# ch2_add_6 - . 8) N (100 , 80) That we have E [ X ] = = 100...

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Normal Approximation to the Binomial Example: 20% of customers at a bakery will buy a brownie. 500 customers arrive at the bakery in a day. Assume that individual customers make their purchases independently. (a) What is the probability that more than 110 customers buy a brownie? (b) How many brownies does the bakery need to have in stock so that the probability of selling out in a day is 1%? Solution: Let X = number of customers who buy a brownie in a day. Assuming that the purchases of individual customers are iid Bernoulli trials with “probability of success” being 20%, X Bin (500 ,. 2). The computations needed to solve the problems given using the Binomial distribution directly are very cumbersome because we have to calculate quantities like 500 X i =111 ± 500 i ² ( . 2) i ( . 8) 500 - i Thus we will use the CLT to approximate the Binomial with a Normal distribution. If X Bin ( n,p ), the CLT tells us that X ˙ N ( np,np (1 - p )) for np > 5 and n (1 - p ) > 5 and that the approximation is very good for np > 20 and n (1 - p ) > 20. Here the approximation is very good and gives X ˙ N (500 × 0 . 2 , 500 × . 2 ×

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Unformatted text preview: . 8) N (100 , 80) That we have E [ X ] = = 100 and V ar [ X ] = 2 = 80. (a) The required probabilty is P ( X &gt; 110) = 1-P ( X 110) 1- 110-100 80 = 1-(1 . 12) = 1-. 86864 = . 1314 from the Normal cdf tables 1 (b) Let t = the number of brownies in stock. The bakery runs out of brownies if X t . To answer the question we need to nd t such that P ( X t ) = . 01 P ( X &gt; t ) = 1-P ( X t-1) 1- t-1-100 80 We set this equal to . 01 and use algebra and the Normal cdf table to nd t : 1- t-1-100 80 = . 01 is equivalent to t-1-100 80 = . 99; thus we nd the 99 th percentile of the normal distribution and equate it to t-1-100 80 . We get 2 . 33 80 = t-1-100 giving t-1 = 100 + 2 . 33 8 . 944272 and thus t = 120 . 8 + 1 122. That is, they need to stock 122 brownies to meet their policy that there is only a 1% chance that they will run out. 2...
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## ch2_add_6 - . 8) N (100 , 80) That we have E [ X ] = = 100...

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