Central Limit Theorem problem. There are 36 white and 64 black balls in the bin. After we have randomly picked some ball we put it back to the bin. How many times we must pick balls from the bin to be sure, that probability of frequency of picking white ball vary from 0.36 at least 0.12 is equal to 0.1.
So:
$P\left ( \left | \frac{X}{n} - 0.36 \right | \geq  0.12 \right )=0.1$
Answer: $n \geq 40 $. How did they get this result?
 A: There are two ways to approach this problem, one is exact, the other using a normal approximation, as alluded to in the title of your question.
If we denote by $X_n$ the number of white balls in the first $n$ draws, then 
$$
\frac{X_n-.36n}{\sqrt{.36\times.64\times n}}
$$
is approximately standard-normally distributed. Consequently the probability that the relative frequency of white balls $X_n/n$ differs from the expected probability $.36$ by at least $.12$ is equal to
$$
P\left(\left|\frac{X_n}{n}-.36\right|\geq .12\right)=P\left(\left|\frac{X_n-.36 n}{\sqrt{.36\times.64\times n}}\right|\geq \frac{.12n}{\sqrt{.36\times.64\times n}} \right)\approx P\left(|N|\geq \frac{.12n}{\sqrt{.36\times.64\times n}}\right),
$$
where $N\sim\mathcal{N}(0,1)$ is a standard normal random variable. In order to find $n$ such that the last probability is equal to $.1$ you can use the cumulative distribution function $\Phi(x)=P(N\leq x)$ of the normal distribution and observe that 
$$
P(|N|\geq x)=1-P(|N|<x)=1-\Phi(x)+\Phi(-x)=0.1 \Rightarrow x=1.64485.
$$
We thus find $n$ by solving
$$
\frac{.12n}{\sqrt{.36\times.64\times n}}=1.64485 \Rightarrow n=43.2887.
$$
Hence, for any $n\leq 43$ (large enough for the normal approximation to be valid), the probability that the relative frequency of white balls $X_n/n$ differs from the expected probability $.36$ by at least $.12$ is at least $.1$
