HW6 - 2 0! = e-2 = . 1353 2 5.60 Solution: λt = 6 ,p ( x ;...

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Section 5.4 Hypergeometric 5.29 Solution: (a) P(exactly 2 of them will be face cards) = ( 12 2 )( 40 5 ) ( 52 7 ) = 0 . 3246 (b) P(at least 1 of them will be a queen) =1-P(none of them will be a queen) = 1 - ( 4 0 )( 48 7 ) ( 52 7 ) = 0 . 4496 5.36 Solution: N=25,n=3,k=the number of defectives containing in a box (a) h (0; 25 , 3 , 3) = ( 3 0 )( 22 3 ) ( 25 3 ) = 77 115 (b) h (1; 25 , 3 , 1) = ( 1 1 )( 24 2 ) ( 25 3 ) = 3 25 5.37 Solution: b ( 0; 3 , 3 25 ) = ( 3 0 )( 3 25 ) 0 ( 22 25 ) 3 = 0 . 6815 5.38 Solution: X followes the hypergeometric distribution h ( x ; N = 10 ,n = 4 ,k = 3) . μ = nk N = 4 × 3 10 = 1 . 2 Section 5.5 Negative binomial 5.51 Solution: p = 0 . 3 ,k = 5 ,x = 10 b * (10; 5 , 0 . 3) = ± 10 - 1 5 - 1 ² (0 . 3) 5 (0 . 7) 5 = ± 9 4 ² (0 . 3) 5 (0 . 7) 5 = 0 . 0515 5.52 Solution: p = 1 6 ,k = 2 ,x = 8 1
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b * (8; 2 , 1 6 ) = ± 8 - 1 2 - 1 ²± 1 6 ² 2 ± 5 6 ² 6 = ± 7 1 ²± 1 6 ² 2 ± 5 6 ² 6 = 0 . 0651 5.54 Solution: (a) p = 1 2 ,k = 3 ,x = 7 b * (7; 3 , 1 2 ) = ± 7 - 1 3 - 1 ²± 1 2 ² 3 ± 1 2 ² 4 = ± 6 2 ²± 1 2 ² 7 = 0 . 1172 (b) p = 1 2 ,k = 1 ,x = 4 b * (4; 1 , 1 2 ) = ± 4 - 1 1 - 1 ²± 1 2 ² 1 ± 1 2 ² 3 = ± 3 0 ²± 1 2 ² 4 = 0 . 0625 Section 5.6 Poisson 5.59 Solution: λt = 2 ,p ( x ; 2) = e - 2 2 x x ! (a) P ( X 4) = 1 - P ( X 3) = 1 - 3 X x =0 P ( x ; 2) = 1 - 0 . 8571 = 0 . 1429 (b) P ( X = 0) = e - 2
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Unformatted text preview: 2 0! = e-2 = . 1353 2 5.60 Solution: λt = 6 ,p ( x ; 6) = e-6 6 x x ! (a) P ( X < 4) = P ( X ≤ 3) = 3 X x =0 P ( x ; 6) = . 1512 (b) P (6 ≤ X ≤ 8) = P ( X ≤ 8)-P ( X ≤ 5) = 8 X x =0 P ( x ; 6)-5 X x =0 P ( x ; 6) = . 8472-. 4457 = . 4015 5.65 Solution: λt = np = (10000)(0 . 001) = 10 ,p ( x ; 10) = e-10 10 x x ! P (6 ≤ X ≤ 8) = P ( X ≤ 8)-P ( X ≤ 5) = 8 X x =0 P ( x ; 10)-5 X x =0 P ( x ; 10) = . 3328-. 0671 = . 2657 5.71 Solution: (a) λt = 14 P ( X > 10) = 1-P ( X ≤ 10) = 1-10 X x =0 P ( x ; 14) = 1-. 1757 = . 8243 (b) λt = 14 3...
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This note was uploaded on 02/04/2009 for the course STATISTIC 3011 taught by Professor Sally during the Fall '08 term at University of Minnesota Crookston.

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HW6 - 2 0! = e-2 = . 1353 2 5.60 Solution: λt = 6 ,p ( x ;...

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