Hw_6_Soln_07 - Andrea Gruber Jery Stedinger Fall 2007...

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Andrea Gruber & Jery Stedinger Fall 2007 HOMEWORK #6 SOLUTIONS: 1. a)  Γ (k+1) =  0 s k e -s ds =  0 s k (-e -s )’ds = s k (-e -s ) 0 - 0 (s k )’(-e -s )ds                      =  0 k(s k-1 ) (e -s )ds = k Γ (k) b)  ( 29 ( 29 0 , 1 / 1 Γ = - - x e x x f x X β α α α β E[X] = 0 x f ( x )d ( 29 - Γ 0 1 dx e x x β α β α = ( 29 - Γ 0 1 β β α β β α x d e x x = ( 29 - Γ 0 ds e s s α α β , β / x s = = ( 29 ( 29 1 + Γ Γ α α β = αβ 2) #80bd Devore 7 th Edition : Section 3.6 (p.124-5): #76bd (Devore 6 th Edition: Section 3.6 (p.137-8) X ~ Poisson ( ν = 8) b) P(6 X 9) = F(9;8) – F(5;8) = .526 d) E(X) = ν = 8, σ X = 2.83 ν = , so P(X > 10.83) = P(X 11) = 1 – P(X 10) = 1 - .816 = .184 3) λ = 6 cars/hr a) The probability that no cars arrive in a 15 minute period? ν = λ t = 5 . 1 60 / 15 6 = for the Poisson process P(X=0| ν =1.5) = 5 . 1 0 ! 0
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