This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
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.
 Fall '08
 AlexeiStepanov

Click to edit the document details