Question

The number of telephone calls that arrive at a phone exchange is often modeled as a

Poisson random variable. Assume that on the average there are 10 calls per hour.

a) What is the probability that there are exactly 5 calls in one hour?

b) What is the probability that there are there are exactly 15 calls in two hours?

c) What is the probability that there are exactly 5 calls in 30 minutes?

Answer 1

for Poisson distribution:

$P(x=k)= \frac{\lambda^ke^{-\lambda}}{k!}$

a. $P(5)=\frac{10^5e^{-10}}{5!}=0.0378$

b.The probability for 15 calls in 2 hours is $P(15,2)=\frac{(2\times 10)^{15}e^{-10\times 2}}{15!}=0.051$

c.The probability for 5 calls in 30 minutes or 0.5 hours is

$P(5,0.5)=\frac{(0.5\times10)^5e^{-10\times 0.5}}{5!}=0.175$