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Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 3 - Problem 92e
Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 3 - Problem 92e

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# Automobiles arrive at a vehicle equipment inspection

ISBN: 9780321629111 32

## Solution for problem 92E Chapter 3

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 92E

Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate ? = 10 per hour. Suppose that with probability .5 an arriving vehicle will have no equipment violations. a. ?What is the probability that exactly ten arrive during the hour and all ten have no violations? b. ?For any fixed y ? 10, what is the probability that y arrive during the hour, of which ten have no violations? c. ?What is the probability that ten “no-violation” cars arrive during the next hour? [Hint: Sum the probabilities in part (b) from y =10 to ?.]

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Answer: Step1: Given, automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate = 10 per hour. Suppose that with probability .5 an arriving vehicle will have no equipment violations. Here, = 10 per hour. The formula for the poisson probability mass function is x P(x;) = ex! for x = 0,1,2,..... Here ‘e’ is the base of the natural logarithm system. Step2: a). To find the probability that exactly ten arrive during the hour and all ten have no violations. = t = 10× 1 = 10 Then, P(10 arrivals in one hour) ×P(all ten have no violation) e ()y P(Y = 10 and no violation) = × P(Y = 10) y! 10 e1(10) 10 = 10! × (0.5) = 0.1251 × 0.000979 = 0.000122. Therefore, the probability that exactly ten arrive during the hour and all ten have no violations is 0.000122. Step3: b). For any fixed y 10. we need to prove that, the probability that y arrive during the hour, of which have no violation. P( y arrive and exactly 10 have no violations) = P( exactly 10 have no violations/y arrive).P(y arrive) 10 10 = P(10 successes in y trail when p = 0.5) * e (10) y 10! y 10 y10 10 (10) = ( )10 (0.5) (0.5) e y! e10(5) = 10!(y10)! 10 y P( y arrive and exactly 10 have no violations) = e (5) . 10!(y10)! Step4: (5)u 5 c). Using the result from part (b) and the fact that u! = e . u = 0 To show that the probability that ten “ no violation” cars arrive during the next hour. e 5 10 And is given by 10! . 10 y (e) (5) P( exactly 10 without a violation) = 10!(y10)! y =10 y10 e10510 (5) = 10! (y10)! y =10 u e10510 (5) = 10! (u)! ( here, u = y - 10) u=0 10 10 = e 5 e5 10! 10 = e55 10! 10 e55 P( exactly 10 without a violation) = 10!

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