Customers arrive at a service center according to a Poisson process with a mean interarrival time of 15 minutes.

Question

Customers arrive at a service center according to a Poisson process with a mean inter-arrival time of 15 minutes.

1. What is the probability that no arrivals occur in the first half hour?
2. What is the expected time until the tenth arrival occurs?
3. What is the probability of more than five arrivals occurring in any half-hour period?
4. If two customers were observed to have arrived in the first hour, what is the probability that both arrived in the last 10 minutes of that hour?
5. If two customers were observed to have arrived in the first hour, what is the probability that at least one arrived in the last 10 minutes of that hour?

Answer 1

(a) mean interval = 15 minutes Hence 1/(15 min) = 60/15 = 4/hour T=1/2 hour f(x) = (e-(4/2)(4/2)^0)/0! = e-2 = 0.1353 (b) P(x≤10) = Sum from x=0 to x=10 ((e-4(4)^x)/x!) = 0.997 (c) P(x≤5) = Sum from x=0 to x=5 ((e-4(4)^x)/x!) = 0.785 P(x>5) = 1 - 0.785 = 0.215