Chapter 5 - 1 Pg 228 Question#1 x3 a = 1 c = 0.75 c(1 x)dx...

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1. Pg 228 Question #1 a) dx x c ) 1 ( 1 1 2 - - = 1 1 1 3 3 - - x x c = 1 c = 0.75 b) Let the cumulative distribution function be F(x). F(x) = - + - = - x x x dy y 1 3 2 3 2 3 4 3 ) 1 ( 4 3 for -1 < x < 1 1
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Pg 228 Question #2 We first need to determine the value of c in the density of X. 1 0 2 = - dx xe c x (1) To solve the above integral we need to employ “integration by parts”: 2 2 2 4 2 x x x e xe dx xe - - - - - = (2) So from (1) and (2), we get that c = 0.25. P{X > 5} = 0.25 + = - - - 2 5 2 5 5 2 4 10 4 1 4 1 e e dx xe x = 14/4 * exp{-5/2} ≈ 0.2873. 2
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Pg 228 Question #4 a) P{X > 20} = 5 . 0 10 10 20 20 2 = - = x dx x b) Let the cumulative distribution function be F(y). F(y) = 10 , 10 1 10 10 2 - = y y dx x y F(y) = 0 for y < 10 c) P{X = @ least 4 will function for at least 15 hours} = i i i i - = 6 6 3 3 1 3 2 6 since P{X > 15} = 3 2 10 10 15 15 2 = - = x dx x . The assumption that we are making is the independence of the events that the devices exceed 15 hours. 4
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Pg 229 Question #10 a) P{Customer goes to destination A} = P{5<X<15 or 20<X<30 or 35<X<45 or 50<X<60} ‘‘ = P{5<X<15} + P{20<X<30} + P{35<X<45} + P{50<X<60} P{a < X < b} = 60 60 60 1 b a x dx a b b a - = = since X ~ Uniform(0, 60) Therefore, P{Customer goes to destination A} = 2/3.
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