{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW 17 Answers (5.2) - January N t is a Poisson process with...

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

Math331, Spring 2008 Instructor: David Anderson Section 5.2 Homework Answers Homework: pgs. 211 - 212, #’s 3, 7, 11, 15. 3. n , the number of people in the theater, is 80. The probability that any one of them is an illegal immigrant is p = . 025. Since n >> 1 and p << 1 and np = 2, if X is the RV giving the number of people in the theater that are illegal immigrants, we may assume X is Poisson with parameter λ = 2. Thus, P { X 2 } = 1 - P { X 1 } ≈ 1 - e - 2 - e - 2 2 1 = . 594 . 7. X is a Poisson RV. Therefore, for some λ > 0 we have that for any x ∈ { 0 , 1 , 2 , ... } p ( x ) = e - λ λ x x ! . We are told that p (1) = p (3). Therefore, e - λ λ = e - λ λ 3 3! = e - λ λ 3 6 = λ 2 = 6 = λ = 6 . Thus, p (5) = e - 6 6 5 / 2 5! 0 . 0634 . 11. Let our time units be hours. Then, if N ( t ) is the number of shooting stars that have arrived by time t , N ( t ) is a Poisson process with rate λ = 5. We are interested in P ( N (1 / 2) = 3). N (1 / 2) is a Poisson random variable with parameter 5 * 1 / 2. Therefore, P ( N (1 / 2) = 3) = e - 5 * 1 / 2 (5 * 1 / 2) 3 3! 0 . 2138 . 15. Let N ( t ) be the number of accidents by time t , where t is the number of days into

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.

Unformatted text preview: January. N ( t ) is a Poisson process with rate 3. Therefore, P ( N (1) = 0) = e-3 ≈ . 0498 . By the independence of increments of Poisson processes, the probability of no accidents on any given day is equal to 0.0498. Therefore, if Y gives the number of days in January for which there are no accidents, Y is binomial with parameters n = 31 and p = 0 . 0498. Therefore, P ( Y = 3) = p 31 3 P . 0498 3 (1-. 0498) 2 8 = 31 * 30 * 29 6 . 0498 3 . 9502 2 8 = 0 . 1328 . 1 Another way to do this is to consider Y as approximately Poisson. This can be (barely) justifed because n is moderately large, p is small and np = 31 * . 0498 = 1 . 5438. Thus, Y is approximately Poisson with λ = 1 . 5438 and so P ( Y = 3) ≈ e-1 . 5438 (1 . 5438) 3 / 6 ≈ . 1310 . Note that these answers are mildly diFerent. 2...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern