Question

Suppose that the number of calls coming per minute into an airline reservation center follows a Poisson distribution. Assume that the mean number of calls is 3 calls per minute. The probability that no calls are received in a given one-minute period is _______.

Answer 1

Answer:

0.0497 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean number of calls = 3 calls per minute

The number of calls coming per minute is following a Poisson distribution.

Formula:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

We have to evaluate

$P( x =0) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\P(x = 0) = \frac{3^0e^{-3}}{0!} = 0.0497$

0.0497 is the probability that no calls are received in a given one-minute period.