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Unformatted text preview: = ⎜ ⎜ ⎛ 24 ⎞⎛12 ⎞ ⎟/ ⎟⎜ ⎟ ⎟⎜ ⎝ x ⎠⎝ 3 − x ⎠ ⎛ 36 ⎞ ⎜⎟ ⎜3 ⎟ ⎝⎠ (b) µ=E(X) = np= 3*24/36=2 V(X)= np(1p)(Nn)/(N1) =2*(124/36)(363)/(361)=0.629 (c) P(X≤2) =1P(X=3) =0.717 319 3101. Let X denote the number of men who carry the marker on the male chromosome for an increased risk for high blood pressure. N=800, K=240 n=10 a) n=10 ( )( ) = ( )( ) = 0.1201 P( X = 1) = ()
240 560 1 9 800 10 240! 560! 1!239! 9!551! 800! 10!790! b) n=10 P ( X > 1) = 1 − P ( X ≤ 1) = 1 − [ P ( X = 0) + P ( X = 1)] P( X = 0) = ( )( ) = ( ()
240 560 0 10 800 10 P ( X > 1) = 1 − P ( X ≤ 1) = 1 − [0.0276 + 0.1201] = 0.8523 240! 560! 0!240! 10!550! 800! 10!790! )( ) = 0.0276 3102. Let X denote the number of cards in the sample that are defective. a) ( )( ) = P( X = 0) = ()
20 0 120 20 140 20 P( X ≥ 1) = 1 − P( X = 0) 120! 20!100! 140! 20!120! = 0.0356 P( X ≥ 1) = 1 − 0.0356 = 0.9644
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.
 Spring '10
 Feng

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