Question

A customer support center for a computer manufacturer receives an average of 1.3 phone calls every five minutes. Assume the number of calls received follows the Poisson distribution.

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

b) for 5 minutes expected calls =1.3=

hence probability that 3 or more calls will arrive during the next five? minutes=P(X>=3)

=1-P(X<=2)=1-(P(X=0)+P(X=1)+P(X=2))=1-(e^-1.3*1.3⁰/0!+e^-1.3*1.3¹/1!+e^-1.3*1.3²/2!)=0.1429

c)for 5 minutes expected calls =1.3*2=2.6=

probability that 3 calls will arrive during the next ten? minutes =e^-2.6*2.6³/3! =0.2176

d) probability that no more than 2 calls will arrive during the next ten? minutes=P(X<=2)

=P(X=0)+P(X=1)+P(X=2) =e^-2.6*2.6⁰/0!+e^-2.6*2.6¹/1!+e^-2.6*2.6²/2!=0.5184