Asst5-suggested

# Asst5-suggested - ⎜ ⎜ ⎝ ⎛ Following the same...

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Statistics 427: Homework # 5 Solutions to Suggested Problems 3.58 Let p denote the actual proportion of defectives in the batch, and X denote the number of defectives in the sample. a. P(the batch is accepted) = P(X 2) = B(2;10,p) p .01 .05 .10 .20 .25 P(accept) 1.00 .988 .930 .678 .526 b. c. P(the batch is accepted) = P(X 1) = B(1;10,p) p .01 .05 .10 .20 .25 P(accept) .996 .914 .736 .376 .244 d. P(the batch is accepted) = P(X 2) = B(2;15,p) p .01 .05 .10 .20 .25 P(accept) 1.00 .964 .816 .398 .236 e. We want a plan for which P(accept) is high for p .1 and low for p > .1 The plan in d seems most satisfactory in these respects. 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 1.0 0.5 0.0 p P(accept)

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3.71 a. Possible values of X are 5, 6, 7, 8, 9, 10. (In order to have less than 5 of the granite, there would have to be more than 10 of the basaltic). P(X = 5) = h(5; 15,10,20) = 0163 . 15 20 10 10 5 10 =
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Unformatted text preview: ⎜ ⎜ ⎝ ⎛ . Following the same pattern for the other values, we arrive at the pmf, in table form below. x 5 6 7 8 9 10 p(x) .0163 .1354 .3483 .3483 .1354 .0163 b. P(all 10 of one kind or the other) = P(X = 5) + P(X = 10) = .0163 + .0163 = .0326 c. E(X) = 5 . 7 20 10 15 = ⋅ = ⋅ N M n ; V(X) = ( ) 9868 . 20 10 1 5 . 7 19 5 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ; σ x = .9934 μ ± σ = 7.5 ± .9934 = (6.5066, 8.4934), so we want P(X = 7) + P(X = 8) = .3483 + .3483 = .6966 3.74 a. h(x;10,15,50) b. When N is large relative to n, h(x;n,M,N) , , ; ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = N M n x b & so h(x;10,150,500) ( ) 3 ,. 10 ; x b = & c. Using the hypergeometric model, E(X) = 3 500 150 10 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ and V(X) = 06 . 2 ) 1 . 2 ( 982 . ) 7 )(. 3 )(. 10 ( 499 490 = = Using the binomial model, E(X) = (10)(.3) = 3, and V(X) = 10(.3)(.7) = 2.1...
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## This note was uploaded on 04/07/2010 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.

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Asst5-suggested - ⎜ ⎜ ⎝ ⎛ Following the same...

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