Question

Suppose that the mean number of students who call me during by

sleepy hours ( 2pm-3pm) is 5.

What is the probability that in a given hour, exactly two students will call?

And what is the probability that more than two will call in the given hour?

Answer 1

This is a Poisson distribution with $\mu=5.$

$P(X=2)=e^{-5}\frac{5^2}{2!}=0.0842.$

$P(X>2)=1-P(X\le2)=1-P(X=0)-P(X=1)-P(X=2)=$

$=1-e^{-5}(\frac{5^0}{0!}+\frac{5^1}{1!}+\frac{5^2}{2!})=0.8754.$