Waiting Time for c Special Items to Occur

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Now consider a larger case In E(Y ) let N = 1000 and k = 100 Table 713 shows the expected waiting time for c special items to occur

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Table 712 k 10 20 30 40 50 60 70 80 90 100 E(Y ) 910000 476667 322903 244146 196275 164098 140986 123580 110000 991089

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The Meaning of the Mean

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80 60 k 40 20 20 40 60 E(Y) 80 100

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Note that, not surprisingly, if c is some percentage of k, then E(Y ) is approximately the same percentage of N, a result easily seen from the formula for E(Y ) The graph in Figure 79 shows this result as well

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Estimating k

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Above we have assumed that k is known, a dubious assumption at best On the basis of a sample, how can we estimate k To be speci c, suppose that a sample of 100 from a population of 400 shows 5 special items What is the maximum likelihood estimate of k This is not a negative hypergeometric situation but a hypergeometric situation Consider prob[k] =

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400 k 95

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400 100

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The ratio prob[k = 1]/prob[k] can be

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simpli ed to ((k + 1)(305 k))/((k 4)(400 k)) Seeing where this is > 1 gives k < 1905

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Table 713 c 10 20 30 40 50 60 70 80 90 100 E(Y ) 991089 198218 297327 396436 495545 594653 693762 792871 891980 991089

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7

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Waiting Time Problems

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1000 800 c 600 400 200 20 40 60 E(Y) 80 100

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The sample then has 20% special items and that is somewhat greater than the maximum likelihood estimator for the percentage of special items in the population Now suppose the population is of size N, the sample is of size s, and the sample contains sp special items It is easy to show that k the maximum likelihood estimator for k, the unknown number if special items in the population, is k= and further k sp = N s 1+ 1 N 1 N sp (N + 1) s s

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showing that the proportion of the population estimated is very close to the percentage of the special items in the sample, especially as the sample size increases, not really much of a surprise Here we have used the hypergeometric distribution to estimate k We nd exactly the same estimate if we use the negative hypergeometric waiting time distribution

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CONCLUSIONS

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This chapter has contained a wide variety of waiting time problems These are often not considered in introductory courses in probability and statistics and yet they offer interesting situations, both in their mathematical analysis and in their practical applications

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EXPLORATIONS

1 Show that if X1 is the waiting time for the rst binomial success and X2 is the waiting time for the second binomial success, then X1 + X2 has a negative binomial distribution with r = 2 2 Show that the expected waiting time for the pattern HT with a fair coin is four tosses

Explorations

3 (a) Suppose there are 4 people going for lunch as in the text How many lunches could a person who never pays for lunch expect (b) Repeat part (a) for a person who pays for lunch exactly once 4 A lot of 50 items contains 5 defective items Find the waiting times for the second defective item to occure if sampled items are not replaced before the next item is selected 5 Two fair coins are tossed and any that comes up heads is put aside This is repeated until all the coins have come up heads Show that the expected number of (group) tosses is 8/3 6 Professor Banach has two jars of candy on his desk and when a student visits, he or she is asked to select a jar and have a piece of candy At some point, one of the jars is found to be empty On average, how many pieces of candy are in the other jar 7 (a) A coin, loaded to come up heads with probability 06, is thrown until a head appears What is the probability an odd number of tosses is necessary (b) If the coin is fair, explain why the probabilities of odd or even numbers of tosses are not equal 8 Suppose X is a negative binomial random variable with p the probability of success at any trial Suppose the rth success occurs at trial t Find the value of p that makes this event most likely to occur