Homework 4  KEY. Jeff Brenion. June 16, Note: Many problems can be solved in more than one way; we present only a single solution here.


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1 Homework 4  KEY Jeff Brenion June 16, 2004 Note: Many problems can be solved in more than one way; we present only a single solution here. 1 Problem 21 Since there can be anywhere from 0 to 4 aces, the range space of X, that is, the set of values that X can take, is{0, 1, 2, 3, 4}. Every poker hand is equally likely, so we can find our desired probabilities by simply counting the number of possible hands and the number of hands that have a given number of aces. To create a poker hand with k aces, we must choose k cards from the 4 aces, and 5 k cards from the 48 nonace cards. Thus, ( 4 48 P (X = k = k( 5 k ( Problem 22 To find the mean and variance in this case, we simply need to apply the given formulas; no special tricks are necessary. E[X] = = 13 6 E[X 2 ] = = 7 V ar(x = E[X 2 ] E[X] 2 = =
2 3 Problem 215ab For part a, we know that the sum of all the probabilities must be 1. Thus, we have k k k 1 3 = k which implies k = 7. Now that we have k, we can find the mean and 8 variance in the usual manner. 4 Problem 34 E[X] = = 11 7 E[X 2 ] = = 3 V ar(x = E[X 2 ] E[X] 2 = = For part a, we can derive a value of Y for each possible value of X. Since the values of X have probabilities given, the probabilities for the corresponding values of Y will be identical; for instance, since X = 10 with probability 0.1, Y = 2000(12 10 = 4000 with probability 0.1. The full set of values is: (4000, 0.1(2000, 0.3(0, 0.4( 2000, 0.1( 4000, 0.1(else, 0 For part b, we can find the values in the standard manner. However, we can simplify the calculations for Y by taking advantage of the linear properties of mean and variance. E[X] = = 11.8 E[X 2 ] = = V ar(x = E[X 2 ] E[X] 2 = = 1.16 E[Y ] = 2000(12 E[X] = 2000( = 400 V ar(y = V ar(x = =
3 5 Problem 59 We can define the variable X to be equal to the number of the first test that fails; X is distributed geometrically with parameter p = Table 51 on page 124 gives probability functions for a number of distributions; we find that: P (X = 5 = = Problem 511 Since P (X = 5 = p 4 (1 p, the plot should be the graph of p 4 p 5 over the interval [0, 1]. To maximize P (X = 5, we look for critical points; thus, we take the derivative of p 4 p 5 and set it equal to zero. We find that 4p 3 5p 4 = 0, which has solutions p = 0 and p = 0.8. p = 0 is obviously not the solution, and analysis of the second derivative shows that p = 0.8 is indeed a maximum. This also makes some amount of intuitive sense; if 4 out of 5 trials succeed, it seems more likely that the 5th trial is the first to fail. 7 Problem 512 For part a, we can define the variable X to be the number of the trip that a sale occurs on; X is distributed geometrically with parameter p = 0.1. Table 51 on page 124 gives the mean for a geometric random variable as 1 p, from which it follows that we expect a sale after 10 attempts. Thus, it will cost = $46000 to make a sale. For part b, since we expect to spend $46,000 and will only profit $15,000, the trips should not be taken. For part c, we wish to know P (X > 20, since we can afford to make 20 trips. An easy way to do this is to find the probability of 20 failures, which is simply , or Problem 534 For this problem, we define the random variable X to be the number of orders. 3
4 For part a, we wish to know P (X > 3. The easiest way to do this is to look up our parameters in the Appendix, Table I, where we find that P (X 3 = Thus, the probability that the journey must be made is = For part b, the expected value of X is simply λ = 2, again by Table 51. For part c, we wish to find n such that P (X n > 0.9. Again, checking the table in the Appendix gives us n = 4. For part d, we cannot simply invoke Table 51 again. Any situation where there would have been a demand for 4 or more parts must be counted as 3, since only three crews can be serviced in that event. Thus, we have: E[services] = 0 P (X = P (X = P (X = P (X 3 = = For part e, since we expect to have 2 requests per day, and expect to service at the tool crib, it follow that the expected number of journeys per day is their difference, or Problem 4A We can consider X to be the sum of n Bernoulli random variables, each with parameter p, since we are observing each individual event and then adding 1 to our total for each success. Again by Table 51, the variance of each Bernoulli random variable is p(1 p, so their sum is np(1 p. 10 Problem 4B The easiest way to find the probability mass function for X is to treat this as a counting problem. There are 2 k 1 possible sequences of flips, since the last flip must be a head. Thus, we must assign the remaining n 1 heads to the remaining k 1 flips, which gives us: ( k 1 n 1 P (X = k = 2 k 1 To find E[X], we can simply find E[X i ] where X i is the expected number of flips to get the ith head after the (i 1st head is observed, and then sum 4
5 the E[X i ]s. Each E[X i ] is a geometric random variable with p = 0.5, so we expect each one to take 2 flips. Since there are n such E[X i ]s, their sum will be 2n. 11 Problem 4C In both cases, we must sum P (X = 0, P (X = 1, P (X = 2, and P (X = 3. For the binomial random variable, we have ( 100 P (X = k = (.01 k ( k k The poisson random variable has probability P (X = k = 1 e k! Summing up the probabilities for each one gives us and respectively, which are indeed very close. 5
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