Question

A customer support center for a computer manufacturer receives an average of 2.7 phone calls every five minutes. Assume the number of calls received follows the Poisson distribution.

a. What is the probability that no calls will arrive during the next five​ minutes?

b. What is the probability that 3 or more calls will arrive during the next five​ minutes?

c. What is the probability that 3 calls will arrive during the next ten​ minutes?

d. What is the probability that no more than 2 calls will arrive during the next ten​ minutes?

Answer 1

µ = 2.7 calls per 5 minutes

This is a poisson distribution

P(X = x) = e^-µ * µ^x / x!

a) P(X = 0) = e^-2.7 * 2.7⁰ / 0! = 0.0672

b) P(X > 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))

                   = 1 - (e^-2.7 * 2.7⁰ / 0! + e^-2.7 * 2.7¹ / 1! + e^-2.7 * 2.7² / 2! )

                   = 1 - 0.4936

                   = 0.5064

c) µ = 2.7 * 2 = 5.4 calls per 10 minutes

P(X = 3) = e^-5.4 * 5.4³ / 3! = 0.1185

d) P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)

                  = e^-5.4 * 5.4⁰ / 0! + e^-5.4 * 5.4¹ / 1! + e^-5.4 * 5.4² / 2!

                  = 0.0948