410Hw06ans - STAT 410 Fall 2009 Homework#6(due Thursday October 8 by 4:00 p.m 1 Suppose the size of largemouth bass in a particular lake is

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Unformatted text preview: STAT 410 Fall 2009 Homework #6 (due Thursday, October 8, by 4:00 p.m.) 1. 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. P ( Y 5 > 7 ) = 1 – P ( Y 5 < 7 ) = 1 – ( 7 / 8 ) 5 ≈ 0.4871 . OR f max X i ( x ) = ( 29 ( 29 ( 29 1 F x f x n n ⋅ ⋅-= 5 ⋅ ( x / 8 ) 4 ⋅ ( 1 / 8 ) = 5 x 4 / 8 5 , 0 < x < 8. P ( Y 5 > 7 ) = ( 29 7 8 5 5 8 7 5 4 8 7 X 8 8 5 max x dx x dx x f i = = ∫ ∫ = 1 – ( 7 / 8 ) 5 ≈ 0.4871 . c) What is the probability that the largest fish weighs between 6 and 7 pounds? P ( 6 < Y 5 < 7 ) = ( 29 6 7 5 5 7 6 5 4 7 6 X 8 8 5 max x dx x dx x f i = = ∫ ∫ = ( 7 / 8 ) 5 – ( 6 / 8 ) 5 ≈ 0.2756 . d) What is the probability that the second largest (fourth smallest) fish weighs between 4 and 6 pounds? Second largest = Fourth smallest P ( 4 < Y 4 < 6 ) = ( 29 ( 29 ∫ ⋅ ⋅ ⋅ ⋅-- - --6 4 4 5 1 4 8 1 8 1 8 4 5 1 4 5 ! ! ! dy y y = ( 29 ( 29 ∫ ∫- =- ⋅ ⋅ ⋅ ⋅ 6 4 4 3 5 6 4 3 5 20 160 8 1 8 20 8 1 dy dy y y y y = ( 29 4 6 5 4 5 4 40 8 1 y y- ⋅ ≈ 0.4453 . OR Let W 6 = number of fish (out of 5) that weigh less than 6 pounds. W 6 has Binomial distribution, n = 5, p = 6 / 8 = 0.75. P ( Y 4 < 6 ) = P ( W 6 ≥ 4 ) = 5 C 4 0.75 4 0.25 1 + 5 C 5 0.75 5 0.25 0 = 0.3955 + 0.2373 = 0.6328. Let W 4 = number of fish (out of 5) that weigh less than 4 pounds. W 4 has Binomial distribution, n = 5, p = 4 / 8 = 0.50. P ( Y 4 < 4 ) = P ( W 4 ≥ 4 ) = 5 C 4 0.50 4 0.50 1 + 5 C 5 0.50 5 0.50 0 = 0.15625 + 0.03125 = 0.1875. P ( 4 < Y 4 < 6 ) = P ( Y 4 < 6 ) – P ( Y 4 < 4 ) = 0.6328 – 0.1875 = 0.4453 . 2. Three actuaries are independently hired to appraise the value of a company. The true value of the company is θ million dollars, and each actuary’s estimate is uniformly distributed between θ – 2 million dollars and θ + 3 million dollars. Find the probability that the actual value of θ lies between the lowest and the highest estimate....
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This note was uploaded on 10/14/2009 for the course STAT 410 taught by Professor Alexeistepanov during the Fall '08 term at University of Illinois at Urbana–Champaign.

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410Hw06ans - STAT 410 Fall 2009 Homework#6(due Thursday October 8 by 4:00 p.m 1 Suppose the size of largemouth bass in a particular lake is

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