STAT_333_Assignment_4_Solutions

STAT_333_Assignment_4_Solutions - STAT 333 Assignment 4...

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Type I Problems 1. #21, page 349 of the text, 9 th edition. E[Time] = E[Time waiting at 1] + E[Service at 1] + E[Time waiting at 2] + E[Service at 2] E[Service at 1] = 1/μ 1 E[Service at 2] = 1/μ 2 E[Time waiting at 1] = 1/μ 1 since we just have to wait for the current customer at 1 to finish E[Time waiting at 2] = 0*P(2 free when finish 1) + E[2 to finish]* P(2 not free when finish 1) = 1/μ 2 1 /(μ 1 2 )) by the alarm clock lemma So E[Time] = 2/μ 1 + 1/μ 2 (1 + μ 1 /(μ 1 2 )) 2. #34, page 351 of the text, 9 th edition. Let A = time until A dies, B = time until B dies, T i = kidney waiting time (for i = kidney 1, 2) a. P(A receives a kidney) = P(T < A) = λ/( λ + μ A ) by the alarm clock lemma b. P(B receives a kidney) = P(B receives the first kidney) + P(B receives the second kidney) = P(A < T 1 < B) + P(T 1 < A and T 1 + T 2 < B) = P(A = min{T 1 , A, B})*P(T 1 = min{T 1 , B}) + P(T1 = min{T 1 , A, B})*P(B>T1+T2|B>T1) = (μ A / (μ A + μ B + λ))*( λ / (μ B + λ)) + (λ / (μ A + μ B + λ))* ( λ / (μ B + λ)) using the alarm clock lemma and the memoryless property = ( λ / (μ B + λ))* ((μ A + λ) / (μ A + μ B + λ)) 3. #59, page 356 of the text, 9 th edition. Let 4000 be the event of a $4000 claim, 1 be the event it was type 1, and 2 be the event it was type 2. We want P(1|4000). Using Bayes’ rule, P(1|4000) = P(4000|1)P(1) / [P(4000|1)P(1) + P(4000|2)P(2) ] Since the Poisson processes for type 1 and 2 claims are independent, the process of all claims is also Poisson but with rate 10 + 1 = 11. Thus P(1) = 10/11 and P(2) = 1/11. The probability a claim is $4000 would theoretically be 0 since we have a continuous distribution of claim size. However, we can think of a $4000 claim being a claim that is between $4000.00 and $4000.01 (or any other similarly small interval.) Thus P(4000|1) ≈ f 1 (4000)(0.01) = (1/1000)e – 4000/1000 0.01 = (0.01/1000)e – 4 And P(4000|2) ≈ f 2 (4000)(0.01) = (1/5000)e – 4000/5000 0.01 = (0.01/5000)e – 0.8 Plugging everything in, the interval length (0.01) cancels out. You could basically ignore the fact that this was a continuous distribution and just use the pdf like a pmf… but I like to be precise. We get P(1|4000) = e
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STAT_333_Assignment_4_Solutions - STAT 333 Assignment 4...

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