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*}