Question

One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analysts generally believe that random phone calls are Poisson distributed. Suppose phone calls to a switchboard arrive at an average rate of 2.2 calls per minute. (Round to 4 decimal places)

a. If an operator wants to take a one-minute break, what is the probability that there will be no calls during a one-minute interval?
b. If an operator can handle at most five calls per minute, what is the probability that the operator will be unable to handle the calls in any one-minute period?
c. What is the probability that exactly three calls will arrive in a two-minute interval?
d. What is the probability that one or fewer calls will arrive in a 30-second interval?

Answer 1

a)

this is Poisson distribution with parameter λ=2.2

probability that there will be no calls during a one-minute interval =P(X=0)=e^-2.2*2.2⁰/0! =0.1108

b)

probability that the operator will be unable to handle the calls in any one-minute period :

probability =

P(X>5)=

1-P(X<=5)=

1-∑x=0x-1   {e-λ*λx/x!}=

0.0249

c)

expected number of calls in 2 minute=2*2.2=4.4

probability that exactly three calls will arrive in a two-minute interval :

P(X=3)=e^-4.44.4³/3! =0.1743

d)expected number of calls in 30 seconds=0.5*2.2=1.1

probability that one or fewer calls will arrive in a 30-second interval

P(X<=1)=P(X=0)+P(X=1)=e^-1.1*1.1⁰/0!+e^-1.1*1.1¹/1!=0.6990