Customers arrive in a single server queue to be serviced acc

Customers arrive in a single server queue to be serviced according to Poisson process with intensity 5 customers an hour. That is, there is a single server who services each customer in the order they arrive while all the customers wait in line.

If the service times of the customers are independent uniform r.v.

Solution

Mean Of Poisson in one hour =5

If T is service time of customer, and T is given to be independent uniform on [0,4]

PDF of T = 0.25,

Arrival of the third customer

The customer can come at any time betwee 8.00 and 9.00 a.m. in one hour

p =0.25 through out

Mean = 0.25(1/2) = 0.125

Variance = 0.25(1/3)-0.125^2 = 0.0833-0.015625

= 0.0677