean and standard deviation of therandom variable X in each of the following binomial distri-butions:a. andb. andc. andd. and

APPLYING THE CONCEPTS5.21 The increase or decrease in the price of a stock be-tween the beginning and the end of a trading day is as-sumed to be an equally likely random event. What is theprobability that a stock will show an increase in its clos-ing price on five consecutive days?

p = 0.50n = 3p = 0.80n = 5p = 0.40n = 4p = 0.10n = 4

P1X = 52?p = 0.83,n = 6 P1X = 82?p = 0.50,n = 10P1X = 92?p = 0.40,n = 10 P1X = 02?p = 0.12,n = 4

X 7 1?X 6 2?X 3?X = 4?

p = 0.40,n = 55.22 The U.S. Department of Transportation reported thatin 2009, Southwest led all domestic airlines in on-time ar-rivals for domestic flights, with a rate of 0.825. Using thebinomial distribution, what is the probability that in thenext six flightsa. four flights will be on time?b. all six flights will be on time?c. at least four flights will be on time?d. What are the mean and standard deviation of the number

of on-time arrivals?e. What assumptions do you need to make in (a) through (c)?

5.23 A student is taking a multiple-choice exam inwhich each question has four choices. Assume that thestudent has no knowledge of the correct answers to any ofthe questions. She has decided on a strategy in which shewill place four balls (marked and D) into a box.She randomly selects one ball for each question and re-places the ball in the box. The marking on the ball will de-termine her answer to the question. There are fivemultiple-choice questions on the exam. What is the prob-ability that she will geta. five questions correct?b. at least four questions correct?c. no questions correct?d. no more than two questions correct?

A, B, C,

The mean number of orders filled correctly in a sample of three orders is 2.67, and thestandard deviation is 0.5419. The probability that all three orders are filled correctly is 0.7050,or 70.50%. The probability that none of the orders are filled correctly is 0.0013, or 0.13%. Theprobability that at least two orders are filled correctly is 0.9664, or 96.64%.

= 0.9664

= 0.2614 + 0.7050

P1X 22 = P1X = 22 + P1X = 32= 310.89210.89210.112 = 0.2614=

3!

2!13 - 22! 10.892210.1121P1X = 2|n = 3,p = 0.892 = 3!

2!13 - 22! 10.892211 - 0.8923-2= 111210.11210.11210.112 = 0.0013=

3!

0!13 - 02! 10.892010.1123P1X = 0|n = 3,p = 0.892 = 3!

0!13 - 02! 10.892011 - 0.8923-0

5.4 Poisson Distribution 197

5.24 Investment advisors agree that near-retirees, definedas people aged 55 to 65, should have balanced portfolios.Most advisors suggest that the near-retirees have no morethan 50% of their investments in stocks. However, duringthe huge decline in the stock market in 2008, 22% of near-retirees had 90% or more of their investments in stocks(P. Regnier, What I Learned from the Crash, Money, May2009, p. 114). Suppose you have a random sample of 10people who would have been labeled as near-retirees in2008. What is the probability that during 2008a. none had 90% or more of their investment in stocks?b. exactly one had 90% or more of his or her investment in

stocks?c. two or fewer had 90% or more of their investment in

stocks?d. three or more had 90% or more of their investment in

stocks?

5.25 When a customer places an order with Rudys On-Line Office Supplies, a computerized accounting informa-tion system (AIS) automatically checks to see if thecustomer has exceeded his or her credit limit. Past recordsindicate that the probability of customers exceeding theircredit limit is 0.05. Suppose that, on a given day, 20 cus-tomers place orders. Assume that the number of customersthat the AIS detects as having exceeded their credit limit isdistributed as a binomial random variable.a. What are the mean and standard deviation of the number

of customers exceeding their credit limits?

b. What is the probability that zero customers will exceedtheir limits?

c. What is the probability that one customer will exceed hisor her limit?

d. What is the probability that two or more customers willexceed their limits?

5.26 In Example 5.4 on page 195, you and twofriends decided to go to Wendys. Now, suppose that

instead you go to Popeyes, which last month filled approxi-mately 84.8% of orders correctly. What is the probability thata. all three orders will be filled correctly?b. none of the three will be filled correctly?c. at least two of the three will be filled correctly?d. What are the mean and standard deviation of the binomial

distribution used in (a) through (c)? Interpret these values.

5.27 In Example 5.4 on page 195, you and two friends de-cided to go to Wendys. Now, suppose that instead you go toMcDonalds, which last month filled approximately 90.1%of the orders correctly. What is the probability thata. all three orders will be filled correctly?b. none of the three will be filled correctly?c. at least two of the three will be filled correctly?d. What are the mean and standard deviation of the binomial

distribution used in (a) through (c)? Interpret these values.e. Compare the result of (a) through (d) with those of

Popeyes in Problem 5.26 and Wendys in Example 5.4on page 195.

SELFTest

5.4 Poisson DistributionMany studies are based on counts of the times a particular event occurs in a given area ofopportunity. An area of opportunity is a continuous unit or interval of time, volume, or anyphysical area in which there can be more than one occurrence of an event. Examples of vari-ables that follow the Poisson distribution are the surface defects on a new refrigerator, thenumber of network failures in a day, the number of people arriving at a bank, and the numberof fleas on the body of a dog. You can use the Poisson distribution to calculate probabilities insituations such as these if the following properties hold:

You are interested in counting the number of times a particular event occurs in a given areaof opportunity. The area of opportunity is defined by time, length, surface area, and so forth.

The probability that an event occurs in a given area of opportunity is the same for all theareas of opportunity.

The number of events that occur in one area of opportunity is independent of the num-ber of events that occur in any other area of opportunity.

The probability that two or more events will occur in an area of opportunity approacheszero as the area of opportunity becomes smaller.

Consider the number of customers arriving during the lunch hour at a bank located in thecentral business district in a large city. You are interested in the number of customers who arriveeach minute. Does this situation match the four properties of the Poisson distribution given ear-lier? First, the event of interest is a customer arriving, and the given area of opportunity is de-fined as a one-minute interval. Will zero customers arrive, one customer arrive, two customersarrive, and so on? Second, it is reasonable to assume that the probability that a customer arrivesduring a particular one-minute interval is the same as the probability for all the other one-minute intervals. Third, the arrival of one customer in any one-minute interval has no effect


POISSON DISTRIBUTION

(5.14)

where

probability that events in an area of opportunitygiven

number of events

constant approximated by 2.71828

of events 1x = 0, 1, 2, , q2x = numbere = mathematical

l = expected

l

X = xP1X = x|l2 = the

P1X = x|l2 = e-llxx!

on (i.e., is independent of) the arrival of any other customer in any other one-minute interval.Finally, the probability that two or more customers will arrive in a given time period approacheszero as the time interval becomes small. For example, the probability is virtually zero that twocustomers will arrive in a time interval of 0.01 second. Thus, you can use the Poisson distribu-tion to determine probabilities involving the number of customers arriving at the bank in a one-minute time interval during the lunch hour.

The Poisson distribution has one characteristic, called (the Greek lowercase letterlambda), which is the mean or expected number of events per unit. The variance of a Poissondistribution is also equal to and the standard deviation is equal to The number ofevents, X, of the Poisson random variable ranges from 0 to infinity

Equation (5.14) is the mathematical expression for the Poisson distribution for computingthe probability of X = x events, given that events are expected.l

1q2. 1l.l,l

To illustrate an application of the Poisson distribution, suppose that the mean number ofcustomers who arrive per minute at the bank during the noon-to-1 P.M. hour is equal to 3.0.What is the probability that in a given minute, exactly two customers will arrive? And what isthe probability that more than two customers will arrive in a given minute?

Using Equation (5.14) and the probability that in a given minute exactly two cus-tomers will arrive is

To determine the probability that in any given minute more than two customers will arrive,

Because in a probability distribution, all the probabilities must sum to 1, the terms on the rightside of the equation also represent the complement of the probability that X is lessthan or equal to 2 [i.e., ]. Thus,

Now, using Equation (5.14),

Thus, there is a 57.68% chance that more than two customers will arrive in the same minute.

= 1 - 0.4232 = 0.5768

= 1 - [0.0498 + 0.1494 + 0.2240]

P1X 7 22 = 1 - c e-3.013.020

0!+

e-3.013.0211!

+e-3.013.022

2!d

P1X 7 22 = 1 - P1X 22 = 1 - [P1X = 02 + P1X = 12 + P1X = 22]1 - P1X 22P1X 7 22

P1X 7 22 = P1X = 32 + P1X = 42 + + P1X = q 2

P1X = 2|l = 32 = e-3.013.022

2!=

9

12.7182823122 = 0.2240

l = 3,

5.4 Poisson Distribution 199

Computing Poisson probabilities can be tedious. Figure 5.4 shows how Poissonprobabilities can be computed by Excel (left) and Minitab (right). Poisson probabilities canalso be looked up in a table of probabilities, as discussed in the Poisson online topic avail-able on this books companion website. (See Appendix C to learn how to access onlinetopics.)

F I G U R E 5 . 4Excel worksheet andMinitab results forcomputing Poissonprobabilities with l = 3

The number of work-related injuries per month in a manufacturing plant is known to follow aPoisson distribution with a mean of 2.5 work-related injuries a month. What is the probabilitythat in a given month, no work-related injuries occur? That at least one work-related injuryoccurs?

SOLUTION Using Equation (5.14) on page 198 with (or Excel, Minitab, or a Pois-son table lookup), the probability that in a given month no work-related injuries occur is

The probability that there will be no work-related injuries in a given month is 0.0821, or8.21%. Thus,

The probability that there will be at least one work-related injury is 0.9179, or 91.79%.

= 0.9179

= 1 - 0.0821

P1X 12 = 1 - P1X = 02

P1X = 0|l = 2.52 = e-2.512.5200!

=1

12.7182822.5112 = 0.0821

l = 2.5

EXAMPLE 5.5ComputingPoissonProbabilities


Problems for Section 5.4LEARNING THE BASICS5.28 Assume a Poisson distribution.a. If findb. If findc. If findd. If find

5.29 Assume a Poisson distribution.a. If findb. If findc. If findd. If finde. If find

5.30 Assume a Poisson distribution with What isthe probability thata.b.c.d.

APPLYING THE CONCEPTS5.31 Assume that the number of network errors experi-enced in a day on a local area network (LAN) is distributedas a Poisson random variable. The mean number of networkerrors experienced in a day is 2.4. What is the probabilitythat in any given daya. zero network errors will occur?b. exactly one network error will occur?c. two or more network errors will occur?d. fewer than three network errors will occur?

5.32 The quality control manager of MarilynsCookies is inspecting a batch of chocolate-chip

cookies that has just been baked. If the production process is incontrol, the mean number of chip parts per cookie is 6.0. Whatis the probability that in any particular cookie being inspecteda. fewer than five chip parts will be found?b. exactly five chip parts will be found?c. five or more chip parts will be found?d. either four or five chip parts will be found?

5.33 Refer to Problem 5.32. How many cookies in a batchof 100 should the manager expect to discard if companypolicy requires that all chocolate-chip cookies sold have atleast four chocolate-chip parts?

5.34 The U.S. Department of Transportation maintains sta-tistics for mishandled bags per 1,000 airline passengers. Inthe first nine months of 2009, airlines had mishandled 3.89bags per 1,000 passengers. What is the probability that inthe next 1,000 passengers, airlines will have

SELFTest

X 1?X 7 1?X 6 1?X = 1?

l = 5.0.

P1X 32.l = 5.0, P1X 12.l = 4.0,P1X 12.l = 0.5, P1X 32.l = 8.0,P1X 22.l = 2.0,P1X = 02.l = 3.7, P1X = 12.l = 0.5,P1X = 82.l = 8.0, P1X = 22.l = 2.5,

a. no mishandled bags?b. at least one mishandled bag?c. at least two mishandled bags?

5.35 The U.S. Department of Transportation maintainsstatistics for consumer complaints per 100,000 airline pas-sengers. In the first nine months of 2009, consumer com-plaints were 0.99 per 100,000 passengers. What is theprobability that in the next 100,000 passengers, there will bea. no complaints?b. at least one complaint?c. at least two complaints?

5.36 Based on past experience, it is assumed that thenumber of flaws per foot in rolls of grade 2 paper followsa Poisson distribution with a mean of 1 flaw per 5 feet of pa-per (0.2 flaw per foot). What is the probability that in aa. 1-foot roll, there will be at least 2 flaws?b. 12-foot roll, there will be at least 1 flaw?c. 50-foot roll, there will be more than or equal to 5 flaws

and fewer than or equal to 15 flaws?

5.37 J.D. Power and Associates calculates and publishesvarious statistics concerning car quality. The initial qualityscore measures the number of problems per new car sold.For 2009 model cars, Ford had 1.02 problems per car andDodge had 1.34 problems per car (data extracted fromS. Carty, U.S. Autos Power Forward with Gains in QualitySurvey, USA Today, June 23, 2009, p. 3B). Let the randomvariable X be equal to the number of problems with a newlypurchased 2009 Ford.a. What assumptions must be made in order for X to be dis-

tributed as a Poisson random variable? Are these assump-tions reasonable?

Making the assumptions as in (a), if you purchased a 2009Ford, what is the probability that the new car will haveb. zero problems?c. two or fewer problems?d. Give an operational definition for problem. Why is the

operational definition important in interpreting the initialquality score?

5.38 Refer to Problem 5.37. If you purchased a 2009Dodge, what is the probability that the new car will havea. zero problems?b. two or fewer problems?c. Compare your answers in (a) and (b) to those for the Ford

in Problem 5.37 (b) and (c).

5.39 Refer to Problem 5.37. Another article reported thatin 2008, Ford had 1.12 problems per car and Dodge had1.41 problems per car (data extracted from S. Carty, Ford

5.5 Hypergeometric Distribution 201

Moves Up in Quality Survey, USA Today, June 5, 2008,p. 3B). If you purchased a 2008 Ford, what is the probabilitythat the new car will havea. zero problems?b. two or fewer problems?c. Compare your answers in (a) and (b) to those for the

2009 Ford in Problem 5.37 (b) and (c).

5.40 Refer to Problem 5.39. If you purchased a 2008Dodge, what is the probability that the new car will havea. zero problems?b. two or fewer problems?c. Compare your answers in (a) and (b) to those for the

2009 Dodge in Problem 5.38 (a) and (b).

5.41 A toll-free phone number is available from 9 A.M. to9 P.M. for your customers to register complaints about aproduct purchased from your company. Past history indi-cates that an average of 0.8 calls is received per minute.a. What properties must be true about the situation de-

scribed here in order to use the Poisson distribution tocalculate probabilities concerning the number of phonecalls received in a one-minute period?

Assuming that this situation matches the properties discussedin (a), what is the probability that during a one-minute periodb. zero phone calls will be received?c. three or more phone calls will be received?d. What is the maximum number of phone calls that will be

received in a one-minute period 99.99% of the time?

5.5 Hypergeometric DistributionBoth the binomial distribution and the hypergeometric distribution are concerned with thenumber of events of interest in a sample containing n observations. One of the differences inthese two probability distributions is in the way the samples are selected. For the binomial dis-tribution, the sample data are selected with replacement from a finite population or without re-placement from an infinite population. Thus, the probability of an event of interest, isconstant over all observations, and the outcome of any particular observation is independent ofany other. For the hypergeometric distribution, the sample data are selected without replace-ment from a finite population. Thus, the outcome of one observation is dependent on the out-comes of the previous observations.

Consider a population of size N. Let A represent the total number of events of interest inthe population. The hypergeometric distribution is then used to find the probability of X eventsof interest in a sample of size n, selected without replacement. Equation (5.15) represents themathematical expression of the hypergeometric distribution for finding x events of interest,given a knowledge of and A.n, N,

p,

HYPERGEOMETRIC DISTRIBUTION

(5.15)

where

events of interest, given knowledge of and A

of events of interest in the population

of events that are not of interest in the populationN - A = number

A = number

N = population size

n = sample size

n, N,P1X = x|n, N, A2 = the probability of x

P1X = x|n, N, A2 =aA

xb aN - A

n - xb

aNnb


Because the number of events of interest in the sample, represented by x, cannot be greaterthan the number of events of interest in the population, A, nor can x be greater than the samplesize, n, the range of the hypergeometric random variable is limited to the sample size or to thenumber of events of interest in the population, whichever is smaller.

Equation (5.16) defines the mean of the hypergeometric distribution, and Equation (5.17)defines the standard deviation.

MEAN OF THE HYPERGEOMETRIC DISTRIBUTION

(5.16)

STANDARD DEVIATION OF THE HYPERGEOMETRIC DISTRIBUTION

(5.17)s = AnA1N - A2N 2 AN - nN - 1

m = E1X2 = nAN

of events of interest in the sample

[see Equation (5.10) on page 191]

x n

x A

aAxb = ACxx = number

In Equation (5.17), the expression is a finite population correction factor that

results from sampling without replacement from a finite population.

To illustrate the hypergeometric distribution, suppose that you are forming a team of 8managers from different departments within your company. Your company has a total of 30managers, and 10 of these people are from the finance department. If you are to randomlyselect members of the team, what is the probability that the team will contain 2 managersfrom the finance department? Here, the population of managers within the com-pany is finite. In addition, are from the finance department. A team of mem-bers is to be selected.

Using Equation (5.15),

= 0.298

=a 10!

2!182! b a1202!162!1142! b

a 30!8!1222! b

P1X = 2|n = 8, N = 30, A = 102 =a10

2b a20

6b

a308b

n = 8A = 10N = 30

AN - nN - 1

5.5 Hypergeometric Distribution 203

F I G U R E 5 . 5Excel worksheet and Minitab results for the team member example

You are a financial analyst facing the task of selecting bond mutual funds to purchase for aclients portfolio. You have narrowed the funds to be selected to ten different funds. In order todiversify your clients portfolio, you will recommend the purchase of four different funds. Sixof the funds are short-term corporate bond funds. What is the probability that of the four fundsselected, three are short-term corporate bond funds?

SOLUTION Using Equation (5.15) with and A

The probability that of the four funds selected, three are short-term corporate bond fundsis 0.3810, or 38.10%

= 0.3810

=a 6!

3!132! b a142!112!132! b

a 10!4!162! b

P1X = 3|n = 4, N = 10, A = 62 =a6

3b a4

1b

a104b

= 6,X = 3, n = 4, N = 10,

EXAMPLE 5.6ComputingHypergeometricProbabilities

Thus, the probability that the team will contain two members from the finance departmentis 0.298, or 29.8%.

Computing hypergeometric probabilities can be tedious, especially as N gets large. Figure5.5 shows how the hypergeometric probabilities for the team formation example can be com-puted by Excel (left) and Minitab (right).

Example 5.6 shows an application of the hypergeometric distribution in portfolioselection.


Problems for Section 5.5LEARNING THE BASICS5.42 Determine the following:a. If and findb. If and findc. If and findd. If and find

5.43 Referring to Problem 5.42, compute the mean andstandard deviation for the hypergeometric distributions de-scribed in (a) through (d).

APPLYING THE CONCEPTS5.44 An auditor for the Internal Revenue Serviceis selecting a sample of 6 tax returns for an audit.

If 2 or more of these returns are improper, the entire pop-ulation of 100 tax returns will be audited. What is theprobability that the entire population will be audited if thetrue number of improper returns in the population isa. 25?b. 30?c. 5?d. 10?e. Discuss the differences in your results, depending on the

true number of improper returns in the population.

5.45 The dean of a business school wishes to form anexecutive committee of 5 from among the 40 tenuredfaculty members at the school. The selection is to be ran-dom, and at the school there are 8 tenured faculty membersin accounting. What is the probability that the committeewill containa. none of them?

SELFTest

P1X = 32.A = 3,n = 3, N = 10, P1X = 02.A = 3,n = 5, N = 12,P1X = 12.A = 3,n = 4, N = 6, P1X = 32.A = 5,n = 4, N = 10,

b. at least 1 of them?c. not more than 1 of them?d. What is your answer to (a) if the committee consists of

7 members?

5.46 From an inventory of 30 cars being shipped to a localautomobile dealer, 4 are SUVs. What is the probability thatif 4 cars arrive at a particular dealership,a. all 4 are SUVs?b. none are SUVs?c. at least 1 is an SUV?d. What are your answers to (a) through (c) if 6 cars being

shipped are SUVs?

5.47 A state lottery is conducted in which 6 winningnumbers are selected from a total of 54 numbers. What isthe probability that if 6 numbers are randomly selected,a. all 6 numbers will be winning numbers?b. 5 numbers will be winning numbers?c. none of the numbers will be winning numbers?d. What are your answers to (a) through (c) if the 6 winning

numbers are selected from a total of 40 numbers?

5.48 In Example 5.6 on page 203, a financial analyst wasfacing the task of selecting bond mutual funds to purchasefor a clients portfolio. Suppose that the number of funds hadbeen narrowed to 12 funds instead of the ten funds (still with6 short-term corporate funds) in Example 5.6. What is theprobability that of the four funds selected,a. exactly 1 is a short-term corporate bond funds?b. at least 1 is a short-term corporate bond fund?c. three are short-term corporate bond funds?d. Compare the results of (c) to that of Example 5.6.

5.6 Online Topic: Using the Poisson Distribution to Approximate the Binomial Distribution

Under certain circumstances, you can use the Poisson distribution to approximate the binomialdistribution. To study this topic, read the Section 5.6 online topic file that is available on thisbooks companion website. (See Appendix C to learn how to access the online topic files.)

Key Equations 205

S U M M A R YIn this chapter, you have studied mathematical expectationand three important discrete probability distributions: thebinomial, Poisson, and hypergeometric distributions. In thenext chapter, you will study several important continuousdistributions including the normal distribution.

To help decide what probability distribution to use for aparticular situation, you need to ask the following questions:

Is there a fixed number of observations, n, each ofwhich is classified as an event of interest or not anevent of interest? Or is there an area of opportunity? If

there is a fixed number of observations, n, each ofwhich is classified as an event of interest or not anevent of interest, you use the binomial or hypergeomet-ric distribution. If there is an area of opportunity, youuse the Poisson distribution.

In deciding whether to use the binomial or hypergeo-metric distribution, is the probability of an event ofinterest constant over all trials? If yes, you can use thebinomial distribution. If no, you can use the hypergeo-metric distribution.

In the Saxon Home Improvement scenario at the beginning of thischapter, you were an accountant for the Saxon Home ImprovementCompany. The companys accounting information system automati-cally reviews order forms from online customers for possible mis-takes. Any questionable invoices are tagged and included in a daily

exceptions report. Knowing that the probability that an order will be tagged is 0.10, you wereable to use the binomial distribution to determine the chance of finding a certain number oftagged forms in a sample of size four. There was a 65.6% chance that none of the forms wouldbe tagged, a 29.2% chance that one would be tagged, and a 5.2% chance that two or more wouldbe tagged. You were also able to determine that, on average, you would expect 0.4 forms to betagged, and the standard deviation of the number of tagged order forms would be 0.6. Now thatyou have learned the mechanics of using the binomial distribution for a known probability of0.10 and a sample size of four, you will be able to apply the same approach to any given proba-bility and sample size. Thus, you will be able to make inferences about the online orderingprocess and, more importantly, evaluate any changes or proposed changes to the process.

U S I N G S TAT I S T I C S @ Saxon Home Improvement Revisited

K E Y E Q U AT I O N SExpected Value, of a Discrete Random Variable

(5.1)

Variance of a Discrete Random Variable

(5.2)

Standard Deviation of a Discrete Random Variable

(5.3)s = 2s2 = AaNi=1

[xi - E1X2]2P1X = xi2

s2 = aN

i=1[xi - E1X2]2P1X = xi2

m = E1X2 = aN

i=1xi P1X = xi2m, Covariance

(5.4)

Expected Value of the Sum of Two Random Variables

(5.5)

Variance of the Sum of Two Random Variables

(5.6)

Standard Deviation of the Sum of Two Random Variables

(5.7)sX+Y = 2s2X+YVar1X + Y2 = s2X+Y = s2X + s2Y + 2sXY

E1X + Y2 = E1X2 + E1Y2

sXY = aN

i=1[xi - E1X2][yi - E1Y2] P1xiyi2


K E Y T E R M S

Portfolio Expected Return

(5.8)

Portfolio Risk

(5.9)

Combinations

(5.10)

Binomial Distribution

(5.11)

Mean of the Binomial Distribution

(5.12)

Standard Deviation of the Binomial Distribution

(5.13)s = 2s2 = 2Var1X2 = 2np11 - p2m = E1X2 = np

P1X = x|n,p2 = n!x!1n - x2! px11 - p2n-x

nCx =n!

x!1n - x2!

sp = 2w2s2X + 11 - w22s2Y + 2w11 - w2sXYE1P2 = wE1X2 + 11 - w2E1Y2

Poisson Distribution

(5.14)

Hypergeometric Distribution

(5.15)

Mean of the Hypergeometric Distribution

(5.16)

Standard Deviation of the Hypergeometric Distribution

(5.17)s = AnA1N - A2N2 AN - nN - 1

m = E1X2 = nAN

P1X = x|n, N, A2 =aA

xb aN - A

n - xb

aNnb

P1X = x|l2 = e-llxx!

area of opportunity 197binomial distribution 190covariance, 185expected value 182expected value, of a discrete

random variable 182expected value of the sum of two

random variables 187finite population correction

factor 202

m,

sXY

hypergeometric distribution 201mathematical model 190Poisson distribution 197portfolio 187portfolio expected return 187portfolio risk 187probability distribution for a discrete

random variable 182rule of combinations 191

standard deviation of a discreterandom variable 184

standard deviation of the sum of tworandom variables 187

variance of a discrete random variable 183

variance of the sum of two randomvariables 187

C H A P T E R R E V I E W P R O B L E M SCHECKING YOUR UNDERSTANDING5.49 What is the meaning of the expected value of aprobability distribution?

5.50 What are the four properties that must be present inorder to use the binomial distribution?

5.51 What are the four properties that must be present inorder to use the Poisson distribution?

5.52 When do you use the hypergeometric distribution in-stead of the binomial distribution?

APPLYING THE CONCEPTS5.53 Darwin Head, a 35-year-old sawmill worker, won$1 million and a Chevrolet Malibu Hybrid by scoring15 goals within 24 seconds at the Vancouver CanucksNational Hockey League game (B. Ziemer, Darwin Evolvesinto an Instant Millionaire, Vancouver Sun, February 28,2008, p. 1). Head said he would use the money to pay off hismortgage and provide for his children, and he had no plansto quit his job. The contest was part of the Chevrolet MalibuMillion Dollar Shootout, sponsored by General MotorsCanadian Division. Did GM-Canada risk the $1 million?

Chapter Review Problems 207

No! GM-Canada purchased event insurance from a companyspecializing in promotions at sporting events such as a half-court basketball shot or a hole-in-one giveaway at the localcharity golf outing. The event insurance company estimatesthe probability of a contestant winning the contest, and for amodest charge, insures the event. The promoters pay the in-surance premium but take on no added risk as the insurancecompany will make the large payout in the unlikely eventthat a contestant wins. To see how it works, suppose that theinsurance company estimates that the probability a contest-ant would win a Million Dollar Shootout is 0.001, and thatthe insurance company charges $4,000.a. Calculate the expected value of the profit made by the

insurance company.b. Many call this kind of situation a winwin opportunity

for the insurance company and the promoter. Do youagree? Explain.

5.54 Between 1896 when the Dow Jones Index was createdand 2009, the index rose in 64% of the years (data extractedfrom M. Hulbert, What the Past Cant Tell Investors, TheNew York Times, January 3, 2010, p. BU2). Based on this in-formation, and assuming a binomial distribution, what do youthink is the probability that the stock market will risea. next year?b. the year after next?c. in four of the next five years?d. in none of the next five years?e. For this situation, what assumption of the binomial distri-

bution might not be valid?

5.55 In late 2007, it was reported that 79% of U.S. adultsowned a cell phone (data extracted from E. C. Baig, TipsHelp Navigate Tech-Buying Maze, USA Today, November28, 2007, p. 5B). Suppose that by the end of 2009, that per-centage was 85%. If a sample of 10 U.S. adults is selected,what is the probability thata. 8 own a cell phone?b. at least 8 own a cell phone?c. all 10 own a cell phone?d. If you selected the sample in a particular geographical area

and found that none of the 10 respondents owned a cellphone, what conclusion might you reach about whether thepercentage of cell phone owners in this area was 85%?

5.56 One theory concerning the Dow Jones Industrial Av-erage is that it is likely to increase during U.S. presidentialelection years. From 1964 through 2008, the Dow Jones In-dustrial Average increased in 9 of the 12 U.S. presidentialelection years. Assuming that this indicator is a randomevent with no predictive value, you would expect that the in-dicator would be correct 50% of the time.a. What is the probability of the Dow Jones Industrial Aver-

age increasing in 9 or more of the 12 U.S. presidentialelection years if the probability of an increase in the DowJones Industrial Average is 0.50?

b. What is the probability that the Dow Jones Industrial Aver-age will increase in 9 or more of the 12 U.S. presidential

election years if the probability of an increase in the DowJones Industrial Average in any year is 0.75?

5.57 Errors in a billing process often lead to customer dis-satisfaction and ultimately hurt bottom-line profits. An arti-cle in Quality Progress (L. Tatikonda, A Less CostlyBilling Process, Quality Progress, January 2008, pp.3038) discussed a company where 40% of the bills pre-pared contained errors. If 10 bills are processed, what is theprobability thata. 0 bills will contain errors?b. exactly 1 bill will contain an error?c. 2 or more bills will contain errors?d. What are the mean and the standard deviation of the

probability distribution?

5.58 Refer to Problem 5.57. Suppose that a quality im-provement initiative has reduced the percentage of billscontaining errors to 20%. If 10 bills are processed, what isthe probability thata. 0 bills will contain errors?b. exactly 1 bill will contain an error?c. 2 or more bills will contain errors?d. What are the mean and the standard deviation of the

probability distribution?e. Compare the results of (a) through (c) to those of Prob-

lem 5.57 (a) through (c).

5.59 A study by the Center for Financial Services Innova-tion showed that only 64% of U.S. income earners aged 15and older had bank accounts (A. Carrns, Banks Court a NewClient, The Wall Street Journal, March 16, 2007, p. D1).

If a random sample of 20 U.S. income earners aged 15and older is selected, what is the probability thata. all 20 have bank accounts?b. no more than 15 have bank accounts?c. more than 10 have bank accounts?d. What assumptions did you have to make to answer (a)

through (c)?

5.60 One of the biggest frustrations for the consumerelectronics industry is that customers are accustomed toreturning goods for any reason (C. Lawton, The War onReturns, The Wall Street Journal, May 8, 2008, pp. D1,D6). Recently, it was reported that returns for no troublefound were 68% of all the returns. Consider a sample of20 customers who returned consumer electronics pur-chases. Use the binomial model to answer the followingquestions:a. What is the expected value, or mean, of the binomial dis-

tribution?b. What is the standard deviation of the binomial distribution?c. What is the probability that 15 of the 20 customers made

a return for no trouble found?d. What is the probability that no more than 10 of the cus-

tomers made a return for no trouble found?e. What is the probability that 10 or more of the customers

made a return for no trouble found?


5.61 Refer to Problem 5.60. In the same time period, 27%of the returns were for buyers remorse.a. What is the expected value, or mean, of the binomial

distribution?b. What is the standard deviation of the binomial distribution?c. What is the probability that none of the 20 customers

made a return for buyers remorse?d. What is the probability that no more than 2 of the

customers made a return for buyers remorse?e. What is the probability that 3 or more of the customers

made a return for buyers remorse?

5.62 One theory concerning the S&P 500 Index is that if it in-creases during the first five trading days of the year, it is likelyto increase during the entire year. From 1950 through 2009, theS&P 500 Index had these early gains in 38 years. In 33 of these38 years, the S&P 500 Index increased for the entire year. As-suming that this indicator is a random event with no predictivevalue, you would expect that the indicator would be correct 50%of the time. What is the probability of the S&P 500 Index in-creasing in 33 or more years if the true probability of an increasein the S&P 500 Index isa. 0.50?b. 0.70?c. 0.90?d. Based on the results of (a) through (c), what do you think

is the probability that the S&P 500 Index will increase ifthere is an early gain in the first five trading days of theyear? Explain.

5.63 Spurious correlation refers to the apparent relation-ship between variables that either have no true relationshipor are related to other variables that have not been measured.One widely publicized stock market indicator in the UnitedStates that is an example of spurious correlation is the rela-tionship between the winner of the National Football LeagueSuper Bowl and the performance of the Dow Jones IndustrialAverage in that year. The indicator states that when a teamrepresenting the National Football Conference wins the Su-per Bowl, the Dow Jones Industrial Average will increase inthat year. When a team representing the American FootballConference wins the Super Bowl, the Dow Jones IndustrialAverage will decline in that year. Since the first Super Bowlwas held in 1967 through 2009, the indicator has been cor-rect 33 out of 43 times. Assuming that this indicator is a ran-dom event with no predictive value, you would expect thatthe indicator would be correct 50% of the time.a. What is the probability that the indicator would be cor-

rect 33 or more times in 43 years?b. What does this tell you about the usefulness of this indicator?

5.64 Approximately 300 million golf balls were lost in theUnited States in 2009. Assume that the number of golf ballslost in an 18-hole round is distributed as a Poisson randomvariable with a mean of 5 balls.

a. What assumptions need to be made so that the number ofgolf balls lost in an 18-hole round is distributed as a Pois-son random variable?

Making the assumptions given in (a), what is the probabilitythatb. 0 balls will be lost in an 18-hole round?c. 5 or fewer balls will be lost in an 18-hole round?d. 6 or more balls will be lost in an 18-hole round?

5.65 According to a Virginia Tech survey, college studentsmake an average of 11 cell phone calls per day. Moreover,80% of the students surveyed indicated that their parentspay their cell phone expenses (J. Elliot, ProfessorResearches Cell Phone Usage Among Students, www.physorg.com, February 26, 2007).a. What distribution can you use to model the number of

calls a student makes in a day?b. If you select a student at random, what is the probability

that he or she makes more than 10 calls in a day? Morethan 15? More than 20?

c. If you select a random sample of 10 students, what distri-bution can you use to model the proportion of studentswho have parents who pay their cell phone expenses?

d. Using the distribution selected in (c), what is the proba-bility that all 10 have parents who pay their cell phoneexpenses? At least 9? At least 8?

5.66 Mega Millions is one of the most popular lottery gamesin the United States. Virtually all states participate in Mega Mil-lions. Rules for playing and the list of prizes in most states aregiven below (see megamillions.com).

Rules: Select five numbers from a pool of numbers from 1 to

52 and one Mega Ball number from a second pool ofnumbers from 1 to 52.

Each wager costs $1.

Prizes: Match all five numbers + Mega Ballwin jackpot

(minimum of $12,000,000) Match all five numberswin $250,000 Match four numbers Mega Ballwin $10,000 Match four numberswin $150 Match three numbers Mega Ballwin $150 Match two numbers Mega Ballwin $10 Match three numberswin $7 Match one number Mega Ballwin $3 Match Mega Ballwin $2

Find the probability of winninga. the jackpot.b. the $250,000 prize. (Note that this requires matching all

five numbers but not matching the Mega Ball.)c. $10,000.d. $150.

+

++

+

www.physorg.comwww.physorg.com

Managing Ashland MultiComm Services 209

e. $10.f. $7.g. $3.h. $2.i. nothing.j. All stores selling Mega Millions tickets are required to

have a brochure that gives complete game rules and

probabilities of winning each prize. (The probability ofhaving a losing ticket is not given.) The slogan for all lot-tery games in the state of Ohio is Play Responsibly.Odds Are, Youll Have Fun. Do you think Ohios sloganand the requirement of making available complete gamerules and probabilities of winning is an ethical approachto running the lottery system?

M A N A G I N G A S H L A N D M U LT I C O M M S E R V I C E SThe Ashland MultiComm Services (AMS) marketing depart-ment wants to increase subscriptions for its 3-For-Alltelephone, cable, and Internet combined service. AMSmarketing has been conducting an aggressive direct-market-ing campaign that includes postal and electronic mailings andtelephone solicitations. Feedback from these efforts indicatesthat including premium channels in this combined service isa very important factor for both current and prospective sub-scribers. After several brainstorming sessions, the marketingdepartment has decided to add premium cable channels as ano-cost benefit of subscribing to the 3-For-All service.

The research director, Mona Fields, is planning to con-duct a survey among prospective customers to determine howmany premium channels need to be added to the 3-For-Allservice in order to generate a subscription to the service.Based on past campaigns and on industry-wide data, sheestimates the following:

d. What does this tell you about the previous estimate ofthe proportion of customers who would subscribe tothe 3-For-All service offer?

2. Instead of offering no premium free channels as in Prob-lem 1, suppose that two free premium channels are in-cluded in the 3-For-All service offer, Given past results,what is the probability thata. fewer than 3 customers will subscribe to the 3-For-All

service offer?b. 0 customers or 1 customer will subscribe to the 3-For-

All service offer?c. more than 4 customers will subscribe to the 3-For-All

service offer?d. Compare the results of (a) through (c) to those of 1.

Suppose that in the actual survey of 50 prospective cus-tomers, 6 customers subscribe to the 3-For-All service offer.

e. What does this tell you about the previous estimate ofthe proportion of customers who would subscribe tothe 3-For-All service offer?

f. What do the results in (e) tell you about the effect ofoffering free premium channels on the likelihood ofobtaining subscriptions to the 3-For-All service?

3. Suppose that additional surveys of 50 prospective cus-tomers were conducted in which the number of freepremium channels was varied. The results were asfollows:

Number of FreePremium Channels

Probabilityof Subscriptions

0 0.021 0.042 0.063 0.074 0.085 0.085

1. If a sample of 50 prospective customers is selected and nofree premium channels are included in the 3-For-All serv-ice offer, given past results, what is the probability thata. fewer than 3 customers will subscribe to the 3-For-All

service offer?b. 0 customers or 1 customers will subscribe to the

3-For- All service offer?c. more than 4 customers will subscribe to the 3-For-All

service offer?Suppose that in the actual survey of 50 prospective cus-tomers, 4 customers subscribe to the 3-For-All service offer.

Number of FreePremium Channels

Numberof Subscriptions

1 53 64 65 7

How many free premium channels should the research directorrecommend for inclusion in the 3-For-All service? Explain.


D I G I TA L C A S EApply your knowledge about expected value and the covari-ance in this continuing Digital Case from Chapters 3 and 4.

Open BullsAndBears.pdf, a marketing brochure fromEndRun Financial Services. Read the claims and examinethe supporting data. Then answer the following:

1. Are there any catches about the claims the brochuremakes for the rate of return of Happy Bull and WorriedBear Funds?

2. What subjective data influence the rate-of-return analysesof these funds? Could EndRun be accused of makingfalse and misleading statements? Why or why not?

3. The expected-return analysis seems to show that the Wor-ried Bear Fund has a greater expected return than theHappy Bull Fund. Should a rational investor never investin the Happy Bull Fund? Why or why not?

R E F E R E N C E S1. Bernstein, P. L., Against the Gods: The Remarkable

Story of Risk (New York: Wiley, 1996).2. Emery, D. R., J. D. Finnerty, and J. D. Stowe, Corporate

Financial Management, 3rd ed. (Upper Saddle River,NJ: Prentice Hall, 2007).

3. Kirk, R. L., ed., Statistical Issues: A Reader for the Be-havioral Sciences (Belmont, CA: Wadsworth, 1972).

4. Levine, D. M., P. Ramsey, and R. Smidt, Applied Statis-tics for Engineers and Scientists Using Microsoft Excel

and Minitab (Upper Saddle River, NJ: Prentice Hall,2001).

5. Microsoft Excel 2010 (Redmond, WA: Microsoft Corp.,2010).

6. Minitab Release 16 (State College, PA.: Minitab, Inc.,2010).

7. Moscove, S. A., M. G. Simkin, and N. A. Bagranoff,Core Concepts of Accounting Information Systems,11th ed. (New York: Wiley, 2010).

211

EG5.1 THE PROBABILITY DISTRIBUTIONFOR A DISCRETE RANDOM VARIABLE

In-Depth Excel Use the COMPUTE worksheet of theDiscrete Random Variable workbook as a template for com-puting the expected value, variance, and standard deviation ofa discrete random variable (see Figure EG5.1). The worksheetcontains the data for the Section 5.1 example on page 182 in-volving the number of interruptions per day in a large com-puter network. For other problems, overwrite the X andvalues in columns A and B, respectively. If a problem has moreor fewer than six outcomes, select the cell range A5:E5 and:If the problem has more than six outcomes:

1. Right-click and click Insert from the shortcut menu.

2. If a dialog box appears, click Shift cells down and thenclick OK.

3. Repeat steps 1 and 2 as many times as necessary.

4. Select the formulas in cell range C4:E4 and copy themdown through the new table rows.

5. Enter the new X and values in columns A and B.

If the problem has fewer than six outcomes, right-click andclick Delete from the shortcut menu. If a dialog box ap-pears, click Shift cells up and then click OK. Repeat asmany times as necessary and then enter the new X andvalues in columns A and B.

P1X2

P1X2

P1X2

In the new worksheet (shown in Figure EG5.2 on page 212):

3. Enter the probabilities and outcomes in the table that be-gins in cell B3.

4. Enter 0.5 as the Weight assigned to X.

In-Depth Excel Use the COMPUTE worksheet of thePortfolio workbook, shown in Figure EG5.2, as a templatefor performing portfolio analysis. The worksheet containsthe data for the Section 5.2 investment example on page 186.Overwrite the X and values and the weight assigned tothe X value when you enter data for other problems. If aproblem has more or fewer than three outcomes, first selectrow 5, right-click, and click Insert (or Delete) in the short-cut menu to insert (or delete) rows one at a time. If you

P1X2

C H A P T E R 5 E X C E L G U I D E

F I G U R E E G 5 . 1

Discrete random variable probability worksheet

EG5.2 COVARIANCE AND ITS APPLICATIONIN FINANCE

PHStat2 Use Covariance and Portfolio Analysis toperform portfolio analysis. For example, to create the port-folio analysis for the Section 5.2 investment example onpage 186, select PHStat Decision-Making Covariance and Portfolio Analysis. In the procedures dia-log box (shown below):

1. Enter 5 as the Number of Outcomes.

2. Enter a Title, check Portfolio Management Analysis,and click OK.


insert rows, select the cell range B4:J4 and copy the con-tents of this range down through the new table rows.

The worksheet uses the SUMPRODUCT worksheetfunction to compute the sum of the products of correspondingelements of two cell ranges. The worksheet also contains aCalculations Area that contains various intermediate calcula-tions. Open the COMPUTE_FORMULAS worksheet toexamine all the formulas used in this area.

EG5.3 BINOMIAL DISTRIBUTION

PHStat2 Use Binomial to compute binomial probabilities.For example, to create a binomial probabilities table andhistogram for Example 5.3 on page 193, similar to those inFigures 5.2 and 5.3, select PHStat Probability & Prob.Distributions Binomial. In the procedures dialog box(shown in next column):

1. Enter 4 as the Sample Size.

2. Enter 0.1 as the Prob. of an Event of Interest.

3. Enter 0 as the Outcomes From value and enter 4 as the(Outcomes) To value.

4. Enter a Title, check Histogram, and click OK.

To add columns to the binomial probabilities table forand check Cumulative

Probabilities before clicking OK in step 4.

In-Depth Excel Use the BINOMDIST worksheet functionto compute binomial probabilities. Enter the function asBINOMDIST (X, sample size, cumulative), where X isthe number of events of interest, is the probability of anevent of interest, and cumulative is a True or False value.(When cumulative is True, the function computes the prob-ability of X or fewer events of interest; when cumulative isFalse, the function computes the probability of exactly Xevents of interest.)

Use the COMPUTE worksheet of the Binomial work-book, shown in Figure 5.2 on page 194, as a template forcomputing binomial probabilities. The worksheet containsthe data for the Section 5.3 tagged orders example.Overwrite these values and adjust the table of probabilitiesfor other problems. To create a histogram of the probabilitydistribution, use the instructions in Appendix Section F.5.

EG5.4 POISSON DISTRIBUTION

PHStat2 Use Poisson to compute Poisson probabilities.For example, to create a Poisson probabilities table similarto Figure 5.4 on page 199, select PHStat Probability &Prob. Distributions Poisson. In this procedures dialogbox (shown the top of the next page):

1. Enter 3 as the

Chapter 5 Discrete Probability Distributions.pdf

Documents

Transcript of Chapter 5 Discrete Probability Distributions.pdf