hw4solu_427 - P(X≥2) = 1 – P(X≤1) = 1 – [P(X=0) +...

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HW 4 Solutions 48. a. P(X 2) = B(2;25,.05) = .873 d. P(X = 0) = P(X 0) = .277 e. E(X) = np = (25)(.05) = 1.25 V(X) = np(1 – p) = (25)(.05)(.95) =1.1875 x = 1.0897 59. a. P(rejecting claim when p = .8) = B(15;25,.8) = .017 b. P(not rejecting claim when p = .7) = P(X 16 when p = .7) = 1 - B(15;25,.7) = 1 - .189 = .811; for p = .6, this probability is = 1 - B(15;25,.6) = 1 - .575 = .425. 60. h(X) = 1 X + 2.25(25 – X) = 62.5 – 1.5X, so E(h(X)) = 62.5 – 1.5E(X) = 62.5 – 1.5np = 62.5 – (1.5)(25)(.6) = $40.00 68. a. X Hypergeometric N=15, n=5, M=6 b. P(X=2) = 420 . 3003 1260 5 15 3 9 2 6 P(X≤2) = P(X=0) + P(X=1) + P(X=2) 713 . 3003 2142 3003 1260 756 126 3003 1260 5 15 4 9 1 6 5 15 5 9
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Unformatted text preview: P(X≥2) = 1 – P(X≤1) = 1 – [P(X=0) + P(X=1)] = 706 . 3003 756 126 1 c. E(X) = 2 15 6 5 ; V(X) = 857 . 15 6 1 15 6 5 14 5 15 ; 926 . ) ( X V 82. a. P(X = 1) = F(1;.2) – F(0;.2) = .982 - .819 = .163 b. P(X 2) = 1 – P(X 1) = 1 – F(1;.2) = 1 - .982 = .018 c. P(1 st doesn’t 2 nd doesn’t) = P(1 st doesn’t) P(2 nd doesn’t) = (.819)(.819) = .671 89. = 1/(mean time between occurrences) = 2 5 . 1 a. t = (2)(2) = 4 b. P(X > 5 ) 1 – P(X 5) = 1 - .785 = .215 c. Solve for t , given = 2: .1 = e- t ln(.1) = - t t = 15 . 1 2 3026 . 2 years...
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This note was uploaded on 03/22/2011 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.

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