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Unformatted text preview: STAT 410 Summer 2009 Homework #5 (due Friday, July 10, by 4:00 p.m.) 1. Let X and Y be independent random variables, each geometrically distributed with the probability of success p , 0 < p < 1. That is, p X ( k ) = p Y ( k ) = ( 29 1 1 k p p , k = 1, 2, 3, , a) Find P ( X > Y ). [ Hint: You may find it helpful to find P ( X = Y ) first. ] P ( X = Y ) = ( 29 ( 29 = 1 Y X k k p k p = ( 29 ( 29  = 1 1 1 1 1 k k k p p p p = ( 29 [ ]  = 1 1 2 2 1 k k p p = ( 29 [ ]  = 2 2 1 n n p p = ( 29 2 2 1 1 p p = p p 2 . P ( X > Y ) + P ( X = Y ) + P ( X < Y ) = 1. Since P ( X > Y ) = P ( X < Y ), P ( X > Y ) = ( 29 ( 29 Y X P 1 2 1 = =  2 1 2 1 p p = p p 2 1 . OR P ( X > Y ) = ( 29 ( 29  = + = 1 1 1 1 1 1 y y x y x p p p p = ( 29 ( 29  = + = 1 1 1 1 2 1 1 y y x x y p p p = ( 29 ( 29 ( 29  = 1 1 2 1 1 1 1 y y y p p p p = ( 29  = 1 1 2 1 y y p p = ( 29 ( 29 [ ]  = 2 1 1 n n p p p = ( 29 ( 29 2 1 1 1 p p p = p p 2 1 . b) Find P ( X + Y = n ), n = 2, 3, 4, , and P ( X = k  X + Y = n ), k = 1, 2, 3, , n 1. P ( X + Y = n ) = ( 29 ( 29  = = = 1 1 Y P X P n k k n k = ( 29 ( 29  = 1 1 1 1 1 1 n k k n k p p p p = ( 29  = 1 1 2 2 1 n k n p p = ( 29 ( 29 2 2 1 1 n p p n , n = 2, 3, 4, . Since X and Y both have Geometric ( p ) distribution and are independent, X + Y has Negative Binomial distribution with r = 2. P ( X = k  X + Y = n ) = ( 29 ( 29 Y X P Y X X P n n k = + = + = = ( 29 ( 29 Y X P Y X P n k n k = + = = = ( 29 ( 29 ( 29 ( 29 2 2 1 1 1 1 1 1 n k n k p p n p p p p = 1 1 n , k = 1, 2, 3, , n 1. 2. Suppose the size of largemouth bass in a particular lake is uniformly distributed over the interval 0 to 8 pounds. A fisherman catches (a random sample of) 5 fish. X 1 , X 2 , X 3 , X 4 , X 5 Y k = k th smallest. First find F X ( x ) = P ( X x ) F X ( x ) = 8 8 1 x x dy = , 0 < x < 8. a) What is the probability that the smallest fish weighs less than 2 pounds? P ( Y 1 < 2 ) = 1 P ( Y 1 > 2 ). P ( Y 1 > 2 ) = P ( X 1 > 2, X 2 > 2, X 3 > 2, X 4 > 2, X 5 > 2 ) = ( 6 / 8 ) 5 0.2373. P ( Y 1 < 2 ) = 1 P ( Y 1 > 2 ) = 1 ( 6 / 8 ) 5 0.7627 . OR f min X i ( x ) = ( 29 ( 29 ( 29 1 F 1 x f x n n  = 5 ( 1 x / 8 ) 4 ( 1 / 8 ) , 0 < x < 8. P ( Y 1 < 2 ) = ( 29 2 X min dx x f i . b) What is the probability that the largest fish weighs over 7 pounds? P ( Y 5 > 7 ) = 1 P ( Y 5 < 7 ). P ( Y 5 < 7 ) = P ( X 1 < 7, X 2 < 7, X 3 < 7, X 4 < 7, X 5 < 7 ) = ( 7 / 8 ) 5 0.5129....
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This note was uploaded on 10/29/2009 for the course STAT 420 taught by Professor Stepanov during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 STEPANOV
 Statistics, Probability

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