Suppose a large number of N people flip a coin 100 times, how would you find the percentage of them that would get less than 45 heads? Assume that the probability of 100 people is big enough that the trial can be approximated by normal distribution.
In the problem, N is unknown. 
I think you would need to use the 68-95-99.9 rule, but I'm not sure exactly how/what formula I would use. Help is appreciated. Thanks!

Let $X$ denote the number of heads we obtain in $100$ flips.
Normal approximation without continuity correction:
$$\mu=np$$
$$\sigma^2=npq$$
where $$\frac{X-\mu}{\sqrt{npq}}\sim N(0,1)$$
Then we have
$$\begin{align*}
P(Z\lt 45)
&=\Phi\left({\frac{45-50}{\sqrt{100\cdot0.5\cdot0.5}}}\right)\\\\
&=\Phi(-1)\\\\
&\approx0.1587
\end{align*}$$
With continuity correction:
$$\begin{align*}
P(Z\lt 45)
&=\Phi\left({\frac{44.5-50}{\sqrt{100\cdot0.5\cdot0.5}}}\right)\\\\
&=\Phi(-1.1)\\\\
&\approx0.1335
\end{align*}$$
Exact probability using binomial distribution:
$$\begin{align*}
P(X<45)
&=\sum_{k=0}^{44} {n\choose k}{0.5^
k}{0.5^{n-k}}\\\\
&=\sum_{k=0}^{44} {100 \choose k}{0.5^
k}{0.5^{100-k}}\\\\
&=0.1356265
\end{align*}$$
This can be calculated in R:
> sum(dbinom(0:44,100,.5))
[1] 0.1356265

