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When you toss a coin $2n$ times, the expected number of heads is $n$. Given the probability density function for a standard normal random variable $\phi(x)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$, approximate the probability that you observe $n+j$ heads when you toss the coin $2n$ times using the density function of a standard random variable $\phi$ and the standard deviation $\sigma$. Hint: use Stirling's formula: $n!\sim \sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}$.

I know that the exact probability that I get $n+j$ heads is $\binom{2n}{n+j}\cdot\dfrac{1}{2^n}$ and I tried to use Stirling's formula on the binomial but it just made it more complicated.

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Hint: Let $X$ be the number of heads in $2n$ tosses of a fair coin. Then $X$ has binomial$(2n, 1/2)$ distribution, so $E(X)=n$ and $V(X)=n/2$. To calculate $P(X=n+j)$, use the continuity correction to express $P(X=n+j)$ as $$ \textstyle P(X=n+j)=P(n+j-\frac12\le X\le n+j+\frac12).\tag1$$ The RHS of (1) is algebraically equal to $$ P\left({j-\frac12\over\sqrt{n/2}}\le {X-n\over\sqrt{n/2}}\le{j+\frac12\over\sqrt{n/2}}\right).\tag2 $$ We recognize that the middle quantity in (2) has approximately standard normal distribution (by the normal approx to the binomial, i.e., the de Moivre-Laplace theorem). The desired prob is therefore approximately the area under the standard normal density between the two endpoints $L:={j-\frac12\over\sqrt{n/2}}$ and $R:={j+\frac12\over\sqrt{n/2}}$. Approximate this area by a rectangle. The rectangle will have height $\phi(z)$, where $z={j\over\sqrt{n/2}}$ is the midpoint of $L$ and $R$. The width of the rectangle will be the difference between the two endpoints $L$ and $R$.

Note: The point of this exercise is to avoid Stirling's approximation, no?

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