What is the probability to have exactly 55 heads out of 100 coin flips ?
The exact answer is $100\choose{55}$$\frac{1}{2^{100}} \approx 0.0484$
We can see this game as a repetition of Bernoulli($\frac{1}{2}$).
Using CLT : $$P(S_{100}) = 1 - (P(S_{100} \leq 54)+P(S_{100} \geq 56))$$ $$=P(S_{100} \leq 56)-P(S_{100} \leq 54)$$ $$=P(\tilde{S_{100}} \leq \frac{56-50}{5})-P(\tilde{S_{100}} \leq \frac{54-50}{5})$$ $$\approx \Phi(1.2) - \Phi(0.8) \approx 0.097$$
but it's false.
So I tried something else : $$P(S_{100}) = P(54.5 \leq S_{100} \leq 55.5)$$ $$\vdots$$ $$\approx \Phi(1.1) - \Phi(0.9) = 0.0484$$ and it works.
So I tried again but instead of ($55 \pm\frac{1}{2}$) I tried ($55 \pm 0.8$) and I found $0.0775$.
Why $P(S_{100}) = P(54.5 \leq S_{100} \leq 55.5)$ worked and not the others ? (It should have worked because we are in a discrete case no ?)
This is called a "continuity correction". Using a range of $\pm 0.5$ around the discrete points is a good rule.
$\pm 0.8$ is the wrong scale - it's an interval of width $1.6$ rather than $1.$ If you divide by $1.6$ you'll get roughly the same answer: $0.0775/1.6\approx 0.0484.$ At the other extreme you can use the density of the pdf, suitable scaled: $1/5\sqrt{2\pi e}\approx 0.0484.$