Changes in airport procedures require considerable planning....

Changes in airport procedures require considerable planning. Arrival rates of aircraft are important factors that must be taken into account. Suppose small aircraft arrive at a certain airport, according to a Poisson process, at the rate of 6 per hour. Thus, the Poisson parameter for arrivals over a period of hours is $\mu=6 t$ (a) What is the probability that exactly 4 small aircraft arrive during a 1 -hour period? (b) What is the probability that at least 4 arrive during a 1 -hour period? (c) If we define a working day as 12 hours, what is the probability that at least 75 small aircraft arrive during a working day?

Transcript

00:01 Let's understand the concept of the poison distribution distribution is one parametric new discrete probability distribution in this event occurred within a given period of time or a fixed interval of time the mean and the variance are equal in the poison distribution variance are equal to the parameter of the present distribution which is lambda now in this question we have to find in part a that probability that exactly four small aircraft arrived during a one period of time therefore let x represent the number of aircraft arrive the number of aircraft arrive during one hour one hour period therefore it is given that mu equals to gt equals to six therefore p x equals to four is equals to e to the power lambda lambda to the power x by x factorial where x equals to 0 1 2 3 and so on therefore p x equals to 4 which equals to e to the power minus 6 6 to the power 4 by 4 factorial therefore which is equals to 0 .002479 into 1296 by 242479 which is equals to 0 .139 therefore the required probability that the exactly four small aircraft arrived during a one hour period is 0 .139 now moving to the part b here we have to find the probability that at least four arrived during a one period therefore p x x greater than equals to 4 it equals to 1 minus p x less than equals to 4 therefore 1 minus p x equals to 0 plus p x equals to 1 plus p x equals to 2 plus p x equals to 3 now which is equals to 1 minus e to the power minus 6 6 to 6 to the the power 0 by 0 factorial minus e to the power minus 6 6 to the power 1 factorial minus e to the power 6 6 to the power 2 by 2 factorial minus e to the power 6 6 to the power 3 by 3 factorial which is equals to 1 minus e to the power minus 6 by 1 plus 6 plus 36 plus 206 by 6 therefore equals to 1 minus 0 .002479 into 61 which is equals to 1 minus 0 .15104 therefore p x greater than equals to 4 is equals to 0 .15124 therefore p x greater than equals to 4 is equals to 0 .848796.

04:53 Therefore, which is a prox equals to px greater than equals to 4 equals to 0 .8488...

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