Let $X_n\sim B(n,\frac{1}{2})$, where $n=100$. Calculate $P\{|X_n-\frac{n}{2}|\geq\frac{\sqrt{n}}{2}\}$ using following formula:
$$ P\{k_1\leq k \leq k_2\}\approx \Phi(\frac{k_2-np}{\sqrt{npq}})-\Phi(\frac{k_1-np}{\sqrt{npq}}),$$
where $0<p<1$, $q=1-p$ and the function $\Phi$ is defined:
$$\Phi(x)=\frac{1}{\sqrt{2\pi}}\int_0^x e^{-\frac{t^2}{2}}dt, x\in \mathbb{R}. $$
We have
$$P\big\{|X_n-\frac{n}{2}|\geq\frac{\sqrt{n}}{2}\big\}=1-P\big\{|X_n-\frac{n}{2}|<\frac{\sqrt{n}}{2}\big\}.$$
Also, $$|X_n-\frac{n}{2}|<\frac{\sqrt{n}}{2}$$ is the same event as
$$\frac{\sqrt n}2-\frac n2<X_n<\frac{\sqrt n}2+\frac n2.$$
So,$$P\big\{|X_n-\frac{n}{2}|\geq\frac{\sqrt{n}}{2}\big\}=1-P\big \{\frac{\sqrt n}2-\frac n2<X_n<\frac{\sqrt n}2+\frac n2\big\}.$$
Now, you can use the formula given.