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Unformatted text preview: HW #6 Solutions Ang and Tang: 3.30, 3.34, 3.54, 3.58, 3.59 Due: 10/20/2011 3.30 Given: both I and N are Poisson processes, with ν I = 0.01/mi and ν N = 0.05/mi. Hence, along a 50mile section of the highway, I and N have Poisson distributions with respective means λ I = (0.01/mi)(50 mi) = 0.5 and λ N = (0.05/mi)(50 mi) = 2.5 (a) P ( N = 2) = N e λ ( λ N ) 2 / 2! = e –2.5 2.5 2 / 2 2245 0.257 (b) A , the number of accidents of either type, has the combined mean rate of occurrence ν A = ν I + ν N = 0.06, hence, A has a Poison distribution with λ A = (0.06/mi)(50 mi) = 3. Thus, P ( A ≥ 3) = 1 – P ( A = 0) – P ( A = 1) – P ( A = 2) = 1 – e –3 (1 + 3 + 3 2 /2) 2245 0.5768 Note that this part can also be solved without introducing variable A : P ( I + N ≥ 3) = 1 – P ( I + N < 3) = 1 – P ( I + N = 0) – P ( I + N = 1) – P ( I + N = 2) = 1 – P ( I = 0 ∩ N = 0) – [ P ( I = 1 ∩ N = 0) + P ( I = 0 ∩ N = 1)] – [ P ( I = 2 ∩ N = 0) + P ( I = 1 ∩ N = 1) + P ( I = 0 ∩ N = 2)] = 1 – P ( I = 0) P ( N = 0) – P ( I = 1) P ( N = 0) – P ( I = 0) P ( N = 1)] – P ( I = 2) P ( N = 0) – P ( I = 1) P ( N = 1) – P ( I = 0) P ( N = 2) But, P ( I = 0) = [ e –0.5 (0.5) /0!] = e –0.5 P ( N = 0) = [ e –2.5 (2.5) /0!] = e –2.5 P ( I = 1) = [ e –0.5 (0.5) 1 /1!] = e –0.5 /2 P ( N = 1) = [ e –2.5 (2.5) 1 /1!] = 2.5 e –2.5 P ( I = 2) = [ e –0.5 (0.5) 2 /2!] = e –0.5 /8 P ( N = 2) = [ e –2.5 (2.5) 2 /2!] = 3.125 e –2.5 So, P ( I + N ≥ 3) = 1 – e –0.5 e –2.5 – e –0.5 e –2.5 /2 – 2.5 e –0.5 e –2.5 – e –0.5 e –2.5 /8 – 1.25 e –0.5 e –2.5 – 3.125 e –0.5 e –2.5 = 1 – e –0.5 e –2.5 [1 + 0.5 + 2.5 + 0.125 + 1.25 + 3.125] = 1 – e –3 [1 + (0.5 + 2.5) + (0.125 + 1.25 + 3.125)] = 1 – e –3 [1 + 3 + 3 2 /2] 2245 0.5768 (c) Let us model this as an n = 2 binomial problem, where a “trial” corresponds to an accident in which...
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This note was uploaded on 01/02/2012 for the course CIVIL ENG 408 at USC.
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