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Unformatted text preview: X = 1.407 b. X has approximately a Poisson distribution with = np = 2, so P(X 4) = 1 P(X 3) = 1 F(3;2) = 1  .857 = .143 c. P(board works properly) = P(all diodes work) = P(X = 0) = F(0;2) = .135 Let Y = the number among the five boards that work, a binomial r.v. with n = 5 and p = .135. Then P(Y 4) = P(Y = 4 ) + P(Y = 5) = 5 4 ) 865 (. ) 135 (. 5 5 ) 865 (. ) 135 (. 4 5 + = . 00144 + .00004 = .00148 92. a. P(X = 10 and no violations) = P(no violations  X = 10) P(X = 10) = (.5) 10 [F(10;10) F(9;10)] = (.000977)(.125) = .000122...
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 Spring '08
 Johnson

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