Gamma becomes normal as \(r \rightarrow \infty\text{.}\) Assume that the average time till a first success is 12 minutes and that \(r = 8\text{.}\) Then, the mean for the Gamma distribution is \(\mu = 12 \cdot 8 = 96\) and \(\sigma^2 = 8 \cdot 12^2 = 1152\) and so \(\sigma \approx 33.9411\text{.}\)
Using the Gamma formulas,
\begin{align*}
P( 90 \le X \le 100 ) & = \int_{90}^{100} f(x) dx \\
& = 0.59252 - 0.47536 = 0.11716.
\end{align*}
Using the normal distribution,
\begin{equation*}
P( 90 \le X \le 100) \approx normalcdf(90,100,96,33.9411) = 0.11707.
\end{equation*}
Amazingly, these are also very close.