mont4e_sm_ch03_sec08

# 8 n 100 96 n n v x np 1 p 40208 06206

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Unformatted text preview: ) = ( ) 3921225 b.) ( )( ) = 4845(1) = 0.001236 c.) P ( X = 4) = ( ) 3921225 20 80 4 0 100 4 P ( X = 6) = 0 , the sample size is only 4 d.) E ( X ) K ⎛ 20 ⎞ = 4⎜ ⎟ = 0 .8 N ⎝ 100 ⎠ ⎛ 96 ⎞ ⎛ N −n⎞ V ( X ) = np (1 − p )⎜ ⎟ = 4(0.2)(0.8)⎜ ⎟ = 0.6206 ⎝ 99 ⎠ ⎝ N −1 ⎠ = np = n 3-18 3-98. ( )( ) = (4 ×16 ×15 ×14) / 6 = 0.4623 ( ) (20 ×19 ×18 ×17) / 24 ( )( ) = 1 b) P ( X = 4 ) = = 0.00021 ( ) (20 ×19 ×18 ×17) / 24 a) P ( X = 1) = 4 1 16 3 20 4 4 16 40 20 4 c) P( X ≤ 2) = P ( X = 0) + P ( X = 1) + P ( X = 2) ( )( ) + ( )( ) + ( )( ) = () () () 4 0 16 4 20 4 4 1 16 3 20 4 4 2 16 2 20 4 = ⎛ 16×15×14×13 4×16×15×14 6×16×15 ⎞ + + ⎜ ⎟ 24 6 2 ⎝ ⎠ ⎛ 20×19×18×17 ⎞ ⎜ ⎟ 24 ⎝ ⎠ = 0.9866 d) E(X) = 4(4/20) = 0.8 V(X) = 4(0.2)(0.8)(16/19) = 0.539 3-99. N=10, n=3 and K=4 0.5 0.4 0.3 P(x) 0.2 0.1 0.0 0 1 2 3 x 3-100. (a) f(x)...
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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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