mont4e_sm_ch03_sec08

29312 07069 00607 46 541 48 543 4643 08005

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Unformatted text preview: denote the number of blades in the sample that are dull. a) P ( X ≥ 1) = 1 − P( X = 0) P ( X = 0) = ( )( ) = () 10 0 38 5 48 5 38! 5!33! 48! 5!43! = 38!43! = 0.2931 48!33! P ( X ≥ 1) = 1 − P( X = 0) = 0.7069 b) Let Y denote the number of days needed to replace the assembly. ( )( ) = c) On the first day, P ( X = 0) = () ( )( ) = On the second day, P ( X = 0) = () 2 0 46 5 48 5 6 0 42 5 48 5 P(Y = 3) = 0.29312 (0.7069) = 0.0607 46! 5!41! 48! 5!43! = 46!43! = 0.8005 48!41! = 42!43! = 0.4968 48!37! 42! 5!37! 48! 5!43! On the third day, P(X = 0) = 0.2931 from part a. Therefore, P(Y = 3) = 0.8005(0.4968)(1-0.2931) = 0.2811. 3-106. a) For Exercise 3-97, the finite population correction is 96/99. For Exercise 3-98, the finite population correction is 16/19. Because the finite population correction for Exercise 3-97 is closer to one, the binomial approximation to the distribution of X should be b...
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This note was uploaded on 02/17/2010 for the course IE 33214 taught by Professor Feng during the Spring '10 term at University of Illinois at Urbana–Champaign.

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