So suppose we flip a coin $n$ times, if we define $X= \text{ the number of tails in those $n$ flips}$, which is the smallest value of $n$ that satisfies that $P[49 \leq X \leq 51]\geq 0.95$.
I got the hint to use a normal approximation i.e:
$$P[49 \leq X \leq 51] = P \left[ \frac{49-\frac{n}{2}}{\sqrt{\frac{n}{4}}}\leq \frac{X-\frac{n}{2}}{\sqrt{\frac{n}{4}}}\leq \frac{51-\frac{n}{2}}{\sqrt{\frac{n}{4}}} \right]$$
And then you have to find $n$ such that
$$\int^{\left.\left(51-\frac{n}{2}\right)\right/\sqrt{\frac{n}{4}}}_{\left.\left(49-\frac{n}{2}\right)\right/\sqrt{\frac{n}{4}}} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx \geq 0.95$$
I'm stuck in that part could you help me? Thanks a lot
What you want is $.49 n\leq X \leq .51 n$. It is "well known" that 95 % of a normal distribution happens for $\pm$ 1.96 standard deviations. Therefore you want $$ {.51\,n - {n\over2} \over \sqrt{n\over 4}}\geq 1.96\rightarrow n\geq 9604. $$