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! } 0 1 2 3 4 5 6 7 8 9 = 1 - (e 8 ){ (8) + (8) + (8) + (8) + (8) + (8) + (8) + (8) + (8) + (8) } 0! 1! 2! 3! 4! 5! 6! 7! 8! 9! 1 8 64 512 4096 32768 262144 2097152 = 1 - (0.000335){ + + 1 1 2 + 6 + 24 + 120 + 720 + 5040 + 16777216 134217728 40320 + 362880 } = 1 - (0.000335){1+8+32+85.34+170.67+273.07+364.09+416.10 +416.10+369.87} = 1 - (0.000335){2136.24} = 1 - 0.7156 = 0.2843. Therefore,Therefore, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2843. Step3: b). 1).Given that time period t = 90 min. = 1.5 hrs. We have = 8t = 8(1.5) = 12 hrs. Therefore, Expected value of X is E(X) = 12 hrs. 2).For poisson distribution variance is = 12 hrs. 3).Standard deviation is given = 12 = 3.4641. Therefore,Standard deviation is 3.4641. Step4: c). Given that time period t = 2.5 hrs.. We have = 8t = 8(2.5) = 20 hrs. 1).P(at least 20 small aircraft arrive during a 2.5-hour period) = P(X 20) Consider, P(X 20) = 1 - P(X 19) Here, P(X 19) is obtained from Excel by using the formula “=poisson(x,,FALSE)” “=poisson(0 to 19,20,FALSE)” x P(X 19) 0 2.06115E-09 1 4.12231E-08 2 4.12231E-07 3 2.7482E-06 4 1.3741E-05 5 5.49641E-05 6 0.000183214 7 0.000523468 8 0.001308669 9 0.002908153 10 0.005816307 11 0.010575103 12 0.017625171 13 0.027115648 14 0.03873664 15 0.051648854 16 0.064561067 17 0.075954196 18 0.084393552 19 0.088835317 sum 0.470257267 Hence, P(X 20) = 1 - P(X 19) = 1 - 0.4702 = 0.5298. Therefore,The probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298. 2).P t most 10 arrive during this period) = P(X 10) P(X 10) is obtained from Excel by using the fo =poisson(x,,FALSE)” “=poisson(0 to 10,20,FALSE)” x P(X 10) 0 2.06115E-09 1 4.12231E-08 2 4.12231E-07 3 2.7482E-06 4 1.3741E-05 5 5.49641E-05 6 0.000183214 7 0.000523468 8 0.001308669 9 0.002908153 10 0.005816307 Sum 0.010812 Hence, P(X 10) = 0.011. Therefore, probability that at most 10 arrive during this period is 0.011.