# 10s418hw4sol - 22. (a) P(0.900 – 0.005 < X < 0.900...

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HW4 – MATH 418 Solutions Wen-Yu Hua Ch 5: 1, 3, 5, 6, 10, 17, 21, 22, 24, 27 1. (a) 1 2 1 3 (1 ) 1 4 c x dx c - - = = (b) 3 2 1 3 3 2 ( ) (1 ) ( ) 4 4 3 3 x x F x c x dx x - = - = - + , -1 < x < 1 3. No, f(5/2) < 0 5. Must choose C so that 0.01 = 1 4 5 5(1 ) (1 ) c x dx c - = - So c = 1-(0.01)^(1/5) 6. (a) EX = 2 2 2 0 0 1 2 2 (3) 4 4 x y x e dx y e dy - - = = Γ = (b) By symmetry of f(x) about x = 0, EX = 0 (c) EX = 5 5 dx x = ∞ 10. (a) P(goes to A) = P(-5<x<15 or 20<x<30 or 35<x<45 or 50<x<60) = 2/3 since x is uniform(0, 60) (b) same as answer from a. 17. E(points) = 10*0.1 + 5*0.2 + 3*0.2 = 2.6 21. Let X = height of 25 yr old man, then X ~ N(71, 6.25) (a) 6’2’’ = 74 inch, so P(X>74) = 1-(P(x<=74) = 74 71 1 ( ) 0.1151 6.25 - = (b) 6’5’’ = 77 inch, so P(X>77 | X>72) = P(X>77 and X>72)/P(X>72) = P(X>77)/P(X>72) P(X>77) = 1-P(X<=77) = 77 71 1 ( ) 0.0082 6.25 - = P(X>72) = 1-P(X<=72) = 72 71 1 ( ) 0.3446 6.25 - = , so P(X>77|X>72) = 0.0238

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Unformatted text preview: 22. (a) P(0.900 – 0.005 < X < 0.900 + 0.005) = P(-1.67<Z<1.67) =2 (1.67) 1 0.9050 Φ- = Hence 9.5 percent will be defective (that is each will be defective with probability 0.0950) (b) P(-0.005/6 < Z < 0.005/6) = 0.005 2 ( ) 1 0.99 σ Φ- = When 0.005 ( ) 0.995 Φ = => 0.0019 = 23. With C denoting the life of a chip, and Φ the standard normal distribution function, we have (a) P(C<1.8*10^6) = 6 6 5 1.8*10 1.4*10 ( ) (1.33) 0.9082 3*10-Φ = Φ = Thus, if N is the number of the chips whose life is less than 1.8*10^6 then N is a binomial random variable with parameters (100, 0.9082). Hence, P(N>19.5) 19.5 90.82 ( ) 1 ( 24.7) 1 90.82*0.0918-≈ Φ = -Φ -≈ 27. P(X>5799.5) = P(Z > 799.5/sqrt(2500)) = P(Z>15.99) closed to 0...
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## This note was uploaded on 04/15/2010 for the course STAT 418 taught by Professor G.jogeshbabu during the Spring '08 term at Pennsylvania State University, University Park.

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10s418hw4sol - 22. (a) P(0.900 – 0.005 < X < 0.900...

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