STAT_333_Assignment_4_Solutions

# STAT_333_Assignment_4_Solutions - STAT 333 Assignment 4...

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 01/16/2011 for the course STAT 333 taught by Professor Chisholm during the Spring '08 term at Waterloo.

### Page1 / 5

STAT_333_Assignment_4_Solutions - STAT 333 Assignment 4...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online