Answer:
a) 0.1558
b) 0.7983
c) 0.1478
Step-by-step explanation:
If we suppose that small aircraft arrive at the airport according to a Poisson process at the rate of 5.5 per hour and if X is the random variable that measures the number of arrivals in one hour, then the probability of k arrivals in one hour is given by:
[tex]\bf P(X=k)=\displaystyle\frac{(5.5)^ke^{-5.5}}{k!}[/tex]
(a) What is the probability that exactly 4 small aircraft arrive during a 1-hour period?
[tex]\bf P(X=4)=\displaystyle\frac{(5.5)^4e^{-5.5}}{4!}=0.1558[/tex]
(b) What is the probability that at least 4 arrive during a 1-hour period?
[tex]\bf P(X\geq4)=1-P(X<4)=1-(P(X=0)+P(X=1)+P(X=2)+P(X=3))=\\\\=1-\left(e^{-5.5}+(5.5)e^{-5.5}+\displaystyle\frac{(5.5)^2e^{-5.5}}{2!}+\displaystyle\frac{(5.5)^3e^{-5.5}}{3!} \right)=0.7983[/tex]
(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?
If we redefine the time interval as 12 hours instead of one hour, then the rate changes from 5.5 per hour to 12*5.5 = 66 per working day, and the pdf is now
[tex]\bf P(X=k)=\displaystyle\frac{(66)^ke^{-66}}{k!}[/tex]
and we want P(X ≥ 75) = 1-P(X<75). But
[tex]\bf P(X<75)=\displaystyle\sum_{k=0}^{74}\displaystyle\frac{(66)^ke^{-66}}{k!}=0.852[/tex]
hence
P(X ≥ 75) = 1-0.852 = 0.1478
