Solution 8.7.6.1.
Given the average of 50 requests per second and X measuring the number of "successes" in a given second long time interval given a Poission distribution
\begin{equation*}
f(x) = \frac{50^x}{x!}e^{-50}.
\end{equation*}
Then,
\begin{equation*}
P(X \lt 50) = F(49) = \sum_{x=0}^{49} \frac{50^x}{x!} e^{-50}
\end{equation*}
and using the graphing calculator function poissoncdf(50,49) = 0.48119.
For a time interval of one second, the mean is given to be 50 requests. Using the formulas developed above, the standard deviation therefore is \(\sqrt{50}\text{.}\) Therefore
\begin{align*}
P(\mu - 2\sigma \le X \le \mu + 2\sigma) & = P(50 - 2\sqrt{50} \le X \le 50 + 2\sqrt{50})\\
& = P(X \in \{ 36, 37, 38, ..., 62, 63, 64 \}).
\end{align*}
Using the distribution function,
\begin{equation*}
F(64) - F(35) \approx 0.97640 - 0.01621 = 0.96019
\end{equation*}
Finally, notice that the time interval has been adjusted. Since the mean formula is proportional to the interval over which X is measured, using \(\mu = \lambda T\) with \(\lambda = 1\) when the interval is 1 second, then when the interval is one hour, T = 3600 seconds. Hence, we would expect on average \(50 \cdot 3600 = 180,000\) requests in one hour.