If instead you plan to sample from this distribution n=49 times, the Central Limit Theorem implies that you will get a random variable \(\overline{X}\) which has an approximate normal distribution with the same mean but with new variance \(\sigma_{\overline{X}}^2 = \frac{199/20}{49} = \frac{199}{580}\text{.}\) Therefore, expanding the interval to include the boundaries of the corresponding histogram areas,
\begin{equation*}
P( 8 \le \overline{X} \lt 12 ) = P(7.5 \le \overline{X} \le 11.5) \approx normalcdf(7.5,11.5,10.5,0.585750) \approx 0.9561 .
\end{equation*}