Everyone using the internet utilizes a series of "routers" who spend their time waiting for someone to show up and ask for something to be done. Let’s consider one such router which, over time, has been shown to receive on average 1000 such requests in any given 10 minute period during regular working hours. In general, a Poisson process with mean 1000 would seem to fit and therefore the Poisson distribution would be a good model. We will find out below that \(\lambda T = \mu = 1000\) and will use that here to get
\begin{equation*}
f(x) = \frac{1000^x}{x!} e^{-1000}.
\end{equation*}
So, suppose we would like to know the likelihood of receiving exactly 1020 requests in a 10 minute time interval. This means we need
\begin{equation*}
P(X = 1020) = f(1020) = \frac{1000^{1020}}{1020!} e^{-1000}
\end{equation*}
which might be totally impossible to compute directly using a regular calculator. However, many graphing calculators have a built-in function where f(x) = poissonpdf(mu,x) and F(x) = poissoncdf(mu,x). To answer our question,
\begin{equation*}
f(1020) = \text{poissonpdf(1000,1020)} \approx 0.01024.
\end{equation*}
On the other hand, suppose the question is to ask whether 1020 or fewer requests wil be made in the 10 minute interval. If so, then
\begin{equation*}
F(1020) = \text{poissoncdf(1000,1020)} \approx 0.74258.
\end{equation*}