EX 4.22

Suppose that calls to a rape crisis center in a certain county occur according to a Poisson process with rate $\alpha =.5$ call per day. Then the number of days $X$ between successive calls has an exponential distribution with parameter value .5, so the probability that more than 2 days elapse between calls is

$$\begin{array}{rl}& P\left(X>2\right)\\ & =1-P\left(X\le 2\right)\\ & =1-F\left(2;0.5\right)\\ & =e^{-\left(0.5\right)\left(2\right)}\\ & =0.368\end{array}$$

The expected time between successive calls is $1/.5=2$ days.