In: Statistics and Probability
Customers arrive at an establishment according to a Poisson process of frequency = 5 per hour. Since the establishment opens at 9:00 am:
a) What is the probability that exactly a client arrived by 9:45 am?
b) What is the probability of maximum five clients by 11:45 am?
solution:
mean number of arrival = 5 per hour
opening time of establishment = 9:00 am
a)
we have to find probability that one client arrive till 9: 45 am (means one client arrive in 45 minutes)= P(X=1)
so, average number of clilent arrive in 45 minutes = 45/(60/5) = 3.75
using, formula for poisson distribution =
b)
probability that maximum 5 clients by 11:45 am =
time upto 11:45am from 9:00am = 2hour 45 minutes = 165 minutes
average number of arrival in 165 minutes = = 165 / (60/5) = 13.75
= P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
= 0 + 0 + 0.0001 + 0.00046 + 0.00159 + 0.00437 =0.0065