IOE265HW4 - Homework 4 Solutions: Chapter 3 48. Let S =...

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Chapter 3 48. Let S = comes to a complete stop, so p=0.25 and n=20 a. P(X 6) = B(6;20,.25) = .786 b. P(X = 6) = b(6;20,.20) = B(6;20,.25) - B(5;20,.25) = .786 - .617 = .169 c. P(X 6) = 1 – P(X 5) = 1 - B(5;20,.25) = 1 - .617 = .383 d. E(X) = (20)(.25) = 5. We expect 5 of the next 20 to stop. 54. 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 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) 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 (paralleling the success rate we want in the entire batch) The plan in (c) seems most satisfactory in these respects. 1
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This note was uploaded on 03/07/2008 for the course IOE 265 taught by Professor Jin during the Winter '07 term at University of Michigan.

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IOE265HW4 - Homework 4 Solutions: Chapter 3 48. Let S =...

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