You flip a coin $100$ times. What's the probability that you get between $[50 - k, 50 + k]$ heads for some integer $k$? I know we can use the CLT with mean $50$ and standard deviation $5$, and I end up finding:
$$P(X \leq 50 + k) - P(X \leq 50 - k) = P(Z \leq k/5) - P(Z \leq -k/5) = \Phi(k/5) - \Phi(-k/5) = \Phi(k/5) - (1 - \Phi(k/5)), $$
which simplifies to $2\cdot \Phi(k/5) - 1$, which is clearly less than $1$. Can someone please point out what I did wrong?
The random variable for the number of heads ($x$) is binomial distributed like $X\sim\textrm{Bin}\left(100,\frac12\right)$. That means that X is a discrete random variable. Thus $P(50-k\leq X \leq 50+k)=P(X\leq 50+k)-P(X\leq 49-k)$. Now you have two options. Applying central limit theorem with continuity correction factor $(b)$ or without it $(a)$.
$(a) \ $ $P(50-k\leq X \leq 50+k)\approx \Phi\left(\frac{50+k-50}{5}\right)-\Phi\left(\frac{49+k-50}{5}\right)=\Phi\left(\frac{k}{5}\right)-\Phi\left(\frac{k-1}{5}\right)$
$(b) \ $$P(50-k\leq X \leq 50+k)\approx \Phi\left(\frac{50+0.5+k-50}{5}\right)-\Phi\left(\frac{49+0.5+k-50}{5}\right)=\Phi\left(\frac{k+0.5}{5}\right)-\Phi\left(\frac{k-0.5}{5}\right)$