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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 > 110) = 1P ( X 110) 1 110100 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 > t ) = 1P ( X t1) 1 t1100 80 We set this equal to . 01 and use algebra and the Normal cdf table to nd t : 1 t1100 80 = . 01 is equivalent to t1100 80 = . 99; thus we nd the 99 th percentile of the normal distribution and equate it to t1100 80 . We get 2 . 33 80 = t1100 giving t1 = 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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 Spring '11
 Zhou
 Binomial, Probability

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