Using the central limit theorem

We apply the central limit theorem to integer-valued random variables. In that situation, we assign to every integer $k$ it's own little interval $\left(k-\frac{1}{2},k+\frac{1}{2}\right)$. Thus instead of $P(X=5)$ we'd be looking for $P(4.5 \leq X \leq 5.5)$.

More generally, in application to $P(S_n = k)$, where the $X_i$ have mean $\mu$  and standard deviation $\sigma$, we'd first replace the single value $k$ with an interval. We are now looking for $P(a\leq S_n \leq b)$ where $a=k-\frac{1}{2}$ and $b=k+\frac{1}{2}$.

Let's subtract the mean $n\mu $ from the inequality to obtain $$a - n \mu \leq S_n - n\mu \leq b - n\mu. $$ Next we divide by $\sqrt{n \sigma ^2 }$ to obtain $$\frac{a - n \mu}{\sqrt{n\sigma^2}} \leq \frac{S_n - n\mu}{\sqrt{n\sigma^2}} \leq \frac{b - n\mu}{\sqrt{n\sigma^2}}.$$ Thus we look to compute $$P\left(\frac{a - n \mu}{\sqrt{n\sigma^2}} \leq Z \leq \frac{b - n\mu}{\sqrt{n\sigma^2}} \right).$$

See also Example 5 in your book.