Use Central Limit Theorem to estimate the probability of Heads in coin toss Let $S_{100}$ be the number of heads that turn in in 100 tosses of a fair coin. Use Central limit theorem to estimate:
Here is a z-score table: http://gyazo.com/b98bdd329d4c1ac5ade0630073f88c6f
a) P($S_{100} \leq 45$)
b) P($45 \leq S_{100} \leq 55$)
c) P($S_{100} \geq 63$)

When I attempt this question, I get a z-score of 99.1 for the first one and It seems that I don't have the right approach here. I'm using the formula $$P(S_n \leq b) = P\left(Z \leq \frac {b-n\mu}{\sigma\sqrt{n}}\right) $$
For the values, I use:
b = 45, n = 100, $\mu$ = 50 ($\mu$ = $np$ because binomial distribution), $\sigma = \sqrt{np(1-p}$ = 5 
I'm not sure where i'm going wrong. My z-score becomes 99.1 using these values and that doesn't seem right. 
 A: Write $S_{100}$ as a sum of $100$ independent indicator variables, $X_1$ through $X_{100}$, each denoting whether a head occurred in the $i$-th toss. Now, $\mathrm{E}[X_i] = \frac{1}{2}$ and $\mathrm{Var}(X_i) = \frac{1}{4}$.
We can use the central limit theorem as follows:
$$
\begin{align*}
\mathrm{P}(0 \leq S_{100} \leq 45) & = \mathrm{P}\left(0 \leq \sum\limits_{i = 1}^{100}X_i \leq 45\right) \\& = \mathrm{P}\left( \frac{0 - 100 \cdot 0.5}{\sqrt{0.25 \cdot100}} \leq \frac{\sum\limits_{i = 1}^{100}X_i - 100 \cdot 0.5}{\sqrt{0.25 \cdot100}} \leq \frac{45 - 100 \cdot 0.5}{\sqrt{0.25 \cdot100}}\right) \\
& = \mathrm{P}(-10 \leq Z \leq -1) = \Phi(-1) - \Phi(-10) \approx 0.1585
\end{align*}
$$
Where $\Phi$ is the cumulative distribution function of a standard normal distribution.
By the way, this is often called the integral de Moivre–Laplace theorem (or de Moivre–Laplace central limit theorem) when used directly on binomial random variables, as I think it was derived much earlier in time.
