20071ee131A_1_131a5sol

20071ee131A_1_131a5sol - p = 0 . 01, and np = 1: k = 0 k =...

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EE 131A Homework #5 Winter 2007 Solution K. Yao 1. P { the sum is 9 or 7 } = 10 36 = 5 18 = p . P ( X = k ) = 5 k · ( 5 18 ) k (1 - 5 18 ) 5 - k , k = 0 , 1 , 2 , 3 , 4 , 5. P ( X = 0) = 0 . 1965; P ( X = 1) = 0 . 3779; P ( X = 2) = 0 . 2907; P ( X = 3) = 0 . 1118; P ( X = 4) = 0 . 0215; P ( X = 5) = 0 . 0017. 2. P ( σ R 2 σ ) = P ( σ < R 2 σ ) = F R (2 σ ) - F R ( σ ) = (1 - e - 4 σ 2 / 2 σ 2 ) - (1 - e - σ 2 / 2 σ 2 ) = e - 1 / 2 - e - 2 = 0 . 6065 - 0 . 1353 = 0 . 4712. P ( R > 3 σ ) = 1 - P ( R 3 σ ) = 1 - F R (3 σ ) = e - 9 σ 2 / 2 σ 2 = e - 9 / 2 = 0 . 01111. f R ( r ) = dF R ( r ) dr = ( 0 , r < 0 r σ 2 e - r 2 / 2 σ 2 , 0 r < . 3. F X ( x ) = ( 0 , x < 0 1 - e - λx , x 0 . F X ( x k ) = k 5 = 1 - e - λx k , k = 1 , 2 , 3 , 4. 1 5 = 1 - e - λx 1 4 5 = e - λx 1 x 1 = ln 5 4 λ . x 2 = ln 5 3 λ , x 3 = ln 5 2 λ , x 4 = ln5 λ . 4. a. 1 = R 1 0 f X ( x ) dx = c R 1 0 x (1 - x ) dx = c [ x 2 2 - x 3 3 ] 1 0 = c 6 c = 6. b. P [ 1 2 < X 3 4 ] = 6 R 3 / 4 1 / 2 x (1 - x ) dx = 6 [ x 2 2 - x 3 3 ] 3 / 4 1 / 2 = 0 . 34375. c. F X ( x ) = 0 , x < 0 R x 0 f X ( t ) dt = 3 x 2 - 2 x 3 , 0 x 1 1 , 1 < x. 5. P ( X = 0) = e - λ λ 0 / 0! = e - λ = e - 2 . Thus, λ = 2. P ( X > 2) = 1 - P ( X = 0) - P ( X = 1) = 1 - e - 2 (1 + 2 1! + 2 2 2! ) = 1 - 5 e - 2 = 0 . 323. 6. For n = 10, p = 0 . 1, and np = 1: k = 0 k = 1 k = 2 k = 3 Binomial 0 . 3487 0 . 387 0 . 1937 0 . 0574 Poisson 0 . 3679 0 . 3679 0 . 1839 0 . 0613 For n = 20, p = 0 . 05, and np = 1: k = 0 k = 1 k = 2 k = 3 Binomial 0 . 3585 0 . 3774 0 . 1887 0 . 06 Poisson 0 . 3679 0 . 3679 0 . 1839 0 . 0613
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For n = 100,
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Unformatted text preview: p = 0 . 01, and np = 1: k = 0 k = 1 k = 2 k = 3 Binomial 0 . 366 . 3697 0 . 1849 0 . 061 Poisson . 3679 0 . 3679 0 . 1839 0 . 0613 7. a. The rate of people entering the store is 1/2 people per minute. Thus in 5 minutes, the average number of people entering the store is = 2 . 5 people. Probability of no one entering the store in this period is e-2 . 5 / 0! = 0 . 08208 . b. The probability that 4 or more people entering the store is 1- e-2 . 5 / 0!- 1 e-2 . 5 / 1!- 2 e-2 . 5 / 2!- 3 e-2 . 5 / 3! = 0 . 2425 . 2...
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This note was uploaded on 06/09/2010 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.

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20071ee131A_1_131a5sol - p = 0 . 01, and np = 1: k = 0 k =...

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