Suppose small aircraft arrive at a certain airport according to a Poisson proces

Question

Suppose small aircraft arrive at a certain airport according to a Poisson process with rate Poisson rv with parameter -8t. (Round your answers to three decimal places.) 8 per hour, so that the nur er ora als during a me ero of ours a (a) What is the probability that exactly 7 small aircraft arrive during a 1-hour period? What is the probability that at least 7 small aircraft arrive during a 1-hour period? What is the probability that at least 11 small aircraft arrive during a 1-hour period? (b) What is the expected value and standard deviation of the number of small aircraft that arrive during a 75-min period? expected value standard deviation (c) What is the probability that at least 24 small aircraft arrive during a 2.5-hour period? What is the probability that at most 15 small aircraft arrive during a 2.5-hour period? You may need to use the appropriate table in the Appendix of Tables to answer this question. Need Help? Read It Talk to a Tutor Submit Answer Save Progress Practice Another Version

Explanation / Answer

=8 per hour

µ= 8t

P(X=x) = e-8t(8t)x / x!

a)number of arrivals during time period of 1 hour =µ= 8*1 = 8

Probability that exactly 7 small aircrafts arrive during a 1 hour period:

P(X=7) = e-8(8)7 / 7! = 0.139587

Probability that at least 7 small aircrafts arrive during a 1 hour period:

P(X 7) = 1-P(X = 6) = 1 – 0.122138 = 0.8877862 (Using the given formula, P(X=x))

Probability that at least 11 small aircrafts arrive during a 1 hour period:

P(X11) = 1 – P(X = 10) = 1-0.099262 = 0.900738

b)For poisson random variable, mean = variance = 8t ( parameter)

therefore, the expected value and standard deviation of the number of small aircraft that arrives during a 75-min period :

expected value = 8*(1.25)=10

standard deviation = sqrt(10) = 3.16228

c) number of arrivals during time period of 2.5 hour =µ= 8*2.5 = 20

Probability that at least 24 small aircraft arrives during a 2.5 hour period:

P(X24)= 1 – P(X=23) =1 – 0.066881 = 0.933199

Probability that atmost 15 small aircraft arrives during a 2.5 hour period:

P(X15) = 15x=0 e-20(20)x / x! = 0.15651

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