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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
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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
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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.
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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
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184 CHAPTER 5 Discrete Probability Distributions
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.
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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
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186 CHAPTER 5 Discrete Probability Distributions
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
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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
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188 CHAPTER 5 Discrete Probability Distributions
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
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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-
folio would you recommend? Explain.
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.
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190 CHAPTER 5 Discrete Probability Distributions
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-
folio would you recommend? Explain.
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,
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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
Sequence with probability
Sequence with probability
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),
192 CHAPTER 5 Discrete Probability Distributions
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.
-
5.3 Binomial Distribution 193
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
EXAMPLE 5.2Determining
Givenand p = 0.1n = 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
EXAMPLE 5.3Determining
Givenand p = 0.1n = 4
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.)
-
194 CHAPTER 5 Discrete Probability Distributions
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.
-
5.3 Binomial Distribution 195
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
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196 CHAPTER 5 Discrete Probability Distributions
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
-
198 CHAPTER 5 Discrete Probability Distributions
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
-
200 CHAPTER 5 Discrete Probability Distributions
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
-
202 CHAPTER 5 Discrete Probability Distributions
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.
-
204 CHAPTER 5 Discrete Probability Distributions
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
-
206 CHAPTER 5 Discrete Probability Distributions
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?
-
208 CHAPTER 5 Discrete Probability Distributions
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.
-
210 CHAPTER 5 Discrete Probability Distributions
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.
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212 CHAPTER 5 Discrete Probability Distributions
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