Homework_7_Solution

Homework_7_Solution - Stat 414 1. 2.4-20 (5 points)...

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Stat 414 Homework 7 Solution Fall 2008 1. 2.4-20 (5 points) P(A) = .4, P(B) = .5, and P(C) = .1 P(at least one of the five is ineffective | A) = 1 – P(none of the five is ineffective | A) = 1 – (1 – .03) 5 = 1 - .97 5 . P(at least one of the five is ineffective | B) = 1 – P(none of the five is ineffective | B) = 1 – (1 – . 02) 5 = 1 - .98 5 . P(at least one of the five is ineffective | C) = 1 – P(none of the five is ineffective | C) = 1 – (1 – . 05) 5 = 1 - .95 5 . P(at least one of the five is ineffective) = P(at least one of the five is ineffective | A)*P(A) + P(at least one of the five is ineffective | B)*P(B) + P(at least one of the five is ineffective | C)*P(C) = (.4)( 1 - .97 5 ) + (.5)( 1 - .98 5 ) + (.1)( 1 - .95 5 ) P(C | at least one of the five is ineffective) = P(at least one of the five is ineffective | C)*P(C) / P(at least one of the five is ineffective) = .178 2. Using the information on the front inside cover of the text, do 2.5-2, a,b,d (5 points) a. (i) Binomial(5,.7). (ii) μ = 3.5, σ
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Homework_7_Solution - Stat 414 1. 2.4-20 (5 points)...

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