Question

5. A shop has an average of five customers per hour. (a) Assume that the time T between any two customers' arrivals is an exponential random variable. (b) Assume that the number of customers who arrive during a given time period is Poisson. What (c) Let Y be exponential random variables míodeling the tine between the ith and 1st customers' What is the probability that no customer arrives in the next fifteen minutes? is the probability that no customer arrives in the next fifteen minutes? arrivals. Find the probability that exactly two customers arrive in the next twenty minutes.

I got e^(-5/4) for (a) and (b), but I do not know how to do (c). Thank you!

Answer 1

a) We are given an average of 5 customers per hour, therefore the average waiting time between 2 customers is given as 1/5 hour.

Probability here is computed as:

15 515/6/40.2865 15/60

b) By poisson process, average number of customers in 15 minutes is computed as:

= 5*(15/60) = 1.25

Therefore the probability here is computed as:

P(N-0) e-1.25-0.2865

c) Average number of arrivals in 20 minutes is computed as:

= 5*(20/60) = 5/3

Now the required probability here is computed as:

(5/3)- -5/30.2623