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}*1.3^{0}/0!+e^{-1.3}*1.3^{1}/1!+e^{-1.3}*1.3^{2}/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}*2.6^{3}/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}*2.6^{0}/0!+e^{-2.6}*2.6^{1}/1!+e^{-2.6}*2.6^{2}/2!=0.5184

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