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

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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
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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...
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