For Binomial with p = 0.3 and n = 50, \(\mu = n \cdot p = 15\) and \(\sigma^2 = n \cdot p \cdot (1-p) = 10.5\text{.}\) So, \(\sigma = \sqrt{10.5}\text{.}\) Therefore,
\begin{align*}
P(\mu - 2\sigma \le X \le \mu + 2\sigma) & = P(15 - 2 \sqrt{10.5} \le X \le 15 + 2 \sqrt{10.5})\\
& P( X \in \{9, 10, 11, ... , 19, 20, 21 \} )
\end{align*}
Then
\begin{equation*}
F(21) - F(8) \approx 0.97491 - 0.01825 = 0.95666
\end{equation*}
For Negative Binomial with p = 0.3 and r = 2, \(\mu = \frac{2}{0.3} = \frac{20}{3}\) and \(\sigma^2 = 2 \frac{0.7}{0.3^2} = \frac{140}{9}\) and so \(\sigma \approx 3.9\text{.}\) Therefore,
\begin{align*}
P(\mu - 2\sigma \le X \le \mu + 2\sigma) & = P(6.7 - 7.8 \le X \le 6.7 + 7.8)\\
& P( X \in \{2, 3, ... , 14, 15, 16 \} )
\end{align*}
Then,
\begin{equation*}
F(16) \approx 0.973888
\end{equation*}