Example 8.4.6. Router Requests Revisited Again.
For the third time, let’s consider a router which, over time, has been shown to receive on average 1000 requests in any given 10 minute period during regular working hours and you want to know the likelihood that it takes more than 4 seconds in order to receive the 5th request. As you have already seen, it takes on average \(\frac{10}{1000} = \frac{1}{100} = 0.01\) minutes to receive the first request so we use that again here. If X were to measure the time interval until the fifth actual request comes in, then the Gamma distribution would be a good model using
\begin{equation*}
f(x) = \frac{x^{5-1} \cdot e^{- \frac{x}{0.01}}}{\Gamma(5) \cdot 0.01^5}
\end{equation*}
The question above asks for
\begin{equation*}
P(X \gt 4 \text{seconds}) = P(X \gt \frac{4}{60} ) = 1 - F(\frac{4}{60}).
\end{equation*}
Therefore
\begin{equation*}
P(X \gt 4 \text{seconds}) = 1 - F(\frac{4}{60}) \approx 0.205627.
\end{equation*}
Again, since X is a continuous variable you must integrate to compute probabilities. This will require integration by parts or you can use the F(x) from the derivation above. Here, let’s just let Sage do the integration for us noting that \(\Gamma(5) = 4! = 24\text{.}\) You can compute the needed integral using the interactive cell immediately below.