Customers arrive at an establishment according to a Poisson process of frequency  = 5 per...

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?

Solutions

Expert Solution

solution:

mean number of arrival = 5 per hour

opening time of establishment = 9:00 am

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 =

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

orchestra answered 1 year ago

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