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Unformatted text preview: STAT 410 Spring 2008 Homework #5 (due Friday, February 22, by 3:00 p.m.) 1. Let X, Y, and Z be i.i.d. Uniform [ , 1 ] random variables Find the probability distribution of W = X + Y + Z. That is, find ( ) w f W . Hint : If V = X + Y, we know the p.d.f. of V, f V ( v ) ( see Examples for 02/11/2008 ): f V ( v ) = v if 0 < v < 1, f V ( v ) = 2 – v if 1 < v < 2, f V ( v ) = 0 otherwise. Now use convolution formula to find the p.d.f. of W = V + Z. There will be 5 possible cases, 2 of them are "boring". ( ) & ¡ ¢ < < = otherwise 1 1 Z z z f ( ) & ¡ ¢ < < = & ¡ ¢ < < = otherwise 1 1 otherwise 1 1 Z w v w v w v w f ( ) w f W = ( ) ( ) ( ) £ ∞ ∞ = + ⋅ dv v w f v f w f Z V Z V (convolution) Case 1 : w < 0. ( ) w f W = 0. Case 2 : 0 < w < 1. Then w – 1 < 0. ( ) ( ) 2 1 2 W w dv v w f w = = & ⋅ . Case 3 : 1 < w < 2. Then 0 < w – 1 < 1. ( ) ( ) ( ) ( ) 2 3 3 1 2 1 2 1 1 1 W + = + = & & ⋅ ⋅ w w dv v dv v w f w w ( ) ( ) 2 1 2 1 + = ⋅ w w . Case 4 : 2 < w < 3. Then 1 < w – 1 < 2. ( ) ( ) ( ) ( ) 2 3 2 9 3 2 1 2 2 2 2 1 W w w w dv v w f w = + = = & ⋅ . Case 5 : w > 3. Then w – 1 > 2. ( ) w f W = 0. 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 ) = ( ) ( ) ( ) 1 F 1 x f x n n ⋅ ⋅ = 5 ⋅ ( 1 – x / 8 ) 4 ⋅ ( 1 / 8 ) , 0 < x < 8. P ( Y 1 < 2 ) = ( ) & 2 X min dx x f i . b) What is the probability that the largest fish weighs over 7 pounds?...
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 Spring '08
 AlexeiStepanov
 Probability, Probability distribution, Probability theory, y1, Discrete probability distribution, PX PX PX

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