02_11ans - S 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.15 ≈ 0.0566 Geometric p = 0.15 c Binomial n = 15 p = 0.15 P X = 4 = 11 4 85

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STAT 400 Examples for 02/11/2011 Spring 2011 1. a) Yes . Binomial, n = 15, p = 1 / 5 = 0.20. b) E ( X ) = n p = 15 0.20 = 3 . c) P ( X = 3 ) = ( ) ( ) 12 3 80 0 20 0 3 15 . . = 0.2501389 . OR P ( X = 3 ) = PMF @ 3 = 0.250 . OR P ( X = 3 ) = CDF @ 3 – CDF @ 2 = 0.648 – 0.398 = 0.250 . d) P ( X 5 ) = CDF @ 5 = 0.939 . e) P ( X 8 ) = 1 – CDF @ 7 = 1 – 0.996 = 0.004 . f) P ( 4 X 6 ) = CDF @ 6 – CDF @ 3 = 0.982 – 0.648 = 0.334 . OR P ( 4 X 6 ) = PMF @ 4 + PMF @ 5 + PMF @ 6 = 0.188 + 0.103 + 0.043 = 0.334 . 2. X = number of sales. Binomial, n = 5, p = 0.30. P( X = 2 ) = ( ) ( ) 3 2 70 . 0 30 . 0 2 5 = 0.3087 . OR P( X = 2 ) = PMF @ 2 = 0.309 . OR P( X = 2 ) = CDF @ 2 – CDF @ 1 = 0.837 – 0.528 = 0.309 .
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3. a) F F F S 0.85 0.85 0.85 0.15 0.092 . Geometric, p = 0.15. b) F F F F F F
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Unformatted text preview: S 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.15 ≈ 0.0566 . Geometric, p = 0.15. c) Binomial, n = 15, p = 0.15. P ( X = 4 ) = ( ) ( ) 11 4 85 . 15 . 4 15 ⋅ ⋅ ≈ 0.11564 . d) [ 14 trials: 3 S’s & 11 F’s ] S ( ) ( ) ⋅ ⋅ 85 15 3 14 11 3 . . ⋅ 0.15 ≈ 0.030837 . Negative Binomial, k = 4, p = 0.15. e) [ 9 trials: 2 S’s & 7 F’s ] S ( ) ( ) ⋅ ⋅ 7 2 85 . 15 . 2 9 ⋅ 0.15 ≈ 0.03895 . Negative Binomial, k = 3, p = 0.15. f) Binomial, n = 10, p = 0.15. P ( X = 3 ) = ( ) ( ) 7 3 85 . 15 . 3 10 ⋅ ⋅ ≈ 0.1298 ....
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This note was uploaded on 02/24/2011 for the course STAT 400 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.

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02_11ans - S 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.85 ⋅ 0.15 ≈ 0.0566 Geometric p = 0.15 c Binomial n = 15 p = 0.15 P X = 4 = 11 4 85

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