Customers arrive in a certain shop according to an approximate Poisson process at
a mean rate of 20 per hour. What is the probability that the shopkeeper will have
to wait more than 5 minutes for the arrival of the first customer? Let $X$ denote the
waiting time in minutes until the first customer arrives, and note that $\lambda = 1/3$ is the
expected number of arrivals per minute. Thus,
$\theta = \frac{1}{\lambda} = 3$
and
$f(x) = \frac{1}{3}e^{-(1/3)x}$,
$0 \le x < \infty$.
Hence,
$P(X > 5) = \int_{5}^{\infty} \frac{1}{3}e^{-(1/3)x} dx = e^{-5/3} = 0.1889$.
The median time until the first arrival is
$m = -3\ln(0.5) = 2.0794$.
