Answer:
Step-by-step explanation:
Step1:
We have Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α =8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter μ = 8t
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
P(X) = , x = 0,1,2,3,...,n.
Where,
μ(mean of the poisson distribution)
a).
Given that time period t = 1hr.
Then,μ = 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,
P(X = 6) =
=
=
= 0.1219.
Therefore,The probability that exactly 6 small aircraft arrive during a 1-hour period is 0.1219.
1).P(At least 6) = P(X 6)
Consider,
P(X 6) = 1 - P(X5)
= 1 - {+++++}
= 1 - (){+++++}
= 1 - (0.000335){+++++}
= 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,
P(X 10) = 1 - P(X9)
= 1 - {+++++
(There are two sides for the length)
(There are two sides for the width)