Question

9. Customers arrive at a service facility according to a Poisson process with an average rate of 5 per hour. Find 

(a) the probabilities that (G) during 6 hours no customers will arrive, (i) at most twenty five customers will arrive; 

(b) the probabilities that the waiting time between the third and the fourth customers will be (i) greater than 30 min.,(ii) equal to 30 min., (ii)i greater than or equal to 30 min. 

(c) the probability that after the first customer has arrived, the waiting time for the fifth customer will be greater than an hour; 

(d) for the same waiting time-its expected value and the standard deviation. 

10. Answer all questions from Exercise 9 under the condition that the mean interarrival time is 30 min.

Answer 1

The number of arrivals at in time is a random variable can be modeled by Poisson distribution. The mean of inter arrival time is .

The Poisson PMF is

. Here is the Poisson parameter - the average number of events in unit time.

a) The probability,

R command below:

> ppois(25,12)
[1] 0.9996922

b) The inter arrival times are exponentially distributed with . The probability that you have to wait greater than 30 minutes is (wait more than 1/2 hours) is

. (Point probability is 0 for continuous distribution).

c) The waiting time till the th customer has Gamma distribution. . Here .

R command below.

> 1-pgamma(1,shape=4,scale=0.5)
[1] 0.8571235

d) The expected value of (Gamma distribution) is