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sec_4_4_sol

# sec_4_4_sol - APPM 4/5570 Solutions to Problems from...

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APPM 4/5570 Solutions to Problems from Section 4.4 60. The pdf for X is f ( x ) = 0 . 01386 e - 0 . 01386 x , x > 0 and zero otherwise. It will be convenient to find the cdf once and use it to answer all questions. F ( x ) = Z x 0 0 . 01386 e - 0 . 01386 u du = 1 - e - 0 . 01386 x . (This holds for x > 0, for x < 0, F ( x ) = 0.) a. P ( X 100) = F (100) = 1 - e - 0 . 01386(100) 0 . 7499 P ( X 200) = F (200) = 1 - e - 0 . 01386(200) 0 . 9375 P (100 < X < 200) = F (200) - F (100) 0 . 9375 - 0 . 7499 = 0 . 1875 (I used “unrounded” values to compute 0 . 1875. It is okay to give the answer as 0 . 1876.) b. The mean for an exponential random variable with rate λ is 1 . The variance is 1 2 . So, in this case, μ = 1 λ = 1 0 . 01386 72 . 15 σ 1 = 1 λ 2 = 1 0 . 01386 2 5205 . 64 We want P ( X > μ +2 σ ) = P ( X > 216 . 45) = 1 - F (216 . 45) = 1 - h 1 - e - 0 . 01386(216 . 45) i 0 . 05 66. a. We know that for X Γ( α, β ), E [ X ] = αβ and V ar [ X ] = αβ 2 . So, we equate αβ = 20 and αβ 2 = 80 Solving for α and β gives α = 5 and β = 4

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b. We want to transform the problem from one involving a Γ(5 , 4) distribu- tion to one involving a Γ(5 , 1) distribution since the latter has cdf values tabulated in Table A.4.
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