Suppose small aircraft arrive at a certain airport according to a Poisson process with rate ? =8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter ? = 8t. a. ?What is the probability that exactly 6 small aircraft arrive during a 1-hour period? At least 6? At least 10? b. ?What are the expected value and standard deviation of the number of small aircraft that arrive during a 90-min period? c. ?What is the probability that at least 20 small aircraft arrive during a 2.5-hour period? That at most 10 arrive during this period?

Problem 85E Answer: Step1: We have Suppose small aircraft arrive at a certain airport according to a Poisson process with rate =8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter = 8t. We need to find, a. What is the probability that exactly 6 small aircraft arrive during a 1-hour period At least 6 At least 10 b. What are the expected value and standard deviation of the number of small aircraft that arrive during a 90-min period c. What is the probability that at least 20 small aircraft arrive during a 2.5-hour period That at most 10 arrive during this period Step2: Let “X” the number of small aircraft that arrive during time t and it follows poisson distribution parameter “” The probability mass function of poisson distribution is given by e x P(X) = x! , x = 0,1,2,3,...,n. Where, (mean of the poisson distribution) a). Given that time period t = 1hr. Then, = 8t = 8(1) = 8 Now, The probability that exactly 6 small aircraft arrive during a 1-hour period is given by P(exactly 6 small aircraft arrive during a 1-hour period) = P(X = 6) Consider, e (8)6 P(X = 6) = 6! 0.000335(262144) = 720 = 87.8182 720 = 0.1219. Therefore,The probability that exactly 6 small aircraft arrive during a 1-hour period is 0.1219. 1).P( t least = P(X 6) Consider, P(X 6) = 1 - P(X5) e (8)0 e (8)1 e (8)2 e (8)3 e (8)4 e (8)5 = 1 - { + + + + + } 0! 0 1! 1 2!2 33! 4 4! 5 5! 8 (8) (8) (8) (8) (8) (8) = 1 - (e ){ 0! + 1! + 2! + 3! + 4! + 5! } 1 8 64 512 4096 32768 = 1 - (0.000335){ + + 1 1 2 + 6 + 24 + 120 } = 1 - (0.000335){1+8+32+85.34+170.67+273.07} = 1 - (0.000335){570.08} = 1 - 0.1909 = 0.8090. Therefore, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8090. 2).P(At least 10) = P(X 10) Consider, (X 10) = 1 - P(X9) 8 0 8 1 8 2 8 3 8 4 8 5 8 6 8 7 = 1 - { e (8) + e (8) + e (8) + e (8) + e (8) + e (8) + e (8) + e (8) + 0! 1! 2! 3! 4! 5! 6! 7! e (8)8 e (8)9 8! + 9!...