Chapter 7: Random Variables and Discrete Probability Distributions

Transcription

1 Chapter 7: Random Variables and Discrete Probability Distributions 7. Random Variables and Probability Distributions This section introduced the concept of a random variable, which assigns a numerical value to each simple event in a sample space, thereby enabling us to work with numerical-valued outcomes. A random variable is discrete if the number of possible values that it can assume is finite or countably infinite; otherwise, the random variable is continuous. The second important concept introduced was that of a discrete probability distribution, which lists all possible values that a discrete random variable can take on, together with the probability that each value will be assumed. When constructing a discrete probability distribution, bear in mind that the probabilities must sum to. Example 7. The local taxation office claims that 0% of the income tax forms processed this year contain errors. Suppose that two of these forms are selected at random. Find the probability distribution of the random variable X, defined as the number of forms selected that contain an error. Solution A sample space for this random experiment is S = { F F, F F, F F F F }, where, for example, F F indicates that the first form selected contains an error and the second form does not. The random variable X assigns a value of 0,, or to each of these simple events. To find the probability that each of these values will be assumed, we must determine the simple events that result in each of these values being assumed and the probabilities that the simple events will occur. For example, X = if and only if the event F F occurs. Since the occurrence of an error on one form is independent of the occurrence of an error on the other form, Similarly, P ( X P( X = ) = p() = P( F F) = P( F ) P( F ) = (.)(.) =.0 = ) = P( X = 0) = p() = P( F F ) + P( F F ) = P( F ) P( F ) + P( F ) P( F ) = (.)(.9) + (.9)(.) =.8 p(0) = P( F F ) = P( F ) P( F ) = (.9)(.9) =.8 Summarizing the probability distribution of X in tabular form, we obtain 7

2 x p(x) Note that we could have used a probability tree to identify the simple events and their associated probabilities. EXERCISES 7. For each of the following random variables, indicate whether the variable is discrete or continuous and specify the possible values that it can assume. a) X = the number of customers served by a restaurant on a given day b) X = the time in minutes required to complete a particular assembly c) X = the number of accidents at a particular intersection in a given week d) X = the number of questions answered correctly by a randomly selected student who wrote a quiz consisting of 5 multiple choice questions e) X = the weight in ounces of a newborn baby chosen at random 7. The following table gives the probability distribution of X, the number of telephones in a randomly selected home in a certain community. x p(x)

3 4.096 If one home is selected at random, find the probability that it will have: a) no telephone; b) fewer than two telephones; c) at least three telephones; d) between one and three telephones (inclusive). 7.3 The security force on a particular campus estimates that 80% of the cars that are parked illegally on campus are detected and issued a $0 parking ticket. Suppose your car will be parked illegally on two occasions during the coming year. Determine the probability distribution of the random variable X = the total dollar amount of parking tickets that you will receive. 7. Expected Value and Variance This section dealt with the two main descriptive measures of a discrete random variable: its expected value and its variance. As well as being able to calculate the numerical values of these measures, you are expected to have a general understanding of their meanings and to be familiar with the laws of expected value and of variance. 74

4 The expected value (or mean value) of a discrete random variable X is a weighted average of the values that X can assume: E ( X ) = µ = x i p ( x i ) n i= The variance of a discrete random variable X is a weighted average of the squared deviations of the values of X from their mean µ: n ( x ) V ( X ) = = µ p( ) i= i x i When computing the variance of X by hand, it is usually easier to use the shortcut formula: = E X ( ) µ A related measure of the dispersion of the values of X about their mean value µ is the standard deviation of X, defined to be the positive square root of the variance of X. The standard deviation of X, denoted, has the advantage of being expressed in the same units as both the values of X and their mean value µ. Example 7. Let X be a random variable with the following probability distribution: x p(x) a) Find the mean and standard deviation of X. b) Find the mean and standard deviation of Y = X + 4. Solution a) From the definitions of E(X) and V(X): 75

5 E ( X ) = µ = x i p ( x i ) 4 i = = (. ) + 4 (.3 ) + 6 (. 4 ) + 8(. ) = = V ( X ) = = ( x i µ) p ( x i ) i = = ( 5 ) (. ) + ( 4 5) (. 3) + (6 5 ) (.4 ) + ( 8 5 ) (. ) = ( 7) (. ) + ( ) (. 3 ) + () (. 4 ) + ( 3) (. ) = ( 49 )(. ) + ()(. 3 ) + ()(. 4 ) + ( 9)(. ) = 7.4 Therefore, the mean value of X is µ = 5, and the standard deviation of X is = 7.4 =.7. A convenient alternative for computing the expected value and variance of X is to record the necessary computations in a table. In the accompanying table, we have recorded the computations involved in the calculation of E(X) and E(X ). Recall that E(X ) is needed in the shortcut formula for computing the variance of X. x p(x) x p(x) x x p(x) = E(X) 3.4 = E(X ) As before, we obtain E(X) = µ = 5.0 V(X) = E(X ) µ = 3.4 (5) = 7.4 x = 7.4 =.7 b) Using the laws of expected value and variance: E (Y ) = E ( X + 4 ) = E ( X ) + E ( 4) = (5 ) + 4 = 4 V (Y ) = V ( X + 4 ) = V ( X ) = ( ) V ( X ) = 4 (7.4) = 9.6 y = 9.6 =

6 In the next to last line above, a common error is to forget to square the before multiplying by V(X). Example 7.3 A firm believes it has a chance of winning an $80,000 contract if it spends $5,000 on the proposal. If the firm spends twice this amount on the proposal, it feels its chances of winning the contract will improve to 60%. If the firm bases its decision solely on expected value, how much should it spend on the proposal? Solution Let X be the net revenue received from the contract (net of the proposal cost) if $5,000 is spent on the proposal, and let Y be the net revenue received if $0,000 is spent. The probability distributions of X and Y are as follows: x p(x) y p(y) 5, , , ,000.6 If $5,000 is spent on the proposal, the expected net revenue is E(X) = ( 5,000)(.5) + (75,000)(.5) = $35,000 If $0,000 is spent on the proposal, the expected net revenue is E(Y) = ( 0,000)(.4) + (70,000)(.6) = $38,000 The firm should spend $0,000 on the proposal, since that results in the higher expected net revenue. EXERCISES 7.4 According to the Statistics Canada Yearbook (985), the distribution of the number of dependent children involved in divorces in 98 is the following: Number of dependent children Percentage of divorces a) If one divorce case is selected at random, find the probability that the number of dependent children involved is at least. 77

7 b) Find the expected value and the variance of the number of dependent children involved in 98 divorces. c) What is the probability that the number of dependent children involved in a divorce falls within two standard deviations of the mean value? 7.5 Suppose you make a $,000 investment in a risky venture. There is a 40% chance that the payoff from the investment will be $,000, a 50% chance that you will just get your money back, and a 0% chance that you will receive nothing at all from your investment. Find the expected value and standard deviation of the payoff from your investment of $, Refer to Exercise 7.5 a) If X is the payoff from the investment, then the net profit from the investment is Y = X,000. Find the expected value and standard deviation of the net profit. b) If you invest $3,000 in the risky venture instead of $,000 and the possible payoffs triple accordingly, what is the expected value and standard deviation of the net profit from the $3,000 investment? 78

8 7.3 Bivariate Distributions (Optional) This section introduced the notion of a bivariate (or joint) probability distribution of two quantitative variables, from which we can examine the relationship between the two variables. A common measure of the strength of the linear relationship between two variables X and Y is the covariance, given by COV( X,Y ) = all( x, y) ( x µ = E( XY) µ µ x )( y µ y) P( X = x and Y = y) x y A covariance can take on any numerical value between - and +. To obtain a measure more amenable to interpretation, we divide COV(X, Y) by the product x y, to obtain the coefficient of correlation COV( X, Y) ρ = x y A correlation ρ can only assume values between - and +, inclusive. If ρ = +, there is a perfect positive linear relationship between X and Y. If ρ = -, there is a perfect negative linear relationship between X and Y. If ρ = 0, there is no linear relationship between X and Y; in this case, X and Y are said to be independent. Example 7.4 Consider the following bivariate distribution of X and Y. X y a. Find the marginal probabilities. b. Determine the mean, variance, and standard deviation for X and for Y. c. Compute the covariance between X and Y, and determine if X and Y are independent. d.compute the coefficient of correlation. Solution a. The four probabilities in the interior of the table are the joint probabilities p(x, y). In particular, 79

9 p(0, ) =.4 p(0, ) =. Summing these two probabilities, we obtain the marginal probability P(X = 0) =.5 Proceeding similarly to find the other marginal probabilities, we obtain the marginal probability distributions of X and Y: X p(x) y p(y) b. µx = Σxip(xi) = 0(.5) + 0(.5) = 5 µy = Σyip(yi) = (.6) + (.4) =.4 V(X) = Σ(xi - µx) p(xi) = (0-5) (.5) + (0-5) (.5) = 5 x = 5 = 5 V(Y) = Σ(yi - µy) p(yi) = ( -.4) (.6) + (-.4) (.4) =.4 y =. 4 =.49 c. COV ( X, Y ) = ( x µ x )( y µ y ) P ( X = x and Y = y ) all (x, y) = (0-5)( -.4)(.4) + (0-5)( -.4)(.) + (0 5)( -.4)(.) + (0-5)( -.4)(.3) = (-5)(-.4)(.4) + (-5)(.6)(.) + (5)(-.4)(.) + (5)(.6)(.3) = =.0 X and Y are not independent, because COV(X, Y) 0 d. The coefficient of correlation is 80

10 ρ= COV ( X,Y ) x y = =.408 Exercise 7.7 The table below lists the joint probabilities of X and Y. x y a. Find the marginal probabilities b. Determine the mean, variance, and standard deviation for X and for Y. c. Compute the covariance and the coefficient of correlation. d. Are X and Y independent? Explain. 8

11 7.4 Investment Portfolio Diversification (Optional) This section demonstrated how statistical measures can be used to provide insight into the notion of portfolio diversification. To minimize calculations, we will restrict our attention to portfolios consisting solely of two stocks. If R and R denote the rates of return on stocks and, respectively, then R and R are random variables. A portfolio consisting of these two stocks will be well-defined once we specify the proportion of the total portfolio dollar value that is invested in each of the two stocks. Let w and w (called the portfolio weights) represent the proportions of the total dollar investment that is allocated to stock and stock, respectively. (Observe that w + w =, because the two stocks constitute the entire investment.) Then the rate of return on the portfolio is given by Rp = w R + w R Applying the laws of expected value and variance to this expression, we obtain formulas to find the expected value and variance of the portfolio rate of return from the expected values and variances of the rates of return for the two stocks. E(Rp) = we(r) + we(r) V ( R p ) p = w = w + w + w + w w COV ( R, R + w w ρ ) Notice also, from the two previous lines, the often-useful relationship between the covariance and the correlation between the two stock returns: COV(R,R ) = ρ Example 7.5 An investor forms a portfolio by putting 70% of her money into investment and the remainder of her money into investment. Investment has an expected return of 0% with a standard deviation of 5%. Investment has an expected return of 0% with a standard deviation of 35%. a) Find the expected return for the portfolio. b) Find the standard deviation of the portfolio return if the correlation between the returns on the two investments is ρ =.5. c) Repeat part (b) assuming ρ = 0. Explain why the standard deviation in this case is smaller than in part (b). 8

12 Solution a) The portfolio weights are w =.7 and w =.3. The return on the portfolio is therefore Rp =.7R +.3R The expected return is E(Rp) =.7E(R ) +.3E(R ) =.7(.0) +.3(.0) =.3, or 3%. b) The variance of the portfolio return is V(R p ) = w + w + w w ρ Therefore, = (.7) (.5) + (.3) (.35) + (.7)(.3)(.5)(.5)(.35) =.0330 p = (.0330).5 =.87 Notice that the riskiness (as measured by standard deviation) of the portfolio return is less than the riskiness of the investment return. c) The variance of the portfolio return is now V(R p ) = w + w + w w ρ = (.7) (.5) + (.3) (.35) + (.7)(.3)(0)(.5)(.35) =.00 Therefore, p = (.00).5 =.483 The riskiness (as measured by standard deviation) of this portfolio return is less than in part (b) because the correlation between the two investment returns (ρ = 0) is smaller than the correlation (ρ =.5) in part (b), leading to a greater diversification effect. 83

13 Exercise 7.8 An investor forms a portfolio by putting 60% of his money into investment and the remainder of her money into investment. Investment has an expected return of 0% with a standard deviation of 0%. Investment has an expected return of 8% with a standard deviation of 30%. a) Find the expected return for the portfolio. b) Find the standard deviation of the portfolio return if the correlation between the returns on the two investments is ρ =.5. c) Repeat part (b) assuming ρ = 0. Explain why the standard deviation in this case is smaller than in part (b). 7.5 Binomial Distribution The most important discrete probability distribution is the binomial distribution. You should be able to do the following: 84

14 . Recognize situations in which a binomial distribution is applicable.. Define the binomial random variable that is appropriate in a given situation. 3. Compute binomial probabilities using the formula and using Table in Appendix B. The primary characteristic of a binomial experiment is that each trial of the experiment must result in one of only two possible outcomes. More precisely, a binomial experiment possesses the following properties:. The experiment consists of a fixed number, n, of trials.. The result of each trial can be classified into one of two categories: success or failure. 3. The probability, p, of a success remains constant from trial to trial. 4. Each trial of the experiment is independent of the other trials. The binomial random variable, X, counts the number of successes in the n trials of a binomial experiment. The probability of observing x successes in the n trials of a binomial experiment is P ( X = x ) = n! x! ( n x )! p x ( p ) n x, for x = 0,,..., n This probability can also be found using Table in Appendix B, for selected values of n and p. Example 7.6 The owner of a boat used for deep-sea fis hing charters advertises that his clients catch at least one large salmon (over 5 pounds) on 60% of the charters. Suppose you book four charters for the coming year. Find the probability distribution of X, the number of charters on which you catch at least one large salmon. Solution Assuming the result of each charter is independent of the results of other charters, we are dealing with a binomial experiment. A good habit to develop when dealing with any binomial experiment is to begin by clearly identifying the following four characteristics of the experiment: n = 4 success = a charter on which at least one large salmon is caught p =.6 X = the number of charters on which at least one large salmon is caught The binomial random variable X can take any one of the values 0,,, 3, and 4. The probabilities with which these values are assumed can be computed from the formulas for binomial probabilities: p( 0) = 4! 0! 4! (. 6 )0 (.4 ) 4 = 4 3 ()(. 056 ) =

15 p() = 4!! 3! (. 6) (. 4) 3 = 4 3 (. 6 )( ) = p( ) = 4!!! (.6 ) (. 4 ) = 4 3 (. 36 )(. 6 ) =.3456 p( 3) = 4! 3!! (. 6) 3 (. 4 ) = 4! 0 p (4) = (.6) 4 (.4 ) = 4!0! 4 3 (. 6 )(. 4 ) = (.96)() = Hence, the probability distribution of X is as follows: x p(x) Example 7.7 Let X be a binomial random variable with n = 0 and p =.4. Find the following probabilities using Table in Appendix B: a) P(X 0) b) P(X = ) c) P(3 X 5) Solution a) Because the cumulative binomial probabilities tabulated in Table are of the form P(X k), any probability that we want to find must be expressed in terms of probabilities of this form. Notice that the event (X 9) is the complement of the event (X 0). Therefore, applying the complement rule, we obtain P(X 0) = P(X 9) =.755 =.45 b) Expressing P(X = ) in terms of probabilities of the form P(X k), we obtain P(X = ) = P(X = 0,, or ) P(X = 0 or ) = P(X ) P(X ) = =

16 c) It is often helpful to begin by enumerating the x-values of interest before expressing the required probability in terms of probabilities of the form P(X k): P(3 X 5) = P(X = 3, 4, or 5) = P(X = 0,,, 3, 4, or 5) P(X = 0,, or ) = P(X 5) P(X ) = =. 87

17 Example 7.8 The Globe and Mail (3 September 987) has reported that 80% of Wall Street firms that speculate on corporate takeovers before they become public also help to underwrite such deals. A random sample of 0 firms that speculate on corporate takeovers is selected. a) Find the probability that at least half the firms selected underwrite takeovers. b) Find the probability that at most three of the firms selected do not underwrite takeovers. c) Find the expected value and variance of the number of firms selected that underwrite takeovers. Solution a) Summarizing the characteristics of the binomial experiment: n = 0 success = a selected firm underwrites takeovers p =.8 X The required probability is = the number of firms selected that underwrite takeovers P(X 5) = P(X 4) =.006 =.994 b) If we redefine a success to be a selected firm that does not underwrite takeovers, then we still have n = 0, and the probability of a success is now p =.. If Y is the number of firms selected that do not underwrite takeovers, then the required probability is P(Y 3) =.879 c) Using the same notation and definition of success as in part a), the expected value of X is E(X) = µ = np = (0)(.8) = 8 The variance of X is V(X) = = npq = (0)(.8)(.) =.6 EXERCISES 7.9 According to Forbes Magazine, 7% of Americans would like to fly in the Space Shuttle. Three Americans are selected at random and asked if they would like to fly in the Space Shuttle. Find the probability distribution of X, the number of persons selected who respond positively. 88

18 7.0 Consider a binomial random variable, X, with n = 5 and p =.7. Use Table in Appendix B to find the following probabilities: a) P(X 5) b) P(X 8) c) P(X = 4) d) P(8 X 3) 7. Sixty percent of the new cars sold in the United States in 984 were made by General Motors. A random sample of 5 purchases of new 984 cars is selected. a) Find the mean and standard deviation of the number of persons selected who purchased a G.M. car. b) Find the probability that at least 0 of those selected purchased a G.M. car. c) Find the probability that at least five of those selected did not purchase a G.M. car. 7.6 Poisson Distribution The Poisson distribution is a second important discrete distribution. You should be able to do the following:. Recognize situations in which a Poisson distribution is applicable.. Define the Poisson random variable that is appropriate in a given situation. 3. Compute Poisson probabilities using the formula and using Table in Appendix B. 4. Approximate binomial probabilities using the appropriate Poisson distribution. The Poisson distribution is applicable when the events of interest (such as incoming telephone calls) occur randomly, independently of one another, and rarely. By rarely we mean that there is only a very 89

19 small probability that the event of interest will occur within a very small specified time interval or region. The Poisson random variable, X, counts the number of occurrences of an event (i.e., number of successes) in a given time interval or region. A Poisson random variable can therefore take any one of infinitely many possible integer values. In contrast, a binomial random variable (which is also discrete) has only finitely many possible values. The probability of observing exactly x successes is given by P(X = x) = e µ µ x, for x = 0,,,... x! where µ is the average number of successes in the given time interval or region. This probability can also be found using Table in Appendix B, for selected values of µ. It is important to ensure that the intervals specified in the definitions of X and µ are the same. For example, suppose X is the number of successes in a specified -hour period and you know that, on average, there are three successes per hour. Then µ must be defined as the average number of successes in the -hour period, and you set µ = 6. You should not obtain a common time period by defining both X and µ in terms of a -hour period, since your interest lies in the number of successes in the -hour period specified. A binomial distribution having n trials and a probability p of a success on each trial can be approximated by a Poisson distribution if p is very small (p <.05 at least). The appropriate approximating Poisson distribution is the one with µ = np, the mean of the binomial distribution. Example P.M. Records show that there is an average of three accidents each day in a certain city between and a) Use the Poisson probability formula to find the probability that there will be exactly one accident between and 3 P.M. on a particular day. b) Use the table of Poisson probabilities to check your answer to part a). c) Find the probability that there will be at least three accidents between and 3 P.M. on a particular day. d) Find the probability that there will be at least three accidents between and :30 P.M. on a particular day. 90

20 Solution a) Let X be the number of accidents between and 3 P.M. Then X satisfies the requirements for a Poisson random variable, since the accidents of interest occur randomly, independently, and rarely within a specified time period. Knowing that the average number of accidents within the specified time period is µ = 3, we obtain P ( X = ) = e µ µ x x! = e 3 3! = e 3 (3 ) () =.49 b) Before using Table in Appendix B, we must express the required probability in terms of the types of cumulative probabilities tabulated in Table : P(X = ) = P(X ) P(X = 0) = =.49 c) P(X 3) = P(X ) =.43 =.577 d) We must first redefine X and µ in terms of the new time period of interest. Let X be the number of accidents between and :30 P.M. The average number of accidents in this time period is µ = 3/ =.5. Using Table, we obtain P(X 3) = P(X ) =.809 =.9 Example 7.0 Consider a binomial random variable X with n = 00 and p =.005. Approximate the values of the following probabilities: a) P(X = ) b) P(X 4) Solution Binomial distributions with p =.005 are not tabulated in Table in Appendix B. The required binomial probabilities can be approximated, however, using the Poisson distribution with µ = np = (00)(.005) =. a) P(X = ) = P(X ) P(X ) =.84 (from Table ) b) P(X 4) = P(X 3).98 =.09 (from Table ) 9

21 EXERCISES 7. A harried executive claims that his productivity suffers from having to deal with an average of random interruptions each morning between 9 A.M. and noon. a) Use the Poisson probability formula to find the probability that he will be interrupted eight times during a particular morning. b) Use the table of Poisson probabilities to check your answer to part a). c) Find the probability that he will be interrupted at least six times between 9 and 0 A.M. on a particular morning. 7.3 The authors of a textbook claim that only % of the pages in the book contain an error. Suppose 0 pages from the book are selected at random. a) Use the binomial probability tables to find the probability that at least one of the pages selected contains an error. b) Use the Poisson probability tables to approximate the binomial probability found in part a). 9