Answer

The exact answer will be

\[ \begin{aligned} \sum_{k=0}^{21}\binom{36}{k}(\frac{1}{2})^{36} = 0.8785 \end{aligned} \]

But the same can be estimated using the CLT as follows

\[ \begin{aligned} \mu &= np = 18\newline \sigma^{2} &= np(1-p) = 9\newline P(S_{n} \leq 21) &\approx P(\frac{S_{n} - 18}{3} \leq \frac{21-18}{3}) \approx 0.843 \end{aligned} \]

Our estimate is in the rough range of the answer but not quite close. We can do better using the \(\frac{1}{2}\) correction

\[ \begin{aligned} P(S_{n} \leq 21) = P(S_{n} < 22) \quad\text{since $S_{n}$ is an integer}\newline \text{Consider}\quad P(S_{n} <= 21.5) \quad\text{as a compromise between the two}\newline P(S_{n} <= 21.5) = P(\frac{S_{n} - 18}{3} \leq \frac{21.5 - 18}{3}) \approx 0.879 \end{aligned} \]

In a similar manner, \(P(S_{n}=19) = P(18.5 \leq S_{n} \leq 19.5)\) using \(\frac{1}{2}\) correction.