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156 Chapter 4 Discrete Random Variables and Probability Distributions
4.7 J OINTLY D ISTRIBUTED D ISCRETE R ANDOM V ARIABLES
Business and economic applications of statistics are often concerned about the rela- tionships between variables. Products at different quality levels have different prices. Age groups have different preferences for clothing, for automobiles, and for music. The percent returns on two different stocks may tend to be related, and the returns for both may increase when the market is growing. Alternatively, when the return on one stock is growing, the return on the other might be decreasing. When we work with probability models for problems involving relationships between variables, it is important that the effect of these relationships is included in the probability model. For example, assume that a car dealer is selling the following automobiles: (1) a red two-door compact, (2) a blue minivan, and (3) a silver full-size sedan; the probability distribution for purchasing would not be the same for women in their 20s, 30s, and 50s. Thus, it is important that probability models reflect the joint effect of variables on probabilities. In Section 3.4 we discussed bivariate probabilities. We now consider the case where two or more, possibly related, discrete random variables are examined. With a single random variable, the probabilities for all possible outcomes can be summarized in a probability distribution. Now we need to define the probabilities that several random variables of interest simultaneously take specific values. At this point we will concen- trate on two random variables, but the concepts apply to more than two. Consider the following example involving the use of two jointly distributed discrete random variables.
Application Exercises 4.67 A company receives a shipment of 16 items. A ran- dom sample of 4 items is selected, and the ship- ment is rejected if any of these items proves to be defective. a. What is the probability of accepting a shipment containing 4 defective items? b. What is the probability of accepting a shipment containing 1 defective item? c. What is the probability of rejecting a shipment containing 1 defective item? 4.68 A committee of 8 members is to be formed from a group of 8 men and 8 women. If the choice of com- mittee members is made randomly, what is the prob- ability that precisely half of these members will be women?
4.69 A bond analyst was given a list of 12 corporate bonds. From that list she selected 3 whose ratings she felt were in danger of being downgraded in the next year. In actuality, a total of 4 of the 12 bonds on the list had their ratings downgraded in the next year. Suppose that the analyst had simply chosen 3 bonds randomly from this list. What is the probability that at least 2 of the chosen bonds would be among those whose rat- ings were to be downgraded in the next year? 4.70 A bank executive is presented with loan applications from 10 people. The profiles of the applicants are similar, except that 5 are minorities and 5 are not mi- norities. In the end the executive approves 6 of the ap- plications. If these 6 approvals are chosen at random from the 10 applications, what is the probability that fewer than half the approvals will be of applications involving minorities?
Example 4.15 Market Research (Joint Probabilities)
Sally Peterson, a marketing analyst, has been asked to develop a probability model for the relationship between the sale of luxury cookware and age group. This model will be important for developing a marketing campaign for a new line of chef-grade cookware. She believes that purchasing patterns for luxury cookware are different for different age groups.
4.7 Jointly Distributed Discrete Random Variables 157
The probability distributions for the individual random variables are frequently desired when dealing with jointly distributed random variables.
Joint Probability Distribution
Let X and Y be a pair of discrete random variables. Their joint probability dis- tribution expresses the probability that simultaneously X takes the specific value x , and Y takes the value y , as a function of x and y. We note that the dis- cussion here is a direct extension of the material in Section 3.4, where we pre- sented the probability of the intersection of bivariate events, P^1 Ai > Bj^2 Here, we use random variables. The notation used is P^1 x , y^2 , so P^1 x , y^2 = P^1 X = x > Y = y^2
Derivation of the Marginal Probability Distribution
Let X and Y be a pair of jointly distributed random variables. In this context the probability distribution of the random variable X is called its marginal probability distribution and is obtained by summing the joint probabilities over all possible values—that is,
P^1 x^2 = (^) a y
P^1 x , y^2 (4.24)
Similarly, the marginal probability distribution of the random variable Y is as follows:
P^1 y^2 = (^) a x
P^1 x , y^2 (4.25)
An example of these marginal probability distributions is shown in the lower row and the right column in Table 4.6.
Joint probability distributions must have the following properties.
Solution To represent the market, Sally proposes to use three age groups—16 to 25, 26 to 45, and 46 to 65—and two purchasing patterns—buy and not buy. Next, she collects a random sample of persons for the age range 16 to 65 and records their age group and desire to purchase. The result of this data collection is the joint probability distribution contained in Table 4.6. Table 4.6, therefore, provides a summary of the probability of purchase and age group that will be a valuable resource for marketing analysis.
Table 4.6 Joint Probability Distribution of Age Group ( X ) versus Purchase Decision ( Y )
AGE GROUP (X) PURCHASE DECISION ( Y )
1 2 3 (16 to 25) (26 to 45) (46 to 65) P(y) 1 (buy) 0.10 0.20 0.10 0. 2 (not buy) 0.25 0.25 0.10 0. P ( x ) 0.35 0.45 0.20 1.
4.7 Jointly Distributed Discrete Random Variables 159
Similarly, it follows that
P 1 x u y 2 = P 1 x 2 Example 4.16 considers the possible percent returns for two stocks, A and B, illus- trates the computation of marginal probabilities and tests for independence, and finds the means and variances of two jointly distributed random variables.
Example 4.16 Stock Returns, Marginal Probability,
Mean, and Variance (Joint Probabilities)
Suppose that Charlotte King has two stocks, A and B. Let X and Y be random variables of possible percent returns (0%, 5%, 10%, and 15%) for each of these two stocks, with the joint probability distribution given in Table 4.7. a. Find the marginal probabilities. b. Determine if X and Y are independent. c. Find the means and variances of both X and Y.
Table 4.7 Joint Probability Distribution for Random Variables X and Y
Y RETURN X R ETURN 0% 5% 10% 15% 0% 0.0625 0.0625 0.0625 0. 5% 0.0625 0.0625 0.0625 0. 10% 0.0625 0.0625 0.0625 0. 15% 0.0625 0.0625 0.0625 0.
Solution a. This problem is solved using the definitions developed in this chapter. Note that for every combination of values for X and Y , P^1 x , y^2 = 0.0625. That is, there is a 6.25% probability for each possible combination of x and y returns. To find the marginal probability that X has a 0% return, consider the following:
P^1 X = 02 = (^) a y
P^1 0, y^2 = 0.0625 + 0.0625 + 0.0625 + 0.0625 = 0.
Here all the marginal probabilities of X are 25%. Notice that the sum of the mar- ginal probabilities is 1. Similar results can be found for the marginal probabili- ties of Y. b. To test for independence, we need to check if P^1 x , y^2 = P^1 x^2 P^1 y^2 for all pos- sible pairs of values x and y. P^1 x , y^2 = 0.0625 for all possible pairs of values x and y P^1 x^2 = 0.25 and P^1 y^2 = 0.25 for all possible pairs of values x and y P^1 x , y^2 = 0.0625 = 1 0.25^21 0.25^2 = P^1 x^2 P^1 y^2 Therefore, X and Y are independent. c. The mean of X is as follows: m X = E^3 X^4 = (^) a x
xP^1 x^2 = 01 0.25^2 + 0.05^1 0.25^2 + 0.10^1 0.25^2 + 0.15^1 0.25^2
= 0.
160 Chapter 4 Discrete Random Variables and Probability Distributions
Conditional Mean and Variance
The conditional mean is computed using the following: m Y u X = E 3 Y u X 4 = (^) a y
1 y u x 2 P 1 y u x 2
Using the joint probability distribution in Table 4.6, we can compute the expected value of Y given that x = 2 : E 3 Y u x = 24 = (^) a y
1 y u x = 22 P 1 y u x = 22 = 112
Similarly the conditional variance is computed using the following: s^2 Y u X = E 31 Y - m Y u X 22 u X 4 = (^) a y
11 y - m Y u X 22 u x 2 P 1 y u x 2
Using the joint probability distribution in Table 4.6, we can compute the variance of Y given that x = 2 : s^21 Y u x = 22 = (^) a y
11 y - 1.56 22 ) u x = 22 P 1 y u x = 22
Computer Applications
Computation of marginal probabilities, means, and variances for jointly distributed random variables can be developed in Excel or other computer packages. For example, we can com- pute marginal probabilities, means, and variances for the jointly distributed random vari- ables X and Y , from Table 4.7, using an Excel worksheet in the format shown in Figure 4.4.
Similarly, the mean of Y is m Y = E 3 Y 4 = 0.075. The variance of X is s^2 X = (^) a x
1 x - m X 22 P 1 x 2 = (^) a x
1 x - m X 22 P ( x ) = (^) a x
1 x - m X 22 (0.25)
and the standard deviation of X is s X = 1 0.003125 = 0.0559016, or 5.59%. Follow similar steps to find the variance and standard deviation of Y.
Figure 4. Marginal Probabilities, Means, and Variances for X and Y Computed Using Excel
Linear Functions of Random Variables
Previously, the expectation of a function of a single random variable was defined. This definition can now be extended to functions of several random variables.
162 Chapter 4 Discrete Random Variables and Probability Distributions
Covariance
Let X be a random variable with mean m X , and let Y be a random variable with mean m Y. The expected value of 1 X - m X^21 Y - m Y^2 is called the covariance be- tween X and Y , denoted as Cov^1 X , Y^2. For discrete random variables Cov 1 X , Y 2 = E 31 X - m X 21 Y - m Y 24 = (^) a x a y
1 x - m X 21 y - m Y 2 P 1 x , y (^2) (4.33)
An equivalent expression is as follows: Cov^1 X , Y^2 = E^3 XY^4 - m X m Y = (^) a x a y
xyP^1 x , y^2 - m X m Y
Correlation
Although the covariance provides an indication of the direction of the relationship be- tween random variables, the covariance does not have an upper or lower bound, and its size is greatly influenced by the scaling of the numbers. A strong linear relationship is defined as a condition where the individual observation points are close to a straight line. It is difficult to use the covariance to provide a measure of the strength of a linear relation- ship because it is unbounded. A related measure, the correlation coefficient, provides a measure of the strength of the linear relationship between two random variables, with the measure being limited to the range from - 1 to +1.
Correlation
Let X and Y be jointly distributed random variables. The correlation between X and Y is as follows:
r = Corr^1 X , Y^2 =
Cov^1 X , Y^2 s X s Y (4.34)
The correlation is the covariance divided by the standard deviations of the two ran- dom variables. This results in a standardized measure of relationship that varies from - 1 to +1. The following interpretations are important:
1. A correlation of 0 indicates that there is no linear relationship between the two random variables. If the two random variables are independent, the correlation is equal to 0. 2. A positive correlation indicates that if one random variable is high (low), then the other random variable has a higher probability of being high (low), and we say that the variables are positively dependent. Perfect positive linear dependency is indi- cated by a correlation of +1.0. 3. A negative correlation indicates that if one random variable is high (low), then the other random variable has a higher probability of being low (high), and we say that the variables are negatively dependent. Perfect negative linear dependency is indi- cated by a correlation of - 1.0. The correlation is more useful for describing relationships than the covariance. With a cor- relation of + 1 the two random variables have a perfect positive linear relationship, and, there- fore, a specific value of one variable, X , predicts the other variable, Y , exactly. A correlation of
- 1 indicates a perfect negative linear relationship between two variables, with one variable, X , predicting the negative of the other variable, Y. A correlation of 0 indicates no linear relation- ship between the two variables. Intermediate values indicate that variables tend to be related, with stronger relationships occurring as the absolute value of the correlation approaches 1. We also know that correlation is a term that has moved into common usage. In many cases correlation is used to indicate that a relationship exists. However, variables that have nonlinear relationships will not have a correlation coefficient close to 1.0. This distinction is important for us in order to avoid confusion between correlated random variables and those with nonlinear relationships.
4.7 Jointly Distributed Discrete Random Variables 163
Example 4.17 Joint Distribution of Stock Prices
(Compute Covariance and Correlation)
Find the covariance and correlation for the stocks A and B from Example 4.16 with the joint probability distribution in Table 4.7.
Solution The computation of covariance is tedious for even a problem such as this, which is simplified so that all of the joint probabilities, P^1 x , y^2 , are 0.0625 for all pairs of values x and y. By definition, you need to find the following:
Cov^1 X , Y^2 = (^) a x a y
xyP^1 x , y^2 - m X m Y
= 031021 0.0625^2 + 1 0.05^21 0.0625^2 + 1 0.10^21 0.0625^2 + 1 0.15^21 0.0625^24
+ 0.05^31021 0.0625^2 + 1 0.05^21 0.0625^2 + 1 0.10^21 0.0625^2 + 1 0.15^21 0.0625^24
+ 0.10^31021 0.0625^2 + 1 0.05^21 0.0625^2 + 1 0.10^21 0.0625^2 + 1 0.15^21 0.0625^24
+ 0.15^31021 0.0625^2 + 1 0.05^21 0.0625^2 + 1 0.10^21 0.0625^2 + 1 0.15^21 0.0625^24
- 1 0.0752 10.075^2
Thus,
r = Corr 1 X , Y 2 =
Cov^1 X , Y^2 s X s Y
Microsoft Excel can be used for these computations by carefully following the example in Figure 4.5.
Figure 4.5 Covariance Calculation Using Microsoft Excel
Joint Probability Distribution of X and Y Y Return % X Return % 0 0.0 5 0.1 0.1 5 P( x ) E(X) 0 0.0625 0.0625 0.0625 0. 0.0 5 0.0625 0.0625 0.0625 0. 0.1 0.0625 0.0625 0.0625 0. 0.1 5 0.0625 0.0625 0.0625 0. 0.25 0.25 0.25 0. E(Y) 0.
Calculation of Covariance xy P( x,y ) xy P( x,y ) xy P( x,y ) xy P( x,y ) xy P( x,y )
xy P( x,y ) xy P( x,y ) xy P( x,y ) 0 0 0 0 0.000156 0. 0 0.000313 0. 0 0.000469 0.
Sum xy P( x,y ) 0 0.000938 0.
0
0.002813 0.00 5625
Covariance Sum xy P( x,y ) 2 E(X)E(Y) 5 0.00 5625 2 0.00 5625 0
4.7 Jointly Distributed Discrete Random Variables 165
But if the covariance is not 0, then Var 1 X - Y 2 = s^2 X + s^2 Y - 2 Cov 1 X , Y 2 Let X 1 , X 2 , c, X (^) K be K random variables with means m 1 , m 2 , c , m K and variances s^21 , s^22 , c, s K^2. The following properties hold:
5. The expected value of their sum is as follows: E 3 X 1 + X 2 + (^) g + X (^) K 4 = m 1 + m 2 + (^) g + m K (4.39) 6. If the covariance between every pair of these random variables is 0, the variance of their sum is as follows: Var 1 X 1 + X 2 + (^) g + X (^) K 2 = s^21 + s^22 + (^) g + s^2 K (4.40) 7. If the covariance between every pair of these random variables is not 0, the variance of their sum is as follows:
Var 1 X 1 + X 2 + (^) g + X (^) K 2 = (^) a
K i = 1
s^2 i + (^2) a
K - 1 i = 1 a
K j 7 i
Cov 1 X (^) i , Yj 2 (4.41)
Example 4.18 Simple Investment Portfolio (Means
and Variances, Functions of Random Variables)
An investor has $1,000 to invest and two investment opportunities, each requiring a minimum of $500. The profit per $100 from the first can be represented by a random variable X , having the following probability distributions:
P 1 X = - 52 = 0.4 and P 1 X = 202 = 0.
The profit per $100 from the second is given by the random variable Y , whose probabil- ity distributions are as follows:
P 1 Y = 02 = 0.6 and P 1 Y = 252 = 0.
Random variables X and Y are independent. The investor has the following possible strategies:
a. $1,000 in the first investment b. $1,000 in the second investment c. $500 in each investment Find the mean and variance of the profit from each strategy.
Solution Random variable X has mean
m X = E 3 X 4 = (^) a x
xP 1 x 2 = 1 - 521 0.4 2 + 12021 0.6 2 = + 10
and variance
s^2 X = E 31 X - m x 22 4 = (^) a x
1 x - m x 22 P 1 x 2 = 1 - 5 - 10221 0.4 2 + 120 - 10221 0.6 2 = 150
Random variable Y has mean
m Y = E^3 Y^4 = (^) a y
yP^1 y^2 = 1021 0.6^2 + 12521 0.4^2 = + 10
and variance
s^2 Y = E^3 ( Y - m Y^22 4 = (^) a y
(^1) y - m Y (^22) P (^1) y (^2) = 10 - 10221 0.6 (^2) + 125 - 10221 0.4 (^2) = 150
166 Chapter 4 Discrete Random Variables and Probability Distributions
Portfolio Analysis
Investment managers spend considerable effort developing investment portfolios that consist of a set of financial instruments that each have returns defined by a probability distribution. Portfolios are used to obtain a combined investment that has a given ex- pected return and risk. Stock portfolios with a high risk can be constructed by combining several individual stocks whose values tend to increase or decrease together. With such a portfolio an investor will have either large gains or large losses. Stocks whose values move in opposite directions could be combined to create a portfolio with a more stable value, implying less risk. Decreases in one stock price would be balanced by increases in another stock price. This process of portfolio analysis and construction is conducted using probability distributions. The mean value of the portfolio is the linear combination of the mean val- ues of the stocks in the portfolio. The variance of the portfolio value is computed using the sum of the variances and the covariance of the joint distribution of the stock values. We will develop the method using an example with a portfolio consisting of two stocks. Consider a portfolio that consists of a shares of stock A and b shares of stock B. We want to use the mean and variance for the market value, W , of a portfolio, where W is the linear function W = aX + bY. The mean and variance are derived in the chapter appendix.
Strategy (a) has mean profit of E 310 X 4 = 10 E 3 X 4 = + 100 and variance of Var 110 X 2 = 100 Var 1 X 2 = 15, Strategy (b) has mean profit E 310 Y 4 = 10 E 3 Y 4 = + 100 and variance of Var 110 Y 2 = 100 Var 1 Y 2 = 15, Now consider strategy (c): $500 in each investment. The return from strategy (c) is 5 X + 5 Y , which has mean E 35 X + 5 Y 4 = E 35 X 4 + E 35 Y 4 = 5 E 3 X 4 + 5 E 3 Y 4 = + 100 Thus, all three strategies have the same expected profit. However, since X and Y are independent and the covariance is 0, the variance of the return from strategy (c) is as follows: Var 15 X + 5 Y 2 = Var 15 X 2 + Var 15 Y 2 = 25 Var 1 X 2 + 25 Var 1 Y 2 = 7, This is smaller than the variances of the other strategies, reflecting the decrease in risk that follows from diversification in an investment portfolio. Most investors would prefer strategy (c), since it yields the same expected return as the other two, but with lower risk.
The Mean and Variance for the
Market Value of a Portfolio
The random variable X is the price for stock A, and the random variable Y is the price for stock B. The portfolio market value , W , is given by the linear function W = aX + bY where a is the number of shares of stock A, and b is the number of shares of stock B. The mean value for W is as follows: m W = E^3 W^4 = E^3 aX + bY^4 = a m X + b m Y (4.42)
168 Chapter 4 Discrete Random Variables and Probability Distributions
Table 4.9 Probability Distribution for New Portfolio of Stock C and Stock D
S TOCK C PRICE
STOCK D P RICE $40 $50 $60 $ $45 0.003333 0.003333 0.003333 0. $50 0.003333 0.003333 0.24 0. $55 0.003333 0.24 0.003333 0. $60 0.24 0.003333 0.003333 0.
Using the probability distribution in Table 4.9 we computed the means, variances, and covariance for the new stock portfolio. The mean for stock C is $53, the same as for stock A. Similarly, the mean for stock D is $55, the same as for stock B. Thus, the mean value of the portfolio is not changed. The variance for each stock is also the same, but the covariance is now –59.17. Thus, the variance for the new portfolio includes a nega- tive covariance term and is as follows: s^2 W = 52 s^2 X + 102 s^2 Y + 2 * 5 * 10 * Cov^1 X , Y^2 = 52 * 31.3 + 102 * 125 + 2 * 5 * 10 * 1 - 59.17^2 = 7,365. We see that the effect of the negative covariance is to reduce the variance and, hence, to reduce the risk of the portfolio. Figure 4.6 shows how portfolio variance—and, hence, risk—changes with different correlations between stock prices. Note that the portfolio variance is linearly related to the correlation. To help control risk, designers of stock portfolios select stocks based on the correlation between prices.
Figure 4.6 Portfolio Variance Versus Correlation of Stock Prices
As we saw in Example 4.19, the correlation between stock prices, or between any two random variables, has important effects on the portfolio value random variable. A posi- tive correlation indicates that both prices, X and Y , increase or decrease together. Thus, large or small values of the portfolio are magnified, resulting in greater range and vari- ance compared to a zero correlation. Conversely, a negative correlation leads to price in- creases for X matched by price decreases for Y. As a result, the range and variance of the portfolio are decreased compared to a zero correlation. By selecting stocks with particu- lar combinations of correlations, fund managers can control the variance and the risk for portfolios.
Exercises 169
Basic Exercises 4.71 Consider the joint probability distribution:
X 1 2 Y 0 0.20 0. 1 0.30 0.
a. Compute the marginal probability distributions for X and Y. b. Compute the covariance and correlation for X and Y. 4.72 Consider the joint probability distribution:
X 1 2 Y 0 0.25 0. 1 0.25 0.
a. Compute the marginal probability distributions for X and Y. b. Compute the covariance and correlation for X and Y. c. Compute the mean and variance for the linear function W = X + Y. 4.73 Consider the joint probability distribution:
X 1 2 Y 0 0.30 0. 1 0.25 0.
a. Compute the marginal probability distributions for X and Y. b. Compute the covariance and correlation for X and Y. c. Compute the mean and variance for the linear function W = 2 X + Y. 4.74 Consider the joint probability distribution:
X 1 2 Y 0 0.70 0. 1 0.0 0.
a. Compute the marginal probability distributions for X and Y. b. Compute the covariance and correlation for X and Y. c. Compute the mean and variance for the linear function W = 3 X + 4 Y. 4.75 Consider the joint probability distribution:
X 1 2 Y 0 0.0 0. 1 0.40 0.
a. Compute the marginal probability distributions for X and Y.
b. Compute the covariance and correlation for X and Y. c. Compute the mean and variance for the linear function W = 2 X - 4 Y. 4.76 Consider the joint probability distribution:
X 1 2 Y 0 0.70 0. 1 0.0 0.
a. Compute the marginal probability distributions for X and Y. b. Compute the covariance and correlation for X and Y. c. Compute the mean and variance for the linear function W = 10 X - 8 Y.
Application Exercises 4.77 A researcher suspected that the number of between- meal snacks eaten by students in a day during final examinations might depend on the number of tests a student had to take on that day. The accompanying ta- ble shows joint probabilities, estimated from a survey.
Number of Snacks ( Y )
Number of Tests ( X ) 0 1 2 3 0 0.07 0.09 0.06 0. 1 0.07 0.06 0.07 0. 2 0.06 0.07 0.14 0. 3 0.02 0.04 0.16 0.
a. Find the probability distribution of X and compute the mean number of tests taken by students on that day. b. Find the probability distribution of Y and, hence, the mean number of snacks eaten by students on that day. c. Find and interpret the conditional probability dis- tribution of Y , given that X = 3. d. Find the covariance between X and Y. e. Are number of snacks and number of tests indepen- dent of each other? 4.78 A real estate agent is interested in the relationship be- tween the number of lines in a newspaper advertisement for an apartment and the volume of inquiries from po- tential renters. Let volume of inquiries be denoted by the random variable X , with the value 0 for little interest, 1 for moderate interest, and 2 for strong interest. The real estate agent used historical records to compute the joint prob- ability distribution shown in the accompanying table.
Number of Lines ( Y )
Number of Inquiries ( X ) 0 1 2 3 0.09 0.14 0. 4 0.07 0.23 0. 5 0.03 0.10 0.
EXERCISES
Chapter Exercises and Applications 171
KEY W ORDS
- Bernoulli random variable, 140
- binomial distribution, 142
- conditional probability distribution, 158
- continuous random variable, 127
- correlation, 162
- covariance, 162
- cumulative probability distribution, 130
- differences of random variable, 164
- discrete random variable, 127
- expected value, 132
- expected value of functions of random variables, 135 - hypergeometric distribution, 154 - independence of jointly distributed random variables, 158 - joint probability distribution, 157 - marginal probability distribution, 157 - mean, 132 - mean and variance of a binomial, 142 - Poisson approximation to the binomial distribution, 151 - Poisson probability distribution, 147 - portfolio analysis, 166 - portfolio market value, 166 - probability distribution function, 129 - properties of cumulative probability distributions, 131 - properties of joint probability distri- butions, 158 - random variable, 127 - relationship between probability dis- tribution and cumulative probability distribution, 131 - variance of a discrete random variable, 133 - properties for linear functions of a random variable, 135
CHAPTER EXERCISES AND APPLICATIONS
4.85 As an investment advisor, you tell a client that an in- vestment in a mutual fund has (over the next year) a higher expected return than an investment in the money market. The client then asks the following questions: a. Does that imply that the mutual fund will certainly yield a higher return than the money market? b. Does it follow that I should invest in the mutual fund rather than in the money market? How would you reply? 4.86 A contractor estimates the probabilities for the num- ber of days required to complete a certain type of con- struction project as follows: Time (days) 1 2 3 4 5 Probability 0.05 0.20 0.35 0.30 0.
a. What is the probability that a randomly chosen project will take less than 3 days to complete? b. Find the expected time to complete a project. c. Find the standard deviation of time required to complete a project. d. The contractor’s project cost is made up of two parts—a fixed cost of $20,000, plus $2,000 for each day taken to complete the project. Find the mean and standard deviation of total project cost. e. If three projects are undertaken, what is the prob- ability that at least two of them will take at least 4 days to complete, assuming independence of indi- vidual project completion times? 4.87 A car salesperson estimates the following probabilities for the number of cars that she will sell in the next week: Number of cars 0 1 2 3 4 5 Probability 0.10 0.20 0.35 0.16 0.12 0.
a. Find the expected number of cars that will be sold in the week. b. Find the standard deviation of the number of cars that will be sold in the week. c. The salesperson receives a salary of $250 for the week, plus an additional $300 for each car sold. Find the mean and standard deviation of her total salary for the week. d. What is the probability that the salesperson’s sal- ary for the week will be more than $1,000? 4.88 A multiple-choice test has nine questions. For each question there are four possible answers from which to select. One point is awarded for each correct answer, and points are not subtracted for incorrect answers. The instructor awards a bonus point if the students spell their name correctly. A student who has not stud- ied for this test decides to choose an answer for each question at random. a. Find the expected number of correct answers for the student on these nine questions. b. Find the standard deviation of the number of correct answers for the student on these nine questions. c. The student spells his name correctly: i Find the expected total score on the test for this student. ii Find the standard deviation of his total score on the test. 4.89 Develop realistic examples of pairs of random vari- ables for which you would expect to find the following: a. Positive covariance b. Negative covariance c. Zero covariance
172 Chapter 4 Discrete Random Variables and Probability Distributions
4.90 A long-distance taxi service owns four vehicles. These are of different ages and have different repair records. The probabilities that, on any given day, each vehicle will be available for use are 0.95, 0.90, 0.90, and 0.80. Whether one vehicle is available is independent of whether any other vehicle is available. a. Find the probability distribution for the number of vehicles available for use on a given day. b. Find the expected number of vehicles available for use on a given day. c. Find the standard deviation of the number of ve- hicles available for use on a given day. 4.91 Students in a college were classified according to years in school ( X ) and number of visits to a museum in the last year ( Y = 0 for no visits, 1 for one visit, 2 for more than one visit). The joint probabilities in the accompanying table were estimated for these random variables.
Number of Visits ( Y )
Years in School ( X ) 1 2 3 4 0 0.07 0.05 0.03 0. 1 0.13 0.11 0.17 0. 2 0.04 0.04 0.09 0.
a. Find the probability that a randomly chosen stu- dent has not visited a museum in the last year. b. Find the means of the random variables X and Y. c. Find and interpret the covariance between the ran- dom variables X and Y. 4.92 A basketball team’s star 3-point shooter takes six 3-point shots in a game. Historically, she makes 40% of all 3-point shots taken in a game. State at the outset what assumptions you have made. a. Find the probability that she will make at least two shots. b. Find the probability that she will make exactly three shots. c. Find the mean and standard deviation of the num- ber of shots she made. d. Find the mean and standard deviation of the total number of points she scored as a result of these shots. 4.93 It is estimated that 55% of the freshmen entering a particular college will graduate from that college in four years. a. For a random sample of 5 entering freshmen, what is the probability that exactly 3 will graduate in four years? b. For a random sample of 5 entering freshmen, what is the probability that a majority will graduate in four years? c. 80 entering freshmen are chosen at random. Find the mean and standard deviation of the proportion of these 80 that will graduate in four years. 4.94 The World Series of baseball is to be played by team A and team B. The first team to win four games wins the series. Suppose that team A is the better team, in the sense that the probability is 0.6 that team A will win
any specific game. Assume also that the result of any game is independent of that of any other. a. What is the probability that team A will win the series? b. What is the probability that a seventh game will be needed to determine the winner? c. Suppose that, in fact, each team wins two of the first four games. i What is the probability that team A will win the series? ii What is the probability that a seventh game will be needed to determine the winner? 4.95 Using detailed cash-flow information, a financial ana- lyst claims to be able to spot companies that are likely candidates for bankruptcy. The analyst is presented with information on the past records of 15 companies and told that, in fact, 5 of these have failed. He selects as candidates for failure 5 companies from the group of 15. In fact, 3 of the 5 companies selected by the ana- lyst were among those that failed. Evaluate the finan- cial analyst’s performance on this test of his ability to detect failed companies. 4.96 A team of 5 analysts is about to examine the earnings prospects of 20 corporations. Each of the 5 analysts will study 4 of the corporations. These analysts are not equally competent. In fact, one of them is a star, having an excellent record of anticipating changing trends. Ideally, management would like to allocate the 4 corporations whose earnings will deviate most from past trends to this analyst. However, lacking this information, management allocates corporations to analysts randomly. What is the probability that at least 2 of the 4 corporations whose earnings will de- viate most from past trends are allocated to the star analyst? 4.97 On average, 2.4 customers per minute arrive at an air- line check-in desk during the peak period. Assume that the distribution of arrivals is Poisson. a. What is the probability that there will be no arriv- als in a minute? b. What is the probability that there will be more than three arrivals in a minute? 4.98 A recent estimate suggested that, of all individuals and couples reporting income in excess of $200,000, 6.5% either paid no federal tax or paid tax at an ef- fective rate of less than 15%. A random sample of 100 of those reporting income in excess of $200,000 was taken. What is the probability that more than 2 of the sample members either paid no federal tax or paid tax at an effective rate of less than 15%? 4.99 A company has two assembly lines, each of which stalls an average of 2.4 times per week according to a Poisson distribution. Assume that the performances of these assembly lines are independent of one another. What is the probability that at least one line stalls at least once in any given week? 4.100 George Allen has asked you to analyze his stock port- folio, which contains 10 shares of stock D and 5 shares of stock C. The joint probability distribution of the
174 Chapter 4 Discrete Random Variables and Probability Distributions
4.106 Faschip, Ltd., is a new African manufacturer of note- book computers. Their quality target is that 99.999% of the computers they produce will perform exactly as promised in the descriptive literature. In order to monitor their quality performance they include with each computer a large piece of paper that includes a direct—toll-free—phone number to the Senior Vice President of Manufacturing that can be used if the computer does not perform as promised. In the first year Faschip sells 1,000,000 computers.
a. If they are achieving their quality target, what is the probability that they will receive fewer than 5 calls? If this occurs what would be a reasonable conclusion about their quality program? b. If they are achieving their quality target, what is the probability that they will receive more than 15 calls? If this occurs, what would be a reasonable conclusion about their quality program?
Appendix: Verifications
1 V ERIFICATION OF AN A LTERNATIVE F ORMULA
FOR THE V ARIANCE OF A D ISCRETE R ANDOM
V ARIABLE (E QUATION 4.6)
Begin with the original definition of variance: s^2 X = (^) a x
(^1) x - m X (^22) P (^1) x (^2) = (^) a x
(^1) x^2 - 2 m Xx + m^2 X (^2) P (^1) x 2
= (^) a x
x^2 P^1 x^2 - 2 m X (^) a x
xP^1 x^2 + m^2 X a x
P^1 x^2
But we have seen that
a xxP^1 x^^2 =^ m X^ and^ a xP^1 x^^2 =^1
Thus, s^2 X = (^) a x
x^2 P 1 x 2 - 2 m^2 X + m^2 X
and, finally,
s^2 X = (^) a x
x^2 P^1 x^2 - m^2 X
2 V ERIFICATION OF THE M EAN AND V ARIANCE
OF A L INEAR F UNCTION OF A R ANDOM
V ARIABLE (E QUATIONS 4.9 AND 4.10)
It follows from the definition of expectation that if Y takes the values a + bx with prob- abilities PX 1 x 2 , its mean is as follows: E 3 Y 4 = m Y = (^) a x
1 a + bx 2 P 1 x 2 = a (^) a x
P 1 x 2 + b (^) a x
xP 1 x 2
Then, since the first summation on the right-hand side of this equation is 1 and the second summation is the mean of X , we have E^3 Y^4 = a + b m X as in Equation 4.9. Further, the variance of Y is, by definition, s^2 Y = E 31 Y - m Y 22 4 = (^) a X
31 a + bx 2 - m Y 42 P 1 x 2
Appendix: Verifications 175
Substituting a + b m X for m Y then gives
s^2 Y = (^) a x
(^1) bx - b m X (^22) P (^1) x (^2) = b^2 a x
(^1) x - m X (^22) P (^1) x 2
Since the summation on the right-hand side of this equation is, by definition, the variance of X , the result in Equation 4.10 follows:
s^2 W = Var^1 a + bX^2 = b^2 s^2 X
3 E XAMPLE TO D EMONSTRATE E QUATION 4.
Show that in general
E^3 g^1 x^24? g^1 m x^2
Using the results in Table 4.12, we show this result for the nonlinear function
g^1 x^2 = bx^2
Where b is a constant and we see that
E^3 bX^2 4 = 1.2 b? b^1 E^3 X^422 = b^1 0.8 22 = 0.64 b
when
E^3 g^1 x^24? g^1 m x^2
Table 4.
x bx^2 P ( x ) E [ X ] E [ bX^2 ] 0 0 0.40 0 0 1 b 0.40 0.40 0.4 b 2 4 b 0.20 0.40 0.8 b 0.80 1.2 b
4 V ERIFICATION OF THE M EAN AND V ARIANCE
OF THE B INOMIAL D ISTRIBUTION
(E QUATIONS 4.19 AND 4.20)
To find the mean and variance of the binomial distribution, it is convenient to return to the Bernoulli distribution. Consider n independent trials, each with probability of success P , and let X (^) i = 1 if the i th trial results in success and 0 otherwise. The random variables X 1 , X 2 ,... , Xn are, therefore, n independent Bernoulli variables, each with probability of suc- cess P. Moreover, the total number of successes X is as follows:
X = X 1 + X 2 + (^) g + X (^) n
Thus, the binomial random variable can be expressed as the sum of independent Bernoulli random variables. The mean and the variance for Bernoulli random variables can be used to find the mean and variance of the binomial distribution. Using Equations 4.15 and 4.16, we know that
E^3 X (^) i^4 = P and s (^) X^2 i = P^11 - P^2 for all i = 1, 2,... , n
