We have just started our probability course and we got the question: Consider the minimal number of tosses $n$ required such that the percentage of heads in fair coin tossing will be between $0.40$ and $0.60$ with probability at least $0.70$.
We are asked to figure out $n$ using the Central Limit Approximation.
I know that for large numbers the distribution beings to be normal.
But I have no idea how to start and would be glad for some help (I'm new to probability)
The number $X$ of heads is a binomial random variable with parameters $n$ (to be determined) and $p=0.5$. Thus (due to the symmetry of $X$, as expressed by $p=0.5$) $X$ is (well) approximated by the normal distribution as $n$ grows, in symbols $$X \sim N(μ=np, σ^2=np(1-p)) \iff X \sim N(μ=\frac{n}{2}, σ^2=\frac{n}{4})$$ approximately as $n \to +\infty$. Thus you need to solve the equation $$0.70>P(0.4n\le X\le 0.6n)=P\left(\frac{0.4n-0.5n}{\sqrt{0.25n}}\le Z\le \frac{0.6n-0.5n}{\sqrt{0.25n}}\right)$$