# hw5sol - HW5 Solution Section 3.1 1 n = 10 p = 0.12(a P(X =...

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Section 3.1 1. n = 10 , p = 0 . 12 (a) P ( X = 3) = 10 3 ! ( . 12) 3 (1 - . 12) 10 - 3 = . 0847 (b) P ( X = 6) = 10 6 ! ( . 12) 6 (1 - . 12) 10 - 6 = . 0004 (c) P ( X 2) = P ( X = 0) + P ( X = 1) + P ( X = 2) = . 2785 + . 3798 + . 2330 = . 8913 (d) P ( X 7) = 3 . 085 × 10 - 5 (e) E X = np = 10( . 12) = 1 . 2 (f) Var X = np (1 - p ) = 10( . 12)(1 - . 12) = 1 . 056 8. X 1 B ( n 1 , p ), that is, X 1 is the sum of n 1 independent Bernoulli trials. Likewise, X 2 B ( n 2 , p ), X 2 is the sum of n 2 independent Bernoulli trials. As X 1 and X 2 are independent, so X 1 + X 2 is the sum of n 1 + n 2 independent Bernoulli trials. Therefore, X 1 + X 2 B ( n 1 + n 2 , p ). Section 3.2 1 (a) P ( X = 4) = (1 - 0 . 7) 3 × 0 . 7 = 0 . 0189 2 (a) P ( X = 5) = 4 2 ! × (1 - 0 . 6) 2 × (0 . 6)3 = 0 . 2074 Section 3.3 1 (a) P ( X = 4) = ( 6 4 ) × ( 5 3 ) ( 11 7 ) = 5 11 Section 3.4 1 (b) P ( X 3) = P ( X = 0) + P ( X = 1) + P ( X = 2) + P ( X = 3) = e - 3 . 2 × 3 . 2 0 0! + e

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## This note was uploaded on 04/16/2010 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.

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hw5sol - HW5 Solution Section 3.1 1 n = 10 p = 0.12(a P(X =...

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