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HW 3 Solutions

# HW 3 Solutions - HW 3 Solutions 4.136 a Observing that the...

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HW 3 Solutions 4.136 a. Observing that the exponential distribution is a special case of the gamma distribution, we can use the gamma moment–generating function with α = 1 and β = θ: t t m θ - = 1 1 ) ( , t < 1/θ. b. The first two moments are found by 2 ) 1 ( ) ( t t m θ - θ = , E ( Y ) = θ = ) 0 ( m . 3 ) 1 ( 2 ) ( t t m θ - θ = , E ( Y 2 ) = 2 2 ) 0 ( θ = m . So, V(Y) = 2θ 2 – θ 2 = θ 2 . 4.140 a. Gamma with α = 2, β = 4 b. Exponential with β = 3.2 c. Normal with μ = –5, σ 2 = 12 6.37 The mass function for the Bernoulli distribution is y y p p y p - - = 1 ) 1 ( ) ( , y = 0, 1. a. t x ty tY Y pe p y p e e E t m + - = = = = 1 ) ( ) ( ) ( 1 0 1 1 . b. n t n i Y tW W pe p t m e E t m i ] 1 [ ) ( ) ( ) ( 1 + - = = = = c. Since the mgf for W is in the form of a binomial mgf with n trials and success probability p , this is the distribution for W . 6.40 Using Theorem 6.4 we know that is a Chi-Squared Random Variable with 2 degrees of freedom 6.56 Let U = total service time for two cars. Similar to Ex. 6.13, U has a gamma distribution with α = 2, β = 1/2. Then, P ( U > 1.5) = du ue u - 5 . 1 2 4 = .1991.

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