Discrete Random Variables and Probability Distributions

Abstract

In Chaps. 2, 3, and 4, we explored descriptive statistical measures, and we examined probability concepts and techniques in Chap. 5. Here we will build on this foundation as we establish the definitions of discrete and continuous random variables and discuss important discrete probability distributions in terms of specific numerical outcomes.

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

FormalPara Key Terms

Random variable	Hypergeometric distribution
Discrete random variable	Hypergeometric random variable
Continuous random variable	Hypergeometric formula
Probability function	Poisson distribution
Probability distribution	Joint probability function
Probability mass function	Joint probability distribution
Cumulative distribution function	Marginal probability function
Step function	Conditional probability function
Expected value	Conditional probability distribution
Bernoulli process	Covariance
Binomial distribution	Coefficient of correlation
Binomial probability function	Option
Lot acceptance sampling	Random walk

1 Introduction

In Chaps. 2, 3, and 4, we explored descriptive statistical measures, and we examined probability concepts and techniques in Chap. 5. Here we will build on this foundation as we establish the definitions of discrete and continuous random variables and discuss important discrete probability distributions in terms of specific numerical outcomes.

The binomial distribution, hypergeometric distribution, Poisson distribution, and joint probability functions are discussed in detail in this chapter. We also explore the Poisson approximation to the binomial distribution and examine joint probability functions and distributions. Finally, we investigate expected value and variance of the sum of both uncorrelated and correlated random variables.

In Appendix 1, the mean and variance for the binomial distribution are derived. And in Appendix 2, we explain how the binomial distribution can be used in developing the binomial option pricing model.

2 Discrete and Continuous Random Variables

A random experiment generally results in numerical values that can be attached to the possible outcomes. In experiments such as throwing a die or measuring a firm’s net earnings, the outcomes are naturally in numerical form. The possible outcomes of tossing a fair die are 1, 2, 3, 4, 5, and 6, and the corresponding probabilities are $\frac{1}{6}$ for each outcome, as we saw in Chap. 5. The result of a random experiment can be conveniently described by a random variable. A random variable is a variable that assigns a numerical value to each possible outcome of a random experiment. We can think of a random variable as a value or magnitude that changes from occurrence to occurrence in no predictable sequence. A breast cancer screening clinic, for example, has no way of knowing exactly how many women will be screened on any one day. So tomorrow’s number of patients is a random variable. For another example, say a company manufactures TV sets that are sometimes defective. Buyers return the defective sets for repair. A variable used to describe the number of TV sets that will be returned before the warranty runs out is a random variable.

Random variables are either discrete or continuous. A discrete random variable is one that can take on a countable number of values; usually it is an integer. The number of claims on an automobile policy in a particular year is a discrete random variable. Another discrete random variable is the number of defective parts produced in a particular run. Here the discrete random variable can take on the values 0, 1, 2, …, n. If we let X stand for a discrete random variable, then we can use x to represent one of its possible values. In other words, X is a quantity and x a value. For example, before the results of rolling a fair die are observed, the random variable can be used to denote the outcome. This random variable can assume the specific values $x=1,x=2,\ldots,x=6$, and each value has a probability of $\frac{1}{6}$. Other discrete random variables include:

• 1. The number of bids received in a stock offering: x = 0, 1, 2, …

• 2. The number of customers waiting to be served in a bank at a particular time: x = 0, 1, 2, …

• 3. The number of sales made by a salesperson in a given month: x = 0, 1, 2, …

• 4. The number of people in a sample of 800 who favor a particular presidential candidate: x = 0, 1, 2, …, 800

In contrast, a continuous random variable can take on an uncountable number of values within an interval. The amount of rainfall in a given area is a continuous random variable. This number can take on an infinite number of values – 8.01 in. of rain is different from 8.012 in.. Measurement may stop at some number of decimal points, but the variable is theoretically continuous. Although it is impossible to attach a probability to the amount of rain equaling exactly 8.012000 … inches, it is possible to give the probability that the amount of rain will be within an interval. Continuous random variables can also represent the amount of time it takes to fill a food order at a restaurant or the length of a bolt used in the production of an automobile. Other continuous random variables appear in the following examples:

• 1. Let X be the arrival time at an airport between 8:00 and 9:00 a.m.: 8:00 ≤ x ≤ 9:00.

• 2. For a new residential division, the length of time X from completion until a specified number of houses are sold: a ≤ x ≤ b, for b > a.

• 3. Let Y be the amount of orange juice loaded into a 24-oz bottle in a bottling operation: 0 ≤ y ≤ 24.

• 4. The depth at which a successful natural gas drilling venture first strikes natural gas.

• 5. The weight of a bag of rice bought in a supermarket.

3 Probability Distributions for Discrete Random Variables

3.1 Probability Distribution

To analyze a random variable, we must generally know the probability that the variable will take on certain values. The probability function, or the probability distribution, of a discrete random variable is a systematic listing of all possible values a discrete random variable can take on, along with their respective probabilities. The probability that the random variable X will assume the value x is symbolized by P(X = x) or simply P(x). Note that X is a quantity and x a value. Because a discrete probability function takes nonzero values only at discrete points x,it is sometimes called a probability mass function.

Example 6.1 Probability Distribution for the Outcome of Tossing a Fair Coin.

Suppose a fair coin is tossed. Let the random variable X represent the outcome, where 1 denotes heads and 0 denotes tails. The probability that heads appears is P($P(X=1)=.5$, and the probability that tails appears is P($P(X=1)=.0$) = .5. Figure 6.1 shows a probability distribution where the possible outcomes are charted on the horizontal axis and probabilities on the vertical axis. The spikes in the figure place the probability of heads and that of tails at.5. Note that the probabilities for both outcomes (heads and tails) are between 0 and 1 inclusive and that the sum of both probabilities is 1.

Fig. 6.1

Probability distribution for Example 6.1

Example 6.2 Probability Distribution for Section Assignment in a Marketing Course.

Suppose that five sections of a marketing course are offered and each section has a different number of openings (see Table 6.1). If students are assigned randomly to the sections, then a probability distribution can be drawn for section assignment. The probability that a student is assigned to section 1, P($X=1$), is equal to $\frac{23 }{179 }=.128$; the probability that a student is assigned to section 2 P($X=2$), is equal to $\frac{45 }{179 }=.251$. The rest of the probabilities are calculated in the same manner, as indicated in the third column of Table 6.1. Figure 6.2 shows this probability distribution.

Table 6.1 Probability distribution of marketing course openings

Fig. 6.2

Probability distribution for marketing course openings

Example 6.3 Probability Distribution for the Outcome of Rolling a Fair Die.

The probability distribution for the roll of a fair die is shown in Fig. 6.3.Here all the spikes are equal to $\frac{1}{6}$ because $P(X = 1) = P(X = 2) = \cdots =P(X = 6) = \frac{1}{6}$.

Fig. 6.3

Probability function for Example 6.3

Examples 6.1, 6.2, and 6.3 show that the probability of a random variable X taking on the specific value x can be denoted as P(X = x). The probability distribution of a random variable is a representation of the probabilities for all possible outcomes. P(X = x) is the probability function of random variable X denoting the probability that X takes on the value of x. This expression can be rewritten as P(X = x) = P(x) where the function is evaluated at all possible values of X.

For all probability functions of discrete random variables,

• 1. P(x _i ) ≥ 0 for all i

• 2.
  $$ \sum\limits_{i=1}^n {P\left( {{x_i}} \right)} =1\hfill $$

  where x _i is the ith observation of random variable X. Property 1 states that the probabilities cannot be negative.Property 2 implies that the individual probabilities add up to 1.

3.2 Probability Function and Cumulative Distribution Function

For some problems, we need to find the probability that X will assume a value less than or equal to a given number.A function representing such probabilities is called a cumulative distribution function (cdf) and is usually denoted by F(x).If x ₁ , x ₂, …, x _m are the m values of X given in increasing order (i.e., if x ₁ < x ₂ < …, < x _m ), then the cumulative distribution function of x _k , 1 ≤ k ≤ m, is given by

$$ F\left( {{x_k}} \right)=P\left( {X\le {x_k}} \right) $$

(6.1)

In Eq. 6.1, P(X ≤ x _k ) gives us the probability that X will be less than or equal to x _k .The relationship between the probability function P(x) and the cumulative distribution function F(x _k ) can be expressed as follows:

$$ F\left( {{X_k}} \right)=P\left( {{x_1}} \right)+P\left( {{x_2}} \right)+\cdots +P\left( {{x_k}} \right)=\sum\limits_{i=1}^k {P\left( {{x_i}} \right)} $$

(6.2)

Because the values outside the range of X (values smaller than x ₁ or larger than x _m ) occur only with probability equal to zero, we may equally well write

$$ \begin{array}{lll} {F\left( {{x_k}} \right)=\sum\limits_{{i=-\infty}}^k {P\left( {{x_i}} \right)} } & {\mathrm{ for}k\le m} \\\end{array} $$

(6.2a)

The following examples show how to calculate the cumulative distribution function.

Example 6.4 Cumulative Distribution Function for Rolling a Fair Die.

Reviewing Example 6.3, we find that the value of a random variable X and its probability of occurring upon the rolling of a fair die are listed in the first and second columns of Table 6.2, respectively. Because x ₁ < x ₂ < … < x ₆, the cumulative distribution function can be calculated in accordance with Eq. 6.1 as follows:

$$ F(1)=P(X=1)=\tfrac{1}{6} $$

$$ F(2)=P(X=1)+P(X=2)=\tfrac{1}{6}+\tfrac{1}{6}=\tfrac{1}{3} $$

$$ F(3)=P(X=1)+P(X=2)+P(X=3)=\tfrac{1}{6}+\tfrac{1}{6}+\tfrac{1}{6}=\tfrac{1}{2} $$

and so on, as listed in the last column of Table 6.2. The cumulative distribution function is shown in Fig. 6.4.

Table 6.2 Cumulative distribution function for the outcome of tossing a fair die

Fig. 6.4

Cumulative probability distribution for Example 6.4

This graph is a step function: the values change in discrete “steps” at the indicated integral values of the random variable X. Thus, F(x) takes the value 0 to the left of the point x = 1, steps up to $F(x) = \frac{1}{6}$ at x = 1, and so on. The dot at the left of each horizontal line segment indicates the probability for that integral value of x. At these points, the values of the cumulative distribution function are read from the upper line segments.

4 Expected Value and Variance for Discrete Random Variables

Probability distributions tell us a great deal about the probability characteristics of a random variable. Graphical depictions reveal at a glance the central tendency and dispersion of a discrete distribution, but numerical measures of central tendency and dispersion for a probability distribution are also useful. The mean (central location) of a random variable is called the expected value and is denoted by E(X). The expected value of a random variable, which we also denote as μ, is calculated by summing the products of the values of the random variable and their corresponding probabilities:

$$ \mu =E(X)=\sum\limits_{i=1}^N {{x_i}P\left( {{x_i}} \right)} $$

(6.3)

John Kraft, a marketing executive for Computerland, Inc., must decide whether to use a new label on one of the company’s personal computer products. The firm will gain $900,000 if Mr. Kraft adopts the new label and it turns out to be superior to the old label. The firm will lose $600,000 if Mr. Kraft adopts the new label and it proves to be inferior to the old one. In addition, Mr. Kraft feels that there is.60 probability that the new label is superior to the old one and.40 probability that it is not. The expected value of the firm’s gain for adopting the new label is

$$ \begin{array}{lll} E(X) & =(\$ 900,000)(.6)+(-\$600,000)(.4) \\& =\$300,000\end{array}$$

Therefore, Mr. Kraft should consider adopting the new label.

Example 6.5 Expected Value for Earnings per Share.

Suppose a stock analyst derives the following probability distribution for the earnings per share (EPS) of a firm.

EPS ($)	P(x)	EPS ($)	P(x)
1.50	.05	2.25	.15
1.75	.30	2.50	.10
2.00	.35	2.75	.05

To calculate the expected value (the mean of the random variable), we multiply each EPS by its probability and then add the products:

$$ E(X) = 1.50\left( {.05} \right) + 1.75\left( {.30} \right) + 2.00\left( {.35} \right) + 2.25\left( {.15} \right) + 2.50\left( {.10} \right) + 2.75\left( {.05} \right) = 2.025 $$

The expected value for the earnings per share is 2.025.

Example 6.6 Expected Value of Ages of Students.

Suppose the distribution of the ages of students in a class is

Age	P(x)	Age	P(x)
20	.06	24	.10
21	.10	25	.03
22	.28	26	.04
23	.39

The expected age is

$$ E(X) = 20\left( {.06} \right) + 21\left( {.10} \right) + 22\left( {.28} \right) + 23\left( {.39} \right) + 24\left( {.10} \right) + 25\left( {.03} \right) + 26\left( {.04} \right) = 22.62 $$

In addition to calculating expected value for a probability distribution, we can compute the variance and standard deviation as measures of variability. The variance of a distribution is computed similarly to the variance for raw data, which we discussed in Chap. 4. The variance is the summation of the square of the deviations from the mean, multiplied by the corresponding probability:

$$ {\sigma^2}=\sum\limits_{i=1}^N {{{{\left( {{x_i}-\mu } \right)}}^2}P\left( {{x_i}} \right)} $$

(6.4)

where σ ² is the variance of X, μ is the mean of X, and P(x _i ) is the probability function of x _i .If $P\left( {{x_1}} \right)=P\left( {{x_2}} \right)=\cdots =P\left( {{x_N}} \right)=1/N$, then Eq. 6.4 reduces to

$$ {\sigma^2}=\frac{{\sum\limits_{i=1}^N {{{{\left( {{x_i}-\mu } \right)}}^2}} }}{N} $$

(6.4a)

The standard deviation is the square root of the variance. An alternative – and possibly easier – way to calculate the variance is to sum the product of the square of values of the random variables multiplied by the corresponding probabilities and then subtract the expected value squared¹:

$$ {\sigma^2}=\sum\limits_{i=1}^N {x_i^2P({x_i})-{\mu^2}} $$

(6.5)

Example 6.7 Expected Value and Variance: Defective Tires.

Suppose the following table gives the number of defective tires that roll off a production line in a day. Calculate the mean and variance.

Defects	Probability
0	.05
1	.15
2	.20
3	.25
4	.25
5	.10

The expected value is equal to (0)(.05) + (1)(.15) + (2)(.20) + (3)(.25) + (4)(.25) + (5)(.10) = 2.8. Thus, the mean number of defective tires in a production run is 2.8 tires in a day.

The variance is

$$ \begin{array}{lll} & {{\left( {0-2.8} \right)}^2}\left( {.05} \right) + {{\left( {1-2.8} \right)}^2}\left( {.15} \right) + {{\left( {2-2.8} \right)}^2}\left( {.20} \right) \\& +{{\left( {3-2.8} \right)}^2}\left( {.25} \right) + {{\left( {4-2.8} \right)}^2}\left( {.25} \right) + {{\left( {5-2.8} \right)}^2}\left( {.10} \right) = 1.86\end{array}$$

The alternative formula yields the same answer for the variance:

$$ \left[ {\left( {{0^2}} \right)\left( {.05} \right)+\left( {{1^2}} \right)\left( {.15} \right)+\left( {{2^2}} \right)\left( {.20} \right)+\left( {{3^2}} \right)\left( {.25} \right)+\left( {{4^2}} \right)\left( {.25} \right)+\left( {{5^2}} \right)\left( {.10} \right)} \right]-{{\left( {2.8} \right)}^2} = 1.86 $$

Example 6.8 Expected Value and Variance: Commercial Lending Rate.

Returning to the example of commercial lending interest rates in Sect. 5.8, we can tabulate the possible lending rates x and the corresponding probabilities, P(x), as follows:

x	P(x)	x	P(x)
15 %	.100	18 %	.150
17	.075	11	.100
20	.075	13	.075
13	.200	16	.075
15	.150

From formulas for the expected value and the variance for discrete random variables, the mean of X is

$$ \begin{array}{lll} E(X) & =\sum\limits_{i=1}^N {{x_i}P({x_i})=\mu } \\& =(.100)(.15)+(.075)(.17)+(.075)(.20)+(.200)(.13) \\& + (.150)(.15)+(.150)(.18)+(.100)(.11)+(.075)(.13) \\& + (.075)(.16) \\& =15.1\%\end{array}$$

(6.6)

The standard deviation of X can be calculated from

$$ \begin{array}{lll} \sigma & ={{\left[ {\sum\limits_{i=1}^N {{{{\left( {{x_i}-\mu } \right)}}^2}P\left( {{x_i}} \right)} } \right]}^{1/2 }} \\& =\left[ {\left( {.100} \right){{{\left( {15-15.1} \right)}}^2}+\left( {.075} \right){{{\left( {17-15.1} \right)}}^2}+\left( {.075} \right){{{\left( {20-15.1} \right)}}^2}} \right. \\& +\left( {.200} \right){{\left( {13-15.1} \right)}^2}+\left( {.150} \right){{\left( {15-15.1} \right)}^2}+\left( {.150} \right){{\left( {18-15.1} \right)}^2} \\& {{\left. {+\left( {.100} \right){{{\left( {11-15.1} \right)}}^2}\left( {.075} \right){{{\left( {13-15.1} \right)}}^2}+\left( {.075} \right){{{\left( {16-15.1} \right)}}^2}} \right]}^{1/2 }} \\& =2.51\%\end{array}$$

(6.7)

A bank manager may use this information to make lending decisions, which is discussed in Application 7.4 in Chap. 7.

5 The Bernoulli Process and the Binomial Probability Distribution

In this section, we examine first the Bernoulli process and then the binomial probability distribution and its applications.

5.1 The Bernoulli Process

The binomial distribution is based on the concept of a Bernoulli process, which has three important characteristics. First, a Bernoulli process is a repetitive random process consisting of a series of independent trials. This means that the outcome of one trial does not affect the probability of the outcome of another. Second, only two outcomes are possible in each trial: success or failure. The probability of success is equal to p, and the probability of failure is (1 − p). Third, the probabilities of success and failure are the same in each trial. For example, suppose the owner of an oil firm believes that the probability of striking oil is.10. Success is defined as striking oil and failure as not striking oil. If the probability of striking oil is.10 on every trial and all the trials are independent of each other, then this is a Bernoulli process. Note that the events of striking oil and not striking oil are mutually exclusive.

A simple example of a Bernoulli process is the tossing of a fair coin. The outcomes can be classified into the events’ success (e.g., heads) and failure (tails). The outcomes are mutually exclusive, and the probability of success is constant at.5. The MINITAB output of the Bernoulli process for the first four experiments of Table 5.1 is presented in Fig. 6.5. Columns C1, C2, C3, and C4 present the number and sequence of heads and tails occurring for random experiments with N = 10, 20, 30, and 40.

Fig. 6.5

MINITAB output of Bernoulli process for four experiments (N = 10, N = 20, N = 30, and N = 40)

5.2 Binomial Distribution

If n trials of a Bernoulli process are observed, then the total number of successes in the n trials is a random variable, and the associated probability distribution is known as a binomial distribution. The number of successes, the number of trials, and the probability of success on a trial are the three pieces of information we need to generate a binomial distribution.

To develop the binomial distribution, assume that each of the n trials of an experiment will generate one of two outcomes, a success, S, or a failure, F. Suppose the trials generate x successes and (n − x) failures. The probability of success on a particular trial is p, and the probability of failure is (1 − p). Thus, the probability of obtaining a specific sequence of outcomes is

$$ {p^x}{{\left( {1-p} \right)}^{n-x }} $$

(6.8)

Equation 6.8 presents the joint probability of x successes and (n − x) failures occurring simultaneously. Because the n trials are independent of each other, the probability of any particular sequency of outcomes is, by the multiplication rule of probabilities (Sect. 5.6), equal to the product of the probabilities for the individual outcomes.

5.3 Probability Function

There are several ways in which x successes can be arranged among (n − x) failures. Therefore, the probability of x successes in n trials for a binomial random variable X is

$$ \begin{array}{lll} P(X=x) & = \begin{array}{lll}n \hfill \\x \hfill \\ \end{array} {p^x}{{\left( {1-p} \right)}^{n-x }} \\& =\frac{n! }{{x!\left( {n-x} \right)!}}{p^x}{{\left( {1-p} \right)}^{n-x }},x=0,1,\ldots,n,\end{array}$$

(6.9)

where

$\left( {\begin{array}{lll} n \\x \\ \end{array}} \right)=n$ combinations taken x at a time

$$ n! = n(n-1)(n-2)(n-3)\cdots (1) $$

The symbol n! is read “n factorial.” When n = 0, then n! = 0! = 1. Equation 6.9 is the binomial probability function, which gives the probability of x successes in n trials: using this formula, we can evaluate a binomial probability.

Example 6.9 Probability Distribution for JNJ Stock.

Suppose that the price of a share of stock in Johnson & Johnson company in the future will either go up (U) or come down (D) in 1 day with the probabilities.40 and.60, respectively. Calculate the probability of each possible outcome of the stock price 4 days later.²

Using the outcome tree approach discussed in Appendix 1 of Chap. 5, we find the possible outcomes e and probabilities p(e) indicated in Table 6.3.

Table 6.3 Probability distribution of JNJ stock 4 days later

The probability of JNJ stock going up three times and coming down once is the sum of the probabilities associated with e ₂, e ₃, e ₅, and e ₉:.0384 + .0384 + .0384 + .0384 = .1536.

Alternatively, this probability can be calculated in terms of the binomial combination formula (Eq. 6.9):

$$ \left( {\begin{array}{lll} 4 \\3 \\\end{array}} \right){{\left( {.4} \right)}^3}\left( {.6} \right)=\frac{4! }{(4-3)!3!}\left( {.0384} \right)=.1536 $$

Hence, the binomial combination formula can be used to replace the diagram for calculating such a probability.

Example 6.10 Probability Function of Insurance Sales.

Assume that an insurance sales agent believes that the probability of she making a sale is.20. She makes five contacts and, eager to leave nothing to chance, calculates a binomial distribution:

$$ P(0\mathrm{ success})=\frac{5! }{0!5! }{.2^0}{.8^5}=.3277 $$

$$ P(1\mathrm{ success})=\frac{5! }{1!4! }{.2^1}{.8^4}=.4096 $$

$$ P(2\mathrm{ successes})=\frac{5! }{2!3! }{.2^2}{.8^3}=.2048 $$

$$ P(3\mathrm{ successes})=\frac{5! }{3!2! }{.2^3}{.8^2}=.0512 $$

$$ P(4\mathrm{ successes})=\frac{5! }{4!1! }{.2^4}{.8^1}=.0064 $$

$$ P(5\mathrm{ successes})=\frac{5! }{5!0! }{.2^5}{.8^0}=.0003 $$

Alternatively these numbers can be calculated by the MINITAB program as shown here:

MTB > SET INTO Cl

DATA> 0 1 2 3 4 5

DATA> END

MTB > PDF C1;

SUBC> BINOMIAL 5 0.2.

Probability Density Function

Binomial with n = 5 and p = 0.200000

x	P(X = x)
0.00	0.3277
1.00	0.4096
2.00	0.2048
3.00	0.0512
4.00	0.0064
5.00	0.0003

Figure 6.6 gives the probability distribution for this sales agent’s successes. Because the events of the sales agent’s number of successes are mutually exclusive, the probability that she has three or more successes is equal to P(3 successes) + P(4 successes) + P(5 successes) =.0512 + .0064 +.0003 =.0579.

Fig. 6.6

Binomial probability distribution for Example 6.10 (n = 5, p = .2)

Example 6.11 Cumulative Probability Distribution for Insurance Sales.

Suppose the sales agent we met in Example 6.10 wants to determine the probability of making between 1 and 4 sales:

$$ P(1\mathrm{ success})+P(2\mathrm{ success}\mathrm{ es})+P\left( {3\mathrm{ success}\mathrm{ es}} \right)+P(4\mathrm{ success}\mathrm{ es})=.672 $$

Unless the number of trials n is very small, it is easier to determine binomial probabilities by using Table Al in Appendix A of this book. All three variables listed in Eq. 6.9 (n, p, and x) appear in the binomial distribution table extracted from the National Bureau of Standards tables. Using probabilities from this table, we can calculate both individual probabilities and cumulative probabilities.

The individual probabilities drawn for Example 6.11 from the binomial table are listed in Table 6.4. These probabilities are identical to those we found with Eq. 6.9.

Table 6.4 Part of binomial table (n = 5, p = .2)

The cumulative binomial function can be denned as

$$ B(n,p)=\sum\limits_{x=0}^n {\left( \begin{array}{lll} n \hfill \\x \hfill \\ \end{array} \right)}{p^x}{{\left( {1-p} \right)}^{n-x }} $$

(6.10)

Using Table 6.4, we can calculate the cumulative probabilities for the sales agent having two or more successes:

$$ \begin{array}{lll} P(X\ge 2|n=5,p=.2) & =P(X=2)+P(X=3)+P(X=4)+P(X=5) \\& =\sum\limits_{x=2}^5 {\left( \begin{array}{lll} n \hfill \\x \hfill \\ \end{array} \right)} {(.2)^x}{(.8)^{5-x }} \\& =.2048+.0512+.0064+.0003=.2627\end{array}$$

In a nationwide poll of 2,052 adults by the American Association of Retired Persons (USA Today, August 8, 1985), approximately 40 % of those surveyed described the current version of the federal income tax system as fair. Suppose we randomly sample 20 of the 2,052 adults surveyed and record x as the number who think the federal income tax system is fair. To a reasonable degree of approximation, x is a binomial random variable. The probability that x is less than or equal to 10 can be defined as³

$$ \begin{array}{lll} & P(X\le 10|n=20,p=0.4) \\& =\sum\limits_{x=1}^{10 } {\left( \begin{array}{lll} n \hfill \\x \hfill \\ \end{array} \right)} {{\left( {0.4} \right)}^x}{{\left( {0.6} \right)}^{20-x }} \\& =0+.005+.0031+.0123+.0350+.0746+.1244 \\& \quad+.1659+.1797+.1597+.1171=.8725\end{array}$$

Another situation that requires the use of a binomial random variable is lot acceptance sampling, where we must decide, on the basis of sample information about the quality of the lot, whether to accept a lot (batch) of goods delivered from a manufacturer (see Appendix 1 in Chap. 11 for further detail). It is possible to calculate the probability of accepting a shipment with any given proportion of defectives in accordance with Eq. 6.9.

Example 6.12 Cumulative Probability Distribution: A Shipment of Calculator Chips.

A shipment of 800 calculator chips arrives at Century Electronics. The contract specifies that Century will accept this lot if a sample of size 20 drawn from the shipment has no more than one defective chip. What is the probability of accepting the lot by applying this criterion if, in fact, 5 % of the whole lot (40 chips) turns out to be defective? What if 10 % of the lot is defective?

This is a binomial situation where there are n = 20 trials and p = the probability of success (chip is defective) = .05. The shipment is accepted if the number of defectives is either 0 or 1, so the probability of the shipment being accepted is

$$ \begin{array}{lll} P(\mathrm{ shipment}\;\mathrm{ accepted}) & =P(X\le 1) \\& =P(0)+P(1)\end{array}$$

Using Table A1 in Appendix A (n = 20, p = .05), we obtain P(0) = .3585 and P(1) = .3774. Hence, the probability that Century Electronics accepts delivery is

$$ P\left( {\mathrm{ shipment}\ \mathrm{ accepted}} \right)=.3585+.3774=.7359 $$

Similarly, if 10 % of the items in the shipment are defective (i.e., if p = .10), then

$$ P\left( {\mathrm{ shipment}\ \mathrm{ accepted}} \right)=.1216+.2702=.3918 $$

This implies that the higher the proportion of defectives in the shipment, the less likely is acceptance of the delivery. And that’s as it should be.

5.4 Mean and Variance

The expected value (mean) of the binomial distribution is simply the number of trials times the probability of a success:

$$ \mu =np $$

(6.11)

The variance of the binomial distribution is equal to

$$ {\sigma^2}=np(1-p) $$

(6.12)

Thus the standard deviation of the binomial distribution is $\sqrt{np(1-p) }$. The derivation of Eqs. 6.11 and 6.12 can be found in Appendix 1.

Example 6.13 Probability Distribution of Insurance Sales.

In the insurance sales case we discussed in Examples 6.10 and 6.11, the expected number of sales can be calculated in terms of Eq. 6.11 as np = 5(.20) = 1. The variance of the distribution can be calculated in terms of Eq. 6.12 as np(1 − p) = 5(.2)(.8) = .8. Thus, the expected number of sales by the sales agent is equal to 1, and the standard deviation is $\sqrt{.8 }=.894$.

6 The Hypergeometric Distribution (Optional)

In the last section, we described the binomial distribution as the appropriate probability distribution for a situation in which the assumptions of a Bernoulli process are met. A major application of the binomial distribution is in the computation of probability for cases where the trials are independent. If the experiment consists of randomly drawing n elements (samples), with replacement, from a set of N elements, then the trials are independent.

In most practical situations, however, sampling is carried out without replacement, and the number sampled is not extremely small relative to the total number in the population. For example, when a researcher selects a sample of families in a city to estimate the average income of all families in the city, the sampling units are ordinarily not replaced prior to the selection of subsequent ones. That is, the families are not replaced in the original population and thus are not given an opportunity to appear more than once in the sample. Similarly, when a sample of accounts receivable is drawn from a firm’s accounting records for a sample audit, sampling units are ordinarily not replaced before the selection of subsequent units. Sampling without replacement also takes place in quality control sampling and other sampling. Furthermore, if the number sampled is extremely small relative to the total number of items, then the trial is almost independent even if the sampling is without replacement (as in Example 6.12).⁴ Under such circumstances and in sampling with replacement, the binomial distribution can be used in the analysis.

The hypergeometric distribution is the appropriate model for sampling without replacement. To solve the following hypergeometric problems, let’s divide our population (such as a group of people) into two categories: adults and children. For a population of size N, h members are S (successes) and (N − h) members are F (failures). Let sample size = n trials, obtained without replacement. Let x = number of successes out of n trials (a hypergeometric random variable).

Suppose there are h = 60 adults and N − h = 40 children. Thus, there are N = 100 persons. Numbers from 1 to 100 are assigned to these individuals and printed on identical disks, which are placed in a box. If 10 chips are randomly drawn from the box, then the hypergeometric problem involves calculating the probability of there being x adults and (n − x) children in a sample of size 10. If n = 10 people are selected, what is the probability that exactly four adults will be included in the sample?

Because there are h = 60 adults, there are $\left( \begin{array}{lll} h \hfill \\x \hfill \\ \end{array} \right)$ possible ways of selecting x = 4 adults. Of the n = 10 people, n − x = 10 − 4 = 6 are children. Hence, there are $\left( \begin{array}{lll} N-h \hfill \\n-x \hfill \\ \end{array} \right)$ possible ways of selecting n − x = 6 children. Thus, the total number of ways of selecting a group of 10 persons that includes exactly four adults and six children is $\left( \begin{array}{lll} h \hfill \\x \hfill \\ \end{array} \right)\left( \begin{array}{lll} N-h \hfill \\n-x \hfill \\ \end{array} \right)$. There are $\left( \begin{array}{lll} N \hfill \\n \hfill \\ \end{array} \right)=\left( \begin{array}{lll} 100 \hfill \\10 \hfill \\ \end{array} \right)$ possible ways of selecting 10 persons from 100 persons. Thus, the probability of selecting a group of 10 persons that includes x = 4 adults is

$$ \frac{{\left( {\begin{array}{lll} {60} \\{4} \\\end{array}} \right)\left( {\begin{array}{lll} {40} \\{10-4} \\\end{array}} \right)}}{{\left( {\begin{array}{lll} {100} \\{10} \\\end{array}} \right)}} $$

6.1 The Hypergeometric Formula

From the example we just outlined, we can state the general hypergeometric probability function for a hypergeometric variable X as

$$ P(X=x)=P[(x\mathrm{ successes}\ \mathrm{ and} (n-x)\mathrm{ failures})]=\frac{{\left( \begin{array}{lll} h \hfill \\x \hfill \\ \end{array} \right)\left( \begin{array}{lll} N-h \hfill \\n-x \hfill \\ \end{array} \right)}}{{\left( \begin{array}{lll} N \hfill \\n \hfill \\ \end{array} \right)}} $$

(6.13)

The hypergeometric formula gives the probability of x successes when a random sample of n is drawn without replacement from a population of N within which h units have the characteristic denoting success. The number of successes achieved under these circumstances is the hypergeometric random variable.

Example 6.14 Sampling Probability Function of Party Membership.

Consider a group of 10 students in which four are Democrats and six are Republicans. A sample of size six has been selected. What is the probability that there will be only 1 Democrat in this sample?

Using the hypergeometric probability function shown in Eq. 6.13, we have

$$ \begin{array}{lll} P(x=1\mathrm{ and}n-x=5) & =\frac{{\left( \begin{array}{lll} 4 \hfill \\1 \hfill \\ \end{array} \right)\left( \begin{array}{lll} 10-4 \hfill \\6-1 \hfill \\ \end{array} \right)}}{{\left( \begin{array}{lll} 10 \hfill \\6 \hfill \\ \end{array} \right)}}=\frac{\displaystyle{\frac{4! }{1!(4-1)!}\frac{6! }{5!(6-5)! }}}{\displaystyle{\frac{10! }{6!4! }}} \\& =\frac{\displaystyle{\frac{24 }{(1)(6)}\frac{720 }{(120)(1) }}}{\displaystyle{\frac{3,628,800 }{(720)(24) }}}=\frac{(4)(6) }{210 }=.1143\end{array}$$

Similarly, we can calculate other probabilities. All possible probabilities are as follows:

• P(x = 0) = .0048

• P(x = 1) = .1143

• P(x = 2) = .4286

• P(x = 3) = .3809

• P(x = 4) = .0714

The hypergeometric probability distribution is shown in Fig. 6.7.

Fig. 6.7

Probability distribution for Example 6.14 (hypergeometric distribution for N = 10, x = 4, n = 6)

6.2 Mean and Variance

The mean of the hypergeometric probability distribution for Example 6.14 can be calculated by using Eq. 6.3:

$$ \begin{array}{lll} \mu & =\sum\limits_{i=1}^n {{x_i}} P({x_i})=0(.0048)+1(.1143)+2(.4286)+3(.3809)+4(.0714) \\& =2.40\end{array}$$

On average, we expect 2.40 students to be Democrats. Alternatively, it can be shown that the mean of this distribution is

$$ \mu =n(h/N) $$

(6.14)

The ratio h/N is the proportion of successes on the first trial. The product n(h/N) is similar to the mean of the binomial distribution, np. It can be shown that the variance of the hypergeometric distribution is equal to

$$ {\sigma^2}=\left( {\frac{N-n }{N-1 }} \right)\left[ {n\left( {\frac{h}{N}} \right)\left( {1-\frac{h}{N}} \right)} \right] $$

(6.15)

In other words, the variance of the hypergeometric distribution is the variance of the binomial distribution with an adjustment factor, $\left( {\frac{N-n }{N-1 }} \right)$. If the sample size is small relative to the total number of objects N, then $\left( {\frac{N-n }{N-1 }} \right)$ is very close to 1. Consequently, the binomial distribution can be used to replace the hypergeometric distribution.⁵

Example 6.15 Mean and Variance of a Hypergeometric Probability Function.

Using the data of Example 6.14, we can calculate the mean and variance of a hypergeometric function as follows:

$$ \mu =6\left( {\frac{4}{10 }} \right)=2.4 $$

$$ \begin{array}{lll} {\sigma^2} & =\left( {\frac{10-6 }{10-1 }} \right)\left[ {(6)\left( {\frac{4}{10 }} \right)\left( {1-\frac{4}{10 }} \right)} \right] \\& =.64\end{array}$$

7 The Poisson Distribution and Its Approximation to the Binomial Distribution

In the previous two sections, we have discussed two major types of discrete probability distributions, one for binomial random variables and the other for hypergeometric random variables. Both of these random variables were defined in terms of the number of success, and these successes were obtained within a fixed number of trials of some random experiment. In this section, we will discuss a distribution called the Poisson distribution. This distribution can be used to deal with a single type of outcome or “event,” such as number of telephone calls that come through a switchboard and number of accidents. It is also possible to use a Poisson distribution to investigate the probability of, say, a certain number of defective parts in a plant in a 1-year period, a certain number of sales in a given week, and a certain number of customers entering a bank in a day.

7.1 The Poisson Distribution

The Poisson distribution, which is named after the French mathematician Simeon Poisson, is useful for determining the probability that a particular event will occur a certain number of times over a specified period of time or within the space of a particular interval. For example, the number of customer arrivals per hour at a bank or other servicing facility is a random variable with Poisson distribution. Here are some other random variables that may exhibit a Poisson distribution:

• 1. The number of days in a given year in which a 50-point change occurs in the Dow Jones Industrial Average

• 2. The number of defects detected each day by a quality control inspector in a light bulb plant

• 3. The number of breakdowns per month that a supercomputer experiences

• 4. The number of car accidents that occur per month (or week or day) in the city of Princeton, New Jersey

The formula for the Poisson probability distribution is

$$ \begin{array}{lll} {P(X=x)={{{{e^{{-\lambda }}}{\lambda^x}}} \left/ {x! } \right.}}{\rm for}{x=0,1,2,3,\ldots }{\rm and}{\lambda >0} \\\end{array} $$

(6.16)

where X represents the discrete Poisson random variable; x represents the number of rare events in a unit of time, space, or volume; λ is the mean value of x; e is the base of natural logarithms and is approximately equal to 2.71828; and ! is the factorial symbol.

It can be shown that the value of both the mean and the variance of a Poisson random variable X is λ. That is,

$$ E(X)=\lambda $$

(6.17a)

$$ \mathrm{ Var}(X)=\lambda $$

(6.17b)

We will explore this distribution further in Chap. 9 when we discuss the exponential distribution.

In studying a retailer’s supply account at a large US Air Force base, the Poisson probability distribution was used to describe the number of customers (x) in a 7-day lead time period (Management Science, April 1983). Here “lead time” is used to describe the time needed to replenish a stock item.

Items were divided into two categories for individual analysis. The first category was items costing $5 or less and the second category was items costing more than $5. The mean number of customers during lead time for the first category was estimated to be.09. For the second category, the mean was estimated to be.15.

From Eqs. 6.17a and b, the mean and variance for the number x of customers who demand items that cost over $5 during lead time is

$$ E(X)=\mathrm{ Var}(X)=\lambda =.15. $$

From Eq. 6.16, the probability that no customers will demand an item that costs $5 or less during the lead time is

$$ P(X=0)={e^{-0.09 }}{(.09)^0}/0!=.9139 $$

Example 6.16 Customer Arrivals in a Bank.

Suppose the average number of customers entering a bank in a 30-min period is five. The bank wants to determine the probability that four customers enter the bank in a 30-min period. Substituting λ = 5 and X = 4 into Eq. 6.16, we obtain

$$ P(X=4)={{{\left( {{e^{-5 }}} \right)\left( {{5^4}} \right)}} \left/ {4! } \right.} $$

Table A2 in Appendix A of this book is a Poisson probability table that can be used to calculate probabilities. From this table, we find that (e⁻⁵)(5⁴)/4! = .1755. As another example, say we know that the probability that three customers enter the bank is.1404. Using the Poisson probability table, we can calculate the other individual probabilities for X = 0, 1, and 2. Table 6.5 gives the probability function for X = 0, 1, …, 4.

Table 6.5 Probability function for Example 6.16 (λ = 5)

Our calculations can tell us such things as the probability that 0, 1, 2, 3, or 4 customers arrive within a 20-min period. From Table 6.5, we know that the probability that four or fewer individuals enter the bank is.1755 +.1404 + .0842 + .0337 + .0067 = .4405.

We could continue by calculating the probabilities for more than four customers and eventually produce a Poisson probability distribution for this bank. Table 6.6 shows such a distribution. To produce this table, we used Eq. 6.16. The probability of more than four customer arrivals can also be calculated from Table A2.

Table 6.6 Poisson probability distribution of customer arrivals per 3-min period

Alternatively, MINITAB can be used to calculate part of Table 6.6 as follows:

MTB > SET INTO Cl

DATA> 0 1 2 3 4 5 6 7

DATA> END

MTB > PDF Cl;

SUBC> POISSON 5.

KP(X = K)

0.000.0067

1.000.0337

2.000.0842

3.000.1404

4.000.1755

5.000.1755

6.000.1462

7.000.1044

MTB > PAPER

Figure 6.8 uses MINITAB to illustrate graphically the Poisson probability distribution of the number of customer arrivals.

Fig. 6.8

MINITAB output of Poisson probability distribution for the number of customer arrivals

Example 6.17 Defective Spark Plug.

In one day’s work on a spark plug assembly line, the average number of defective parts is 2. The manager is concerned that more than four defectives could occur and wants to estimate the probability of that happening. Using the Poisson distribution table (Table A2 in Appendix A), she determines that the probability of 0–4 defective spark plugs is P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = .1353 + .2707 + .2707 +.1804 + .0902 = .9473. Then the probability of having more than four defective spark plugs is 1 − .9473 = .0527.

7.2 The Poisson Approximation to the Binomial Distribution

The Poisson distribution can sometimes be used to approximate the binomial distribution and avoid tedious calculations.

If the number of trials in a binomial, n, is large, then a Poisson random variable with λ = np will provide a reasonable approximation. This is a good approximation, provided that n is large (n > 20) and p is small (p < .05):

$$ P(X=x)=\frac{{{e^{-np }}{{{\left( {np} \right)}}^x}}}{x! } $$

(6.18)

Example 6.18 Comparison of the Poisson and Binomial Probability Approaches.

Suppose 20 parts are selected from a production process and tested for defects. The manager of the firm wants to determine the probability that three defectives are encountered. Previous experience indicates that the probability of a part being defective is.05. The mean is np = (20)(.05) = 1. Setting λ = 1, we can use the Poisson distribution formula of Eq. 6.18 to calculate the probability:

$$ P(X=3)={1^3}{e^{-1 }}/3!=.0613 $$

If we use the binomial distribution formula of Eq. 6.9, then the probability is

$$ P(x)=\frac{20! }{3!(20-3)! }{(.05)^3}{(.95)^{17 }}=.0596 $$

The difference between.0613 and.05916 is slight (only about.2 %).

8 Jointly Distributed Discrete Random Variables (Optional)

In Sects. 5.4 and 5.5, we discussed conditional, joint, and marginal probabilities in terms of events. We now consider these probabilities for two or more related discrete random variables. For a single random variable, the probabilities for all possible outcomes can be summarized by using a probability function; for two or more possible related discrete random variables, the probability function must define the probabilities that the random variables of interest simultaneously take specific values.

8.1 Joint Probability Function

Suppose we want to know the probability of a worker being a member of a labor union and over age 50. We now concern ourselves with the distribution of random variables, age (X) and membership in a labor union (Y). In notation, the probability that X takes on a value x and that Y takes on a value y is given by

$$ P(x,y)=P(X=x,Y=y) $$

(6.19)

Equation 6.19 represents the joint probability function of X and Y. Joint probabilities are usually presented in tabular form so that the probabilities can be identified easily. Joint probability distributions of discrete random variables are probability distributions of two or more discrete random variables. The next example illustrates the use of Eq. 6.19.

Example 6.19 Joint Probability Distribution for 100 Students Classified by Sex and by Number of Accounting Courses Taken.

Table 6.7 shows the probability function for two random variables, X (the total number of accounting courses a student takes) and Y (the sex of the student, where 1 denotes a male student and 0 a female). The values in the cells of Table 6.7 are joint probabilities of the outcomes denoted by the column and row headings for X and Y. Also displayed in the margins of the table are separate univariate probability distributions of X and Y.

Table 6.7 Joint probability distribution for 100 students classified by sex and number of accounting courses taken

The joint probability that a student is female and takes three courses, P(3, 0) = P(X = 3, Y = 0), is equal to.17. The probability that a student is male and takes four courses, P(4, 1) = P(X = 4, Y = 1), is equal to.12. The probabilities inside the box are all joint probabilities, which are, again, probabilities of the intersection of two events.

The probability distribution of a single discrete random variable is graphed by displaying the value of the random variable along the horizontal axis and the corresponding probability along the vertical axis. In the case of a bivariate distribution, two axes are required for the values of random variables and a third for the probability. A graph of the joint probability of Table 6.7 is shown in Fig. 6.9.

Fig. 6.9

Graph of the bivariate probability distribution shown in Table 6.7

8.2 Marginal Probability Function

The marginal probability can be obtained by summing all the joint probabilities over all possible values. In other words, the probabilities in the margins of the table are the marginal probabilities. These probabilities form marginal probability functions. For example (see Table 6.7 in Example 6.19), the probability that a randomly selected student is female, P(Y = 0), is found by adding the respective probabilities that a female student takes two courses (.14), three courses (.17), four courses (.08), and five courses (.12), for a total of.51. The probability that a randomly selected student is male, P(Y = 1), is therefore.49 (1 − .51). Similarly, the probability that a randomly selected student takes two courses, P(X = 2), is equal to the probability that a female takes two courses (.14), plus the probability that a male takes two courses (.16), for a total of.30. Note that the sum of the marginal probabilities is 1. From these results, we can define marginal probability functions for X and Y as follows:

$$ \begin{array}{lll} {P({x_i})=\sum\limits_{j=1}^m {P({x_i},{y_j})}, } & {i=1,\ldots,n} \\\end{array} $$

(6.20)

$$ \begin{array}{lll} {P({y_j})=\sum\limits_{i=1}^n {P({x_i},{y_j}),} } & {j=1,\ldots,m} \\\end{array} $$

(6.21)

where

• x _i  = the ith observation of the X variable

• y _j  = the jth observation of the Y variable

8.3 Conditional Probability Function

Conditional probability functions can be calculated from joint probabilities. The conditional probability function of X, given Y = y, is

$$ P(x|y)=P(x,y)/P(y) $$

(6.22)

The conditional probability is found by taking the intersection of the probability of X = x and Y = y and dividing by the probability of Y. For example, the probability from Table 6.7 that a student takes four courses, given that the student is female, is P(4|0) = P(4,0)/P(0) = .08/.51 = .16.

Similarly, the probability that a student is male, given that the student takes three courses, is P(1|3) = P(3,1)/P(3) = .20/.37 = .54. The conditional probability distribution for X given Y = 1 is shown in Table 6.8.

Table 6.8 Conditional probability distribution for numbers of accounting courses, given that the student is male (Y = 1)

8.4 Independence

Returning to the terminology of events explained in Chap. 5, we saw in Sect. 5.6 that if two events are statistically independent, then P(B|A) = P(B) and P(B ∩ A) = P(B)P(A). In random variable notation, the analogous statement is that if X and Y are independent random variables, then

$$ \begin{array}{lll} {P(X=x|Y=y)=P(X=x)} & {\mathrm{ for}\mathrm{ all}X} \\{P(Y=y|X=x)=P(Y=y)} & {\mathrm{ and}\mathrm{ all}Y} \\\end{array} $$

(6.23)

Equation 6.23 implies that the conditional probability function of X given Y or of Y given X is the same as the marginal probability of X or Y. We will illustrate this definition of independence by returning to Table 6.7.

Suppose we consider the outcome pair (3, 1) – that is, X = 3 and Y = 1. In this case,

$$ P(X=3|Y=1)=\frac{.20 }{.49 }=.4082 $$

and

$$ P(X=3)=.37 $$

Because P(X = 3|Y = 1) is not equal to P(X = 3), X and Y are not independent.

Example 6.20 Store Satisfaction.

Table 6.9 shows the probability function for two random variables: X, which measures a consumer’s satisfaction with food stores in a particular town, and Y, the number of years the consumer has resided in that town.⁶ Suppose X can take on the value 1, 2, 3, and 4, which reflect a satisfaction level ranging from low to high, and that Y takes on the value 1 if the consumer has lived in the town fewer than 6 years and 2 otherwise. The values in the cells of Table 6.9 are joint probabilities of the respective joint events denoted by the column and row headings for x and y. Also displayed in the margins of the table are separate univariate probability distributions of x and y.

Table 6.9 Joint probability distribution for consumer satisfaction (x) and number of years of residence in a particular town (Y)

The joint probability that a consumer has satisfaction level 3 and has lived in town fewer than 6 years, P(3, 1) = P(X = 3, Y = 1), is.23. The probability that a consumer has satisfaction level 4 and has lived in the town more than 6 years, P(4, 2), is.05. The probabilities inside the box are all joint probabilities, which are, again, the intersections of two events.

The marginal probability is obtained by summing the joint probabilities over all possible values, as discussed in Example 6.19. For example (see Table 6.9), the probability that a consumer has lived in town fewer than 6 years, P(Y = 1), is found by adding the probabilities that a consumer has satisfaction level 1, 2, 3, and 4, a total of.48. The marginal probability that a consumer has lived in town 6 or more years, P(Y = 2), is therefore.52. Similarly, the probability that a randomly selected consumer has satisfaction level 1, P(X = 1), is equal to the probability that a consumer has lived in town fewer than 6 years,.04, plus the probability that a consumer has lived in the town more than 6 years,.07, for a total of. 11. Note that the sum of the marginal probabilities is equal to 1.

Conditional probability functions can be calculated from joint probabilities as discussed in Example 6.19. The conditional probability is found by taking the intersection of the probability of X = x and Y = y and dividing by the probability of Y = y. For example, the probability (from Table 6.9) that a consumer has satisfaction level 4, given that the consumer has lived in town fewer than 6 years, is P(X = 4|Y = 1) = P(X = 4, Y = 1)/P(Y = 1) = .07/.48 = .1458.

Similarly, the probability that a consumer has lived in town 6 or more years, given that the consumer has satisfaction level 3, is P(Y = 2|X = 3) = P(Y = 2, X = 3)/P(X = 3) = .23/.46 = .5. The conditional probability distribution for X given Y = 2 is shown in Table 6.10.

Table 6.10 Conditional probability distribution for satisfaction level for a consumer who has lived in town 6 or more years

9 Expected Value and Variance of the Sum of Random Variables (Optional)

9.1 Covariance and Coefficient of Correlation Between Two Random Variables

The concept of expected value and variance of discrete random variables discussed in Sect. 6.4 can be extended to measure the degree of relationship between two discrete random variables X and Y. Here we will discuss two alternative means of determining the possibility of a linear association between two random variables X and Y. These two measures are covariance and coefficient of correlation.

The covariance is a statistical measure of the linear association between two random variables X and Y. Its sign reflects the direction of the linear association. The covariance is positive if the variables tend to move in the same direction. If the variables tend to move in opposite directions, the covariance is negative. Specifically, the covariance between X and can be defined as

$$ Cov(X,Y)={\sigma_{X,Y }}=E\left[ {\left( {X-{\mu_X}} \right)\left( {Y-{\mu_Y}} \right)} \right] $$

(6.24)

where μ _X and μ _y are the means of X and Y, respectively. For discrete variables, Eq. 6.24 can be defined as

$$ {\sigma_{X,Y }}=\sum\limits_{j=1}^m {\sum\limits_{i=1}^n {\left( {{X_i}-{\mu_X}} \right)\left( {{Y_j}-{\mu_Y}} \right)P\left( {{X_i},{Y_j}} \right)} } $$

(6.25)

Equation 6.25 can be written as a shortcut formula as follows⁷:

$$ \begin{array}{lll} Cov(X,Y) & =E(XY)-{\mu_X}{\mu_Y} \\& =\sum\limits_{j=1}^m {\sum\limits_{i=1}^n {\left( {{X_i}{Y_j}} \right)P\left( {{X_i},{Y_j}} \right)} } -\left[ {\sum\limits_{i=1}^n {{X_i}P\left( {{X_i}} \right)} } \right]\left[ {\sum\limits_{j=1}^m {{Y_j}P\left( {{Y_j}} \right)} } \right]\end{array}$$

(6.26)

To illustrate, we evaluate the covariance between number of years of residence in the town and satisfaction level, as discussed in Example 6.20. Using the probabilities in Table 6.9, we calculate μ _X , μ _y , and E(XY) as

$$ {\mu_X}=\sum\limits_{i=1}^4 {{X_i}P} \left( {{X_i}} \right)=1(.11)+2(.31)+3(.46)+4(.12)=2.59 $$

$$ {\mu_Y}=\sum\limits_{j=1}^2 {{Y_j}P\left( {{Y_j}} \right)} =1(.48)+2(.52)=1.52 $$

$$ \begin{array}{lll} E(XY) & =\sum\limits_{j=1}^m {\sum\limits_{i=1}^n {\left( {{X_i}{Y_j}} \right)} } P\left( {{X_i}{Y_j}} \right) \\& =(1)(1)(.04)+(1)(2)(.14)+(1)(3)(.23)+(1)(4)(.07)+(2)(1)(.07) \\& \quad+(2)(2)(.17)+(2)(3)(.23)+(2)(4)(.05) \\& =3.89\end{array}$$

Substituting this information into Eq. 6.26, we obtain the covariance:

$$ \mathrm{ Cov}(X,Y)=3.89-(2.59)(1.52)=-0.05 $$

The negative value of covariance indicates some tendency toward a negative relationship between number of years of residence in the town and level of satisfaction.

In addition to the direction of the relationship between variables, we may want to measure its strength. We can easily do so by scaling the covariance to obtain the coefficient of correlation.

The coefficient of correlation p between X and Y is equal to the covariance divided by the product of the variables’ standard deviations. That is,

$$ \rho =\frac{{{\sigma_{X,Y }}}}{{{\sigma_X}{\sigma_Y}}} $$

(6.27)

where ρ = coefficient of correlation, σ _x  = standard deviation of X, and σ _Y  = standard deviation of Y.

It can be shown that ρ is always less than or equal to 1.0 and greater than or equal to −10:

$$ -1\le \rho \le 1 $$

Again, let us use data given in Table 6.9 to show how to calculate the correlation coefficient between X and Y. We use Eq. 6.5 to calculate the variances of X and Y:

$$ \begin{array}{lll} \sigma_X^2 & =\sum\limits_{i=1}^4 {X_i^2P\left( {{X_i}} \right)-{{{\left( {{\mu_X}} \right)}}^2}} \\& ={(1)^2}(.11)+{(2)^2}(.31)+{(3)^2}(.46)+{(4)^2}(.12)-{{\left( {2.59} \right)}^2} \\& =.7019\end{array}$$

$$ \begin{array}{lll} \sigma_Y^2 & =\sum\limits_{j=1}^2 {Y_j^2P\left( {{Y_j}} \right)} -{{\left( {{\mu_Y}} \right)}^2} \\& ={(1)^2}(.48)+{(2)^2}(.52)-{{\left( {1.52} \right)}^2} \\& =.2496\end{array}$$

Then we substitute ${\sigma_{X,Y }}=-.15$, ${\sigma_X}=\sqrt{.7019 }=.8378$, and ${\sigma_Y}=\sqrt{.2496 }=.5$ into Eq. 6.27. We obtain

$$ \rho =\frac{-.05 }{(.8378)(.5) }=-.1194 $$

This means the relationship between X and Y is negative, as indicated by the covariance.

As might be expected, the notions of covariance (and coefficient of correlation) and statistical independence are not unrelated. However, the precise relationship between these notions is beyond the scope of this book. Covariance and coefficient of correlation will be discussed in detail in Chaps. 13 and 14.

9.2 Expected Value and Variance of the Summation of Random Variables X and Y

If X and Y are a pair of random variables with means μ _X and μ _Y and variances $\sigma_X^2$ and $\sigma_Y^2$ and the covariance between X and Y is Cov(X, Y) = σ _x.y, then:

• 1. The expected value of their sum (difference) is the sum (difference) of their expected values:

  $$ \begin{array}{lll} {E(X+Y)={\mu_X}+{\mu_Y}} \\{E(X-Y)={\mu_X}-{\mu_Y}} \\\end{array} $$

• 2. The variance of the sum of X and Y, Var(X + Y), or the difference of X and Y, Var(X − Y), is the sum of their variances plus (minus) two times the covariance between X and Y:

  $$ \begin{array}{lll} \mathrm{ Var}(X+Y)=\sigma_X^2+\sigma_Y^2+2{\sigma_{X,Y }} \hfill \\\mathrm{ Var}(X-Y)=\sigma_X^2+\sigma_Y^2-2{\sigma_{X,Y }} \hfill \\ \end{array} $$

  (6.29)

Example 6.21 Rates of Return and Variance for a Portfolio.

Rates of return for stocks A and B are listed in Table 6.11. Let X = rates of return for stock A, and let Y = rates of return for stock B. The worksheet for calculating μ _X , μ _Y , σ _X , σ _Y , ρ _XY , E(X + Y), and Var(X + Y) is presented in Table 6.12.

Table 6.11 Rates of return for stocks A and B

Table 6.12 Worksheet to calculate summary statistics

$$ {\mu_X}=\frac{.25 }{5}=.05 $$

$$ {\mu_Y}=\frac{-.05 }{5}=-.01 $$

Substituting information into related formulas for variance, covariance, and correlation coefficient yields

$$ \sigma_X^2=.025/4=.00625 $$

$$ \sigma_Y^2=.032/4=.008 $$

$$ {\sigma_{X,Y }}=-.015/4=-.00375 $$

$$ {\rho_{X,Y }}=\frac{-.00375 }{{\sqrt{(.00625)(.008) }}}=-.5303 $$

The MINITAB output of these empirical results is presented in Fig. 6.10. To calculate the expected rate of return and the variance of a portfolio which composes W ₁ percent of stock A and W ₂ percent of stock B, we need to modify Eqs. 6.28 and 6.29 as⁸

$$ { Var}(Rp)={ Var}\left( {{W_1}X+{W_2}Y} \right)=W_1^2{Var}(X)+W_2^2\mathrm{ Var}(Y) +2W_1W_2Cov(X,Y)$$

(6.28)

$$E(Rp)=E({W_1}X+{W_2}Y)={W_1}E(X)+{W_2}E(Y)$$

(6.29)

where E(Rp) and variance (Rp) represent the expected rates of return and variance, respectively. In addition, the summation of weights is assumed to be one (W₁ + W ₂ = 1). This assumption is used to guarantee that all available money has been invested in either stock A or stock B.

Fig. 6.10

MINITAB output for Example 6.21

If John has invested 40 % and 60 % of his portfolio in stock A and stock B, respectively, then the expected rate of return and variance of his portfolio can be calculated in accordance with Eqs. 6.28′ and 6.29′ as

$$ E(Rp)=(.6)(.05)+(.4)(-.01)=.026 $$

$$ \begin{array}{lll} \mathrm{ Var}(Rp) & ={{\left( {.6} \right)}^2}(.00625)+{{\left( {.4} \right)}^2}(.008)+2(.6)(.4)(-.00375) \\& =.00173\end{array}$$

These statistics suggest several things:

• 1. The average rate of return for stock A is higher than that for stock B, and the variance of rates of return for stock A is smaller than that for stock B.

• 2. The rates of return for stock A are negatively correlated with those of stock B.

• 3. E(W ₁ X + W ₂ Y) = .026 represents the average rate of return for a portfolio wherein the different percentage of the money is invested in stock A and in stock B.

• 4. Var(W ₁ X + W ₂ Y) = .00173 represents the variance of a portfolio:

9.3 Expected Value and Variance of Sums of Random Variables

For n random variables, X ₁, X ₂, …, X _n , Eq. 6.28 can be generalized as

$$ E\left( {{X_1}+{X_2}+\cdots +{X_n}} \right)=E\left( {{X_1}} \right)+E\left( {{X_2}} \right)+\cdots +E\left( {{X_n}} \right) $$

(6.30)

Thus the expected value of a sum of n random variables is equal to the sum of the expected values of these random variables.

A somewhat analogous relationship exists for variances of uncorrelated random variables.⁹ If X ₁, X ₂, …, X _n are n uncorrelated variables, then

$$ \mathrm{ Var}\left( {{X_1}+{X_2}+\cdots +{X_n}} \right)=\mathrm{ Var}\left( {{X_1}} \right)+\mathrm{ Var}({X_2})+\cdots +\mathrm{ Var}({X_n}) $$

(6.31)

Otherwise covariances are needed, as in Eq. 6.29.

Example 6.22 Rates of Return, Variance, and Covariance for JNJ, MRK, and S&P 500.

Using annual rates of return for Johnson & Johnson, Merck, and S&P 500 during the period 1970–2009, we calculate the average rate of return, variance, covariance, and correlation coefficient by using MINITAB. The MINITAB outputs are presented in Fig. 6.11.

Fig. 6.11

(continued)

If we let X, Y, and Z represent rates of return for JNJ, MRK, and S&P 500, respectively, then we find from Fig. 6.11 that $\bar{X} =.0510,\bar{Y}=.0638,\bar{X}+\bar{Y}=.1148,S_X^2=.1042$. $S_Y^2 =.1190,{S_{XY }}=.0523,\ \mathrm{ and}{\rho_{XY }}=.470$. Means of these annual rates of return can be used to measure the profitability of the investments; variances of these annual rates of return can be used as a measure of the risk, or uncertainty, involved in the different investments.

10 Summary

In this chapter, we discussed basic concepts and properties of probability distributions for discrete random variables. Important discrete distributions such as the binomial, hypergeometric, and Poisson distributions are discussed in detail. Applications of these distributions in business decisions are also examined.

Using the probability distribution for a random variable, we can calculate the probabilities of specific sample observations. If the probabilities are difficult to calculate, then the means and standard deviations can be used as numerical descriptive measures that enable us to visualize the probability distributions and thereby to make some approximate probability statements about sample observations.

In this chapter, we also discussed joint probability of two random variables. The covariance and coefficient of correlation were presented as means of measuring the degree of relationship between two random variables X and Y.

Questions and Problems

• 1. A team of students participates in a project. The results show that all students are able to finish the project in 7 days. The distribution for the finishing time is given in the following table.

  Finishing time, hours	1	2	3	4	5	6	7
  Students	21	43	23	48	31	29	35

  Define x as the finishing time.

  • (a) Obtain P(X = 1), P(X = 2), …, P(X = 7).

  • (b) Draw the probability distribution.

  • (c) Calculate the cumulative function F _x (x).

  • (d) Draw the cumulative function.

• 2. An investment banker estimates the following probability distribution for the earnings per share (EPS) of a firm.

  x(EPS)	2.25	2.50	2.75	3.00	3.25	3.50	3.75
  P(x)	.05	.10	.20	.35	.15	.10	.05

  Calculate the expected value of the EPS.

• 3. The following table gives the number of unpainted machines in a container. Calculate the mean and standard deviation.

  Unpainted	0	1	2	3	4	5	6	7
  Probability	.05	.09	.15	.30	.25	.10	.05	.01

• 4. The following table gives the probability distribution for two random variables: x, which measures the total number of times a person will be ill during a year, and y, the sex of this person, where 1 represents male and 0 female.

  x	l	2	3	4	Total
  Y
  0	.10	.11	.11	.17	.49
  1	.13	.16	.07	.15	.51
  	.23	.27	.18	.32	1.00

  • (a) Calculate the expected value and standard deviation of the number of times a person will be ill during a year, given that the sex of the person is male. Hint: The conditional expectation and conditional standard deviation, which we have not addressed, can be defined as follows:

    $$ F\left( {{X_k}} \right)=P\left( {X\le {X_k}|Y=0} \right) $$
    $$ F\left( {{X_k}} \right)=P\left( {X\le {X_k}|Y=1} \right) $$
    $$ {\mu_0}=\sum {X_k}P\left( {X={X_k}|Y=0} \right) $$
    $$ {\mu_1}=\sum{X_k},P\left( {X={X_k}|Y=1} \right) $$
    $$ {\sigma_0}={{\left[ {\sum\limits_{i=1}^n {{{{\left( {{X_i}-{\mu_0}} \right)}}^2}} P\left( {{X_i}|Y=0} \right)} \right]}^{1/2 }} $$
    $$ {\sigma_1}={{\left[ {\sum\limits_{i=1}^n {{{{\left( {{X_i}-{\mu_1}} \right)}}^2}P\left( {{X_i}|Y=1} \right)} } \right]}^{1/2 }} $$

  • (b) Calculate the mean and standard deviation of the number of times a person will be ill during a year, given that the sex of the person is female.

• 5. The rate of defective items in a production process is 15 %. Assume a random sample of 10 items is drawn from the process. Find the probability that two of them are “defective”. Calculate the expected value and variance.

• 6. Find the mean and standard deviation of the number of successes in binomial distributions characterized as follows:

  • (a) n = 20, p = .5

  • (b) n = 100, p = .09

  • (c) n = 30, p = .7

  • (d) n = 50, p = .4

• 7. A fair die is rolled 10 times. An “ace” means to roll a “6.” Find the probability of getting exactly four aces, of getting five aces, of getting six aces, and of getting four aces or more.

• 8. A fair coin is tossed eight times.

  • (a) Use MINITAB to construct a probability function table.

  • (b) What is the probability that you will have exactly four heads?

• 9. Consider a group of 12 employees of whom five are in management and seven do clerical work. Select at random a sample of size 4. What is the probability that there will be one manager in this sample?

• 10. A survey was conducted. Of 20 questionnaires that were sent, 12 were completed and returned. We know that 8 of the 20 questionnaires were sent to students and 12 to nonstudents. Only two of the returned questionnaires were from students.

  • (a) What is the response rate for each group?

  • (b) Assume we have a response in hand. What is the probability that it comes from a student?

• 11. The number of people arriving at a bank teller’s window is Poisson distributed with a mean rate of.75 persons per minute. What is the probability that two or fewer people will arrive in the next 6 min?

• 12. The Wicker company has one repair specialist who services 200 machines in the shop and repairs machines that break down. The average breakdown rate is λ = .5 machine per day (or 1 breakdown every 2 days). This technician can fix two machines a day.

  • (a) Use MINITAB to construct a probability function, including breakdown frequency from 0 to 10 cases per day. Assume a Poisson distribution.

  • (b) Wicker is interested in determining the probability that there will be more than two breakdowns in a day.

• 13. Two teams are playing each other in seven basketball games. Team A is considered to have a 60–40 % edge over team B. What is the probability that team A will win four or more games?

• 14. A baseball player usually has four at bats each game. Suppose the baseball player is a lifetime 0.25 hitter. Find the probability that this player will have:

  • (a) Two hits out of four at bats

  • (b) No hits out of four at bats

  • (c) At least one hit out of four at bats

• 15. A certain insurance salesman sees an average of five customers in a week. Each time he speaks to a customer, he has a 30 % chance of making a deal. What is the probability that he makes five deals after speaking with five customers in a week?

• 16. A student takes an exam that consists of 10 multiple-choice questions. Each question has five possible answers. Suppose the student knows nothing about the subject and just guesses the answer on each question. What is the probability that this student will answer four out of the 10 questions correctly?

• 17. A hospital has three doctors working on the night shift. These doctors can handle only three emergency cases in a time period of 30 min. On average, $\frac{1}{2}$ an emergency case arises in each 30-min period. What is the probability that four emergency cases will arise in a 30-min period?

• 18. An average of three small businesses go bankrupt each month. What is the probability that five small businesses will go bankrupt in a certain month?

• 19. During each hour, 0.1 % of the total production of paperclips is defective. For a random sample of 500 pieces of the product, what is the chance of finding more than one defective item?

• 20. The local bank manager has found that one out of every 400 bank loans end up in default. Last year the bank made 400 loans. What is the probability that two bank loans will end up in default?

• 21. Every week a truckload of springs is delivered to the warehouse you supervise. Every time the springs arrive, you have to measure the strength of 400 springs. You accept the shipment only when there are fewer than 20 bad springs. One day a truckload of springs arrives that contains 10 % bad springs. What is the probability that you will accept the shipment? (Just set up the question. Do not try to solve it.)

• 22. Returning to question 21, say (1), your company’s policy is to accept the shipment only when fewer than two springs (out of 400 springs examined) are bad, and (2) the proportion of the bad springs in the truck is only 0.0001. Under these conditions, what is the probability that you will accept the shipment?

• 23. Despite your discomfort with statistics, you find yourself employed by a dog food manufacturer to do statistical research for quality control purposes. Your job is to weigh the dog food to determine whether the cans contain the 16 oz of dog food that the label will claim they contain. You pick 25 cans from each hour’s production and weigh them. If there are more than two cans that contain less than 16 oz, you are to discard the production from that hour. If in a certain hour, 5 % of the cans of dog food produced actually contain less than 16 oz, what is the probability that the whole hour’s production will be discarded?

• 24. A medical report shows that 5 % of stock brokers suffer stress and need medical attention. There are 10 brokers working for your brokerage house. What is the probability that three of them will need medical attention as a result of stress?

• 25. There are 38 numbers in the game of roulette. They are 00, 0, 1, 2, …, 36. Each number has an equal chance of being selected. In the game, the winning number is found by a spin of the wheel. Say a gambler bets $1 on the number 35 three times.

  • (a) What is the probability that the gambler will win the second bet?

  • (b) What is the probability that the gambler will win two of the three bets?

• 26. In the game of roulette, a gambler who wins the bet receives $36 for every dollar she or he bet. A gambler who does not win receives nothing. If the gambler bets $1, what is the expected value of the game?

• 27. A company found that on average, on a given day,.5 % of its employees call in sick. Assume a Poisson distribution. What is the probability that fewer than two of 300 employees will call in sick?

• 28. Billings Company is considering leasing a computer for the next 3 years. Two computers are available. The net present value of leasing each computer in the next 3 years, under different business conditions, is summarized in the following table.

  	Business is
  	Good	Bad
  Plan A: big computer	200,000	20,000
  Plan B: small computer	150,000	100,000

  A consulting company estimates that the chances of having good and of having bad business are 20 % and 80 %, respectively. Compute the expected net present value of leasing a big computer. Compute the expected value of leasing a small computer. What are the variances of these two plans?

• 29. The makers of two kinds of cola are having a contest in the local shopping mall. Assume that 60 % of the people in this region prefer brand A and 40 % prefer brand B. Ten local residents were randomly selected to test the colas. What is the probability that five of these 10 testers will prefer brand A?

• 30. In a certain statistics course, the misguided professor is very lenient. He fails about 1 % of the students in the class. Assume that the probability of failing the course follows a Poisson distribution. In a certain year, the professor teaches 400 students. What is the probability that no one fails the course? What is the average number of failing students?

• 31. A factory examines its work injury history and discovers that the chance of there being an accident on a given workday follows a Poisson distribution. The average number of injuries per workday is.01. What is the probability that there will be three work injuries in a given month (30 days)?

• 32. The state highway bureau found that during the rush hour, in the treacherous section of a highway, an average of three accidents occur. The probability of there being an accident follows a Poisson distribution. What is the probability that there is no accident in a given day?

• 33. In question 32, what is the probability that there are no traffic accidents in all five workdays of a week?

• 34. Of seven prominent financial analysts who are attending a meeting, three are pessimistic about the future of the stock market, and four are optimistic. A newspaper reporter interviews two of the seven analysts. What is the probability that one of these interviewees takes an optimistic view and the other a pessimistic view?

• 35. After assembly, a finished TV is left turned on for one full day (24 h) to determine whether the product is reliable. On average, two TVs break down each day. Yesterday 500 TVs were produced. What is the probability that less than one TV broke down?

• 36. A soft drink company argues that its new cola is the favorite soft drink of the next generation. Ten teenagers were picked to test-drink the cola one by one. Assume that five of them liked the new cola and the rest did not.

  • (a) What is the probability that the first test-drinker liked the new cola?

  • (b) What is the probability that the second test-drinker liked the new cola?

  • (c) What is the probability that after five test-drinks, the new product received three yes votes and two no votes?

• 37. Suppose school records reveal that historically, 10 % of the students in Milton High School have dropped out of school. What is the probability that more than two students in a class of 30 will drop out?

• 38. An insurance company found that one of 5,000 50-year-old, nonsmoking males will suffer a heart attack in a given year. The company has 50,000 50-year-old, nonsmoking male policyholders. What is the probability that fewer than three such policyholders will suffer a heart attack this year?

• 39. Suppose that of 40 salespersons in a company, 10 are females and the rest males. Five of them are randomly chosen to attend a seminar. What is the probability that three females and two males are chosen?

• 40. Consider a single toss of a fair coin, and define X as the number of heads that come up on that toss. Then X can be 0 or 1, with a probability of 50 %. The expected value of X is $\frac{1}{2}$. Can we “expect” to get $\frac{1}{2}$ a head when we toss the coin? If not, how should we interpret the concept of the expected value?

• 41. What is a random variable? What is a discrete random variable? What is a continuous random variable? Give some examples of discrete random variables.

• 42. Tell whether each of the following is a discrete or a continuous random variable:

  • (a) The number of beers sold at a bar during a particular week

  • (b) The length of time it takes a person to drive 50 miles

  • (c) The interest rate on 3-month Treasury bills

  • (d) The number of products returned to a store on a particular day

• 43. An analyst calculates the probability of McGregor stock going up in value for any month as.6 and the probability of the same stock going down in any month as.4. Calculate the probability that the stock will go up in value in exactly 7 months during a year. (Assume independence.)

• 44. Using the information from question 43, compute the probability that the stock will go up in at least 7 months during the year.

• 45. What is a Bernoulli trial? Give some examples of a Bernoulli trial related to the binomial distribution.

• 46. What is the Poisson distribution? Give some examples of situations wherein it would be appropriate to use the Poisson distribution. Compare the Poisson approximation to the binomial distribution.

• 47. Suppose Y represents the number of times a homemaker stops by the local convenience store in a week. The probability distribution of Y follows. Find the expected value and variance of Y.

  y	Probability
  0	.15
  1	.25
  2	.25
  3	.20
  4	.15

• 48. The managers of a grocery store are interested in knowing how many people will shop in their store in a given hour. Suppose they collect data and find that the average number of people who enter the store in any 15-min period is 12. Find the probability that eight people will enter the store in any 15-min period. What is the probability that no more than eight people will enter the store in any 15-min period?

• 49. Suppose you are tossing a fair coin 20 times. What is the probability that you will toss exactly five heads? What is the probability that you will toss five or fewer heads?

• 50. Calculate the mean and variance for the distribution given in question 49.

• 51. You are rolling a six-sided fair die eight times. What is the probability that you will roll exactly two sixes? What is the probability that you will roll two or fewer sixes?

• 52. Calculate the mean and variance for the distribution given in question 51.

• 53. Doctors at the Centers for Disease Control estimate that 30 % of the population will catch the Tibetan flu. What is the probability that in a sample of 10 people, exactly three will catch the flu? What is the probability that three or fewer people in this sample will catch the flu? (Assume that the conditions of a Bernoulli process apply.)

• 54. A golfer enters a long-driving contest in which he wins if he drives the golf ball 300 yards or more and loses if he drives it less than 300 yards. Assume that every time he hits a golf ball, he has a 40 % chance of driving it over 300 yards. If the golfer gets to hit four balls and needs only one 300-yard drive to win, what is the probability that he will win?

• 55. A phone marketing company knows that the number of people who answer the phone between 10:00 and 10:15 a.m. has a Poisson distribution. The average number is eight. What is the probability that the phone company will reach exactly 10 people when it calls during this period? What is the probability that it will reach exactly three people?

• 56. A market survey shows that 75 % of all households own a VCR. Suppose 100 households are surveyed.

  • (a) What is the probability that none of the households surveyed owns a VCR?

  • (b) What is the probability that exactly 75 of the households surveyed own a VCR?

  • (c) Suppose X is the number of households that own a VCR. Compute the mean and variance for X.

• 57. You are given the following information about a stock:

  S = $100	Price of stock
  X = $10 l	Exercise price for a call option on the stock
  r = .005	Interest rate per month
  n = 5	Number of months until the option expires
  u = 1.10	Amount of increase if stock goes up
  d = .95	Amount of decrease if stock goes down

  Calculate the value of the call option if the stock goes up in 3 out of the 5 months.

• 58. Answer question 57 when u = 1.20. How does a change in the amount of increase if the stock goes up affect the value of the call option?

• 59. Answer question 57 when d = .85. How does a change in the amount of decrease if the stock goes down affect the value of the call option?

• 60. Answer question 57 when X = $95. How does a change in the exercise price affect the value of the call option?

• 61. Answer question 57 when S = $110. How does a change in the value of the stock affect the value of the call option?

• 62. Answer question 57 again, finding the value of the option if the stock goes up in 4 out of the 5 months.

• 63. Suppose a box is filled with 25 white balls and 32 red halts. Find the probability of drawing six red balls and four white balls in 10 draws without replacement.

• 64. Redo question 63 with replacement.

• 65. You are drawing five cards from a standard deck of cards with replacement. You win if you draw at least three red cards in five draws. What is the probability of your winning?

• 66. Two tennis players are playing in a final-set tie breaker. Player A is considered to have a 70–30 % edge over player B. The player who wins seven of 13 points will win the championship. What is the probability that player B will win the championship?

• 67. A car sales representative sees an average of four customers in a day. Each time she talks to a customer, she has a 25 % chance of making a deal. What is the chance that she will make four deals after talking to four customers in a day?

• 68. Again consider the car sales rep in question 67. What is the probability that she will make at least two sales after speaking to four customers?

• 69. Again consider the car sales rep in question 67. What is the probability that she will make exactly two sales after speaking to four customers?

• 70. The following test is given to people who claim to have extrasensory perception (ESP). Five cards with different shapes on them are hidden from the person. A card is randomly drawn, and the person is then supposed to guess (or use ESP to determine) the shape on the card. Suppose that this test is administered 10 times with replacement. What is the probability that the subject will get five correct? Do you think getting more than five out of 10 correct supports the subject’s claim to be endowed with ESP?

• 71. Again consider the test in question 70. What is the probability that a person taking the test will get eight out of 20 correct? Do you believe that a person who gets eight out of 20 correct has ESP?

• 72. Say we toss two six-sided dice and let the random variable be the total number of dots observed.

  • (a) Calculate both the probability and the cumulative probability distributions.

  • (b) Draw a graph associated with probability distribution you obtained in part (a).

• 73.

  • (a) Use the monthly rates of return for both GM and Ford listed in Fig. 6.11 to calculate the correlation coefficient between the monthly rates of return for these two companies.

  • (b) Using the results you obtained in part (a) and some other statistics listed in Fig. 6.11, discuss how these two securities are related and how this information can be used to make investment decisions.

• 74. The following table exhibits the monthly rate of return of S&P 500 and American Express. Use the MINITAB program to:

  • (a) Calculate the mean and standard deviation of both returns.

  • (b) Calculate the correlation coefficient between these two sets of monthly returns.

  • (c) Explain the results you get from the two questions above.

  		S&P 500	AMEX
  1989	9	−.65	−2.69
  	10	−2.52	12.73
  	11	16.54	−2.97
  	12	21.42	−1.11
  1990	1	−6.88	−15.25
  	2	8.54	−2.10
  	3	24.26	−11.59
  	4	−2.69	6.23
  	5	9.20	6.93
  	6	−.89	5.91

• 75. A production process produces 0.05 % defective parts. A sample of 10,000 parts from the production is selected. In (a) and (b), what is the probability that:

  • (a) The sample contains exactly two defective parts?

  • (b) The sample contains no defective parts?

  • (c) Find the expected number of defective parts.

  • (d) Find the standard deviation for the number of defective parts.

• 76. The results of a survey of married couples and the number of children they had are shown below.

  Number of children	Probability
  0	0.150
  1	0.125
  2	0.500
  3	0.175
  4	0.050

  Determine the expected number of children and the standard deviation for the number of children.

• 77. The average number of calls received by an operator in a 30-min period is 12.

  • (a) What is the probability that between 17:00 and 17:30 the operator will receive exactly eight calls?

  • (b) What is the probability that between 17:00 and 17:30 the operator will receive more than nine calls but fewer than 15 calls?

  • (c) What is the probability that between 17:00 and 17:30 the operator will receive no calls?

• 78. In a lot of 200 parts, 50 of them are defective. Suppose a sample of 10 parts is selected at random, what is the probability that two of them are defective? What is the expected number of defective parts? What is the standard deviation of the number of defective parts?

Appendices

Appendix 1: The Mean and Variance of the Binomial Distribution

Let X _i represent the Bernoulli random variable; then the random variables X ₁, X ₂, …, X _n are independent Bernoulli variables. From Eqs. 6.3 and 6.4, we know that the mean of a Bernoulli variable is

$$ E({X_i})=\sum\limits_{i=1}^2 {{x_i}} P\left( {{x_i}} \right)=0(1-p)+1(p)=p $$

(6.32)

and that the variance is

$$ \begin{array}{lll} \mathrm{ Var}\left( {{X_i}} \right) & =\sum\limits_{i=1}^2 {{{{\left( {{x_i}-{\mu_X}} \right)}}^2}} P\left( {{x_i}} \right) \\& ={{\left( {0-p} \right)}^2}(1-p)+{{\left( {1-p} \right)}^2}p=p(1-p)\end{array}$$

(6.33)

To find the mean and variance of the binomial distribution, we use the fact that the binomial random variable can be expressed as the sum of independent Bernoulli random variables (X _i ):

$$ X={X_1}+{X_2}+\cdots +{X_n} $$

(6.34)

From Eqs. 6.30 and 6.32, we can derive Eq. 6.11 as

$$ E(X)=\mu =E\left( {{X_1}+{X_2}+\cdots +{X_n}} \right)=E\left( {{X_1}} \right)+E\left( {{X_2}} \right)+\cdots +E\left( {{X_n}} \right)=np $$

(6.35)

Because the X _i variables are statistically independent of one another, the variance of their sum is equal to the sum of their variances. Therefore, following Eq. 6.31, we can derive Eq. 6.12 as

$$ {\sigma^2}=\mathrm{ Var}\left( {\sum\limits_{i=1}^n {{X_i}} } \right)=\mathrm{ Var}\left( {{X_1}} \right)+\mathrm{ Var}\left( {{X_2}} \right)+\cdots +\mathrm{ Var}\left( {{X_n}} \right)=np(1-p) $$

Appendix 2: Applications of the Binomial Distribution to Evaluate Call Options

In this appendix, we show how the binomial distribution is combined with some basic finance concepts to generate a model for determining the price of stock options.

2.1 What Is an Option?

In the most basic sense, an option is a contract conveying the right to buy or sell a designated security at a stipulated price. The contract normally expires at a predetermined date. The most important aspect of an option contract is that the purchaser is under no obligation to buy; it is, indeed, an “option.” This attribute of an option contract distinguishes it from other financial contracts. For instance, whereas the holder of an option may let his or her claim expire unused if he or she so desires, other financial contracts (such as future and forward contracts) obligate their parties to fulfill certain conditions.

A cad option gives its owner the right to buy the underlying security a put option the right to sell. The price at which the stock can be bought (for a call option) or sold (for a put option) is known as the exercise price.

2.2 The Simple Binomial Option Pricing Model

Before discussing the binomial option pricing model, we must recognize its two major underlying assumptions. First, the binomial approach assumes that trading takes place in discrete time – that is, on a period-by-period basis. Second, it is assumed that the stock price (the price of the underlying asset) can take on only two possible values each period; it can go up or go down.

Say we have a stock whose current price per share S can advance or decline during the next period by a factor of either u (up) or d (down). This price either will increase by the proportion u − 1 ≥ 0 or will decrease by the proportion 1 − d, 0 < d < 1. Therefore, the value S in the next period will be either uS or dS. Next, suppose that a call option exists on this stock with a current price per share of C and an exercise price per share of X and that the option has one period left to maturity. This option’s value at expiration is determined by the price of uS underlying stock and the exercise price X. The value is either

$$ {C_u}=Max(0,uS-X) $$

(6.36)

or

$$ {C_d}=Max(0,dS-X) $$

(6.37)

Why is the call worth Max(0, uS − X) if the stock price is uS? The option holder is not obliged to purchase the stock at the exercise price of X, so she or he will exercise the option only when it is beneficial to do so. This means the option can never have a negative value. When is it beneficial for the option holder to exercise the option? When the price per share of the stock is greater than the price per share at which he or she can purchase the stock by using the option, which is the exercise price X? Thus, if the stock price uS exceeds the exercise price X, the investor can exercise the option and buy the stock. Then he or she can immediately sell it for uS, making a profit of uS − X (ignoring commission). Likewise, if the stock price declines to dS, the call is worth Max(0, dS − X).

Also for the moment, we will assume that the risk-free interest rate for both borrowing and lending is equal to r percent over the one time period and that the exercise price of the option is equal to X.

To intuitively grasp the underlying concept of option pricing, we must set up a risk-free portfolio – a combination of assets that produces the same return in every state of the world over our chosen investment horizon. The investment horizon is assumed to be one period (the duration of this period can be any length of time, such as an hour, a day, and a week). To do this, we buy h shares of the stock and sell the call option at its current price of C.¹⁰ Moreover, we choose the value of h such that our portfolio will yield the same payoff whether the stock goes up or down:

$$ h(uS)-{C_u}=h(dS)-{C_d} $$

(6.38)

By solving for h, we can obtain the number of shares of stock we should buy for each call option we sell:

$$ h=\frac{{{C_u}-{C_d}}}{(u-d)S } $$

(6.39)

Here h is called the hedge ratio. Because our portfolio yields the same return under either of the two possible states for the stock, it is without risk and therefore should yield the risk-free rate of return, r percent, which is equal to the risk-free borrowing and lending rate. The condition must be true; otherwise, it would be possible to earn a risk-free profit without using any money. Therefore, the ending portfolio value must be equal to (1 + r) times the beginning portfolio value, hS − C:

$$ (1+r)(hS-C)=h(uS)-{C_u}=h(dS)-{C_d} $$

(6.40)

Note that S and C represent the beginning values of the stock price and the option price, respectively.

Setting R = 1 + r, rearranging to solve for C, and using the value of h from Eq. 6.39, we get

$$ C={{\left[ {\left( {\frac{R-d }{u-d }} \right){C_u}+\left( {\frac{u-R }{u-d }} \right){C_d}} \right]}} \bigg/ {R} $$

(6.41)

where d < r < u. To simplify this equation, we set

$$ p=\frac{R-d }{u-d}\text{so}1-p=\left\{ {\frac{u-R }{u-d }} \right\} $$

(6.42)

Thus we get the option’s value with one period to expiration:

$$ C={{{\left[ {p{C_u}+(1-p){C_d}} \right]}} \left/ {R} \right.} $$

(6.43)

This is the binomial call option valuation formula in its most basic form. In other words, this is the binomial option valuation formula with one period to expiration of the option.

Table 6.13 Possible option values at maturity

To illustrate the model’s qualities, let’s plug in the following values while assuming the option has one period to expiration. Let

$$ \begin{array}{lll} X & =\$100 \\S & =\$100 \\u & =(1.10),\text{so}uS=\$110 \\d & =(.90),\text{so}dS=\$r90 \\R & =1+r=1+.07=1.07\end{array}$$

First we need to determine the two possible option values at maturity, as indicated in Table 6.13.

Next we calculate the value of p as indicated in Eq. 6.42:

$$ \begin{array}{lll} p=\displaystyle{\frac{1.07-.90 }{1.10-.90 }=.85} & {\mathrm{ so}} & {1-p=\frac{1.10-1.07 }{1.10-.90 }=.15} \\ \end{array} $$

Solving the binomial valuation equation as indicated in Eq. 6.43, we get

$$ \begin{array}{lll} C & ={{{\left[ {.85(10)+.15(0)} \right]}} \left/ {1.07 } \right.} \\& =\$7.94\end{array}$$

The correct value for this particular call option today, under the specified conditions, is $7.94. If the call option does not sell for $7.94, it will be possible to earn arbitrage profits. That is, it will be possible for the investor to earn a risk-free profit while using none of his or her own money. Clearly, this type of opportunity cannot continue to exist indefinitely.

2.3 The Generalized Binomial Option Pricing Model

Suppose we are interested in the case where there is more than one period until the option expires.¹¹ We can extend the one-period binomial model to consideration of two or more periods. Because we are assuming that the stock follows a binomial process, from one period to the next, it can only go up by a factor of u or go down by a factor of d. After one period the stock’s price is either uS or dS. Between the first and second periods, the stock’s price can once again go up by u or down by d, so the possible prices for the stock two periods from now are uuS, udS, and ddS. This process is demonstrated in tree diagram form (Fig. 6.12) in Example 6.23 later in this appendix.

Note that the option’s price at expiration, two periods from now, is a function of the same relationship that determined its expiration price in the one-period model. More specifically, the call option’s maturity value is always

$$ {C_T}={\rm{ Max}}\left[ {0,{S_T}-X} \right] $$

(6.44)

where T designates the maturity date of the option.

To derive the option’s price with two periods to go (T = 2), it is helpful as an intermediate step to derive the value of C _u and C _d with one period to expiration when the stock price is uS and dS, respectively:

$$ {C_u}={{{\left[ {p{C_{uu }}+(1-p){C_{ud }}} \right]}} \left/ {R} \right.} $$

(6.45)

$$ {C_d}=\left[ {p{C_{du }}+(1-p){C_{dd }}} \right]/R $$

(6.46)

Equation 6.45 tells us that if the value of the option after one period is C _u , the option will be worth either C _uu (if the stock price goes up) or C _ud (if stock price goes down) after one more period (at its expiration date). Similarly, Eq. 6.46 shows that if the value of the option is C _d after one period, the option will be worth either C _du or C _dd at the end of the second period. Replacing C _u and C _d in Eq. 6.43 with their expressions in Eqs. 6.45 and 6.46, respectively, we can simplify the resulting equation to yield the two-period equivalent of the one-period binomial pricing formula, which is

$$ C=\left[ {{p^2}{C_{uu }}+2p(1-p){C_{ud }}+{(1-p)^2}{C_{dd }}} \right]/{R^2} $$

(6.47)

In Eq. 6.47, we used the fact that C _ud  = C _du because the price will be the same in either case.

We know the values of the parameters S and X. If we assume that R, u, and d will remain constant over time, the possible maturity values for the option can be determined exactly. Thus, deriving the option’s fair value with two periods to maturity is a relatively simple process of working backward from the possible maturity values.

Using this same procedure of going from a one-period model to a two-period model, we can extend the binomial approach to its more generalized form, with n periods to maturity:

$$ C=\frac{1}{{{R^n}}}\sum\limits_{k=0}^n {\frac{n! }{{k!\left( {n-k} \right)!}}} {p^k}{{\left( {1-p} \right)}^{n-k }}\rm Max\left[ {0,{\it u^k}{\it d^{n-k }}\it S-X} \right] $$

(6.48)

To actually get this form of the binomial model, we could extend the two-period model to three periods, then from three periods to four periods, and so on. Equation 6.48 would be the result of these efforts. To show how Eq. 6.48 can be used to assess a call option’s value, we modify the example as follows: S = $100, X = $100, R = 1.07, n = 3, u = 1.1, and d = .90.

First we calculate the value of p from Eq. 6.42 as.85, so 1 − p is. 15. Next we calculate the four possible ending values for the call option after three periods in terms of Max[0, u ^k d ^n−k S − X]:

$$ \begin{array}{lll} {C_1} & =\rm Max\left[ {0,{{{\left( {1.1} \right)}}^3}{{{\left( {.90} \right)}}^0}\left( {100} \right)-100} \right]=33.10 \\{C_2} & =\rm Max\left[ {0,{{{\left( {1.1} \right)}}^2}\left( {.90} \right)\left( {100} \right)-100} \right]=8.90 \\{C_3} & =\rm Max\left[ {0,\left( {1.1} \right){{{\left( {.90} \right)}}^2}\left( {100} \right)-100} \right]=0 \\{C_4} & =\rm Max\left[ {0,{{{\left( {1.1} \right)}}^0}{{{\left( {.90} \right)}}^3}\left( {100} \right)-100} \right]=0\end{array}$$

Now we insert these numbers (C ₁, C ₂, C ₃, and C ₄) into the model and sum the terms:

$$\begin{array}{lll} C= \frac{1}{{{(1.07)^3}}}\left[ {\frac{3! }{0!3! }} \right.{(.85)^0}{(.15)^3}\times 0+\frac{3! }{1!2! }{(.85)^1}{(.15)^2}\times 0 \hfill \\ \left. {+\frac{3! }{2!1! }{(.85)^2}{(.15)^1}\times 8.90+\frac{3! }{3!0! }{(.85)^3}{(.15)^0}\times 33.10} \right] \hfill \\= \frac{1}{1.225}\left[ {0+0+\frac{{3\times 2\times 1}}{{2\times 1\times 1}}} \right.(.7225)(.15)(8.90) \hfill \\ + \frac{{3\times 2\times 1}}{{3\times 2\times 1\times 1}}\times (.61413)(1)(33.10) \hfill \\= \frac{1}{1.225 }[(.32513\times 8.90)+(.61413\times 33.10)] \hfill \\= \$18.96 \hfill \end{array} $$

As this example suggests, working out a multiple-period problem by hand with this formula can become laborious as the number of periods increases. Fortunately, programming this model into a computer is not too difficult.

Now let’s derive a binomial option pricing model in terms of the cumulative binomial density function. As a first step, we can rewrite Eq. 6.48 as

$$ \begin{array}{llll}C=S\left[ {\sum\limits_{k=m}^n {\frac{n! }{{k!\left( {n-k} \right)!}}{p^k}{{{\left( {1-p} \right)}}^{n-k }}\frac{{{u^k}{d^{n-k }}}}{{{R^n}}}} } \right]-\frac{X}{{{R^n}}}\left[ {\sum\limits_{k=m}^n {\frac{n! }{{k!\left( {n-k} \right)!}}{p^k}{{{\left( {1-p} \right)}}^{n-k }}} } \right]\end{array} $$

(6.49)

This formula is identical to Eq. 6.48 except that we have removed the Max operator. In order to remove the Max operator, we need to make u ^k d ^n−k S − X positive, which we can do by changing the counter in the summation from k = 0 to k = m. What is m? It is the minimum number of upward stock movements necessary for the option to terminate “in the money” (i.e., u ^k d ^n−k S − X > 0). How can we interpret Eq. 6.49? Consider the second term in brackets; it is just a cumulative binomial distribution with parameters of n and p.¹² Likewise, via a small algebraic manipulation, we can show that the first term in the brackets is also a cumulative binomial distribution. This can be done by defining p′ ≡ (u/R)p and $1-{p}^{\prime}\equiv \left( {d/R} \right)\left( {1-p} \right).$ ¹³ Thus

$$ {p^k}{{\left( {1-p} \right)}^{n-k }}\frac{{{u^k}{d^{n-k }}}}{{{R^n}}}={{p^{\prime}}^k}{{\left( {1-{p}^{\prime}} \right)}^{n-k }} $$

Therefore, the first term in brackets is also a cumulative binomial distribution with parameters of n and ${p}^{\prime}$. Using Eq. 6.10 in the text, we can write the binomial call option model as

$$ C=S{B_1}(n,{p}^{\prime},m)-\frac{X}{{{R^n}}}{B_2}(n,p,m) $$

(6.50)

where

$$ {B_1}\left( {n,{p}^{\prime},m} \right)=\sum\limits_{k=m}^n {_n{C_k}} {{p^{\prime}}^k}{{\left( {1-{p}^{\prime}} \right)}^{n-k }} $$

$$ {B_2}(n,p,m)=\sum\limits_{k=m}^n {_n{C_k}{p^k}{{{\left( {1-p} \right)}}^{n-k }}} $$

and m is the minimum amount of time the stock has to go up for the investor to finish in the money (i.e., for the stock price to become larger than the exercise price).

In this appendix, we showed that by employing the definition of a call option and by making some simplifying assumptions, we can use the binomial distribution to find the value of a call option. In the next chapter, we will show how the binomial distribution is related to the normal distribution and how this relationship can be used to derive one of the most famous valuation equations in finance, the Black–Scholes option pricing model.

Example 6.23 A Decision Tree Approach to Analyzing Future Stock Price.

By making some simplifying assumptions about how a stock’s price can change from one period to the next, it is possible to forecast the future price of the stock by means of a decision tree. To illustrate this point, let’s consider the following example.

Suppose the price of company A’s stock is currently $100. Now let’s assume that from one period to the next, the stock can go up by 17.5 % or go down by 15 %. In addition, let us assume that there is a 50 % chance that the stock will go up and a 50 % chance that the stock will go down. It is also assumed that the price movement of a stock (or of the stock market) today is completely independent of its movement in the past; in other words, the price will rise or fall today by a random amount. A sequence of these random increases and decreases is known as a random walk. Method of testing the randomness of stock rates of return will be discussed in Sect. 17.8 of Chap. 17.

Given this information, we can lay out the paths that the stock’s price may take. Figure 6.12 shows the possible stock prices for company A for four periods.

Fig. 6.12

Price path of underlying stock (Source: R. J. Rendelman, Jr., and B. J. Bartter(1979). “Two-State Option Pricing,” Journal of Finance 34 (December), 1096)

Note that in period 1 there are two possible outcomes: the stock can go up in value by 17.5 % to $117.50 or down by 15 % to $85.00. In period 2, there are four possible outcomes. If the stock went up in the first period, it can go up again to $138.06 or down in the second period to $99.88. Likewise, if the stock went down in the first period, it can go down again to $72.25 or up in the second period to $99.88. Using the same argument, we can trace the path of the stock’s price for all four periods.

If we are interested in forecasting the stock’s price at the end of period 4, we can find the average price of the stock for the 16 possible outcomes that can occur in period 4:

$$ \begin{array}{lll} \bar{P} & =\frac{{\sum\limits_{i=1}^{16 } {{P_i}} }}{16 }=\frac{{190.61+137.89+\cdots +52.20}}{16 } \\& =\$105.09\end{array}$$

We can also find the standard deviation for the stock’ return:

$$ \begin{array}{lll} {\sigma_P} & ={{\left[ {\frac{{{{{\left( {190.61-105.09} \right)}}^2}+\cdots +{{{\left( {52.20-105.09} \right)}}^2}}}{16 }} \right]}^{1/2 }} \\& =\$34.39\end{array}$$

$\bar{p}$ and σ _P can be used to predict the future price of stock A.