 # Approximation to Binomial Probabilities. 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)=1/(sqrt{2pi}) e^{(-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. cochetezgh 2022-09-11 Answered
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 $\varphi \left(x\right)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{{x}^{2}}{2}}$, 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 $\varphi$ 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 $\left(\genfrac{}{}{0}{}{2n}{n+j}\right)\cdot \frac{1}{{2}^{n}}$ and I tried to use Stirling's formula on the binomial but it just made it more complicated.
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Step 1
Let X be the number of heads in 2n tosses of a fair coin. Then X has binomial(2n,1/2) distribution, so $E\left(X\right)=n$ and $V\left(X\right)=n/2$. To calculate $P\left(X=n+j\right)$, use the continuity correction to express $P\left(X=n+j\right)$ as
$\begin{array}{}\text{(1)}& P\left(X=n+j\right)=P\left(n+j-\frac{1}{2}\le X\le n+j+\frac{1}{2}\right).\end{array}$
Step 2
The RHS of (1) is algebraically equal to
$\begin{array}{}\text{(2)}& P\left(\frac{j-\frac{1}{2}}{\sqrt{n/2}}\le \frac{X-n}{\sqrt{n/2}}\le \frac{j+\frac{1}{2}}{\sqrt{n/2}}\right).\end{array}$
Step 2
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 $L:=\frac{j-\frac{1}{2}}{\sqrt{n/2}}$ and $R:=\frac{j+\frac{1}{2}}{\sqrt{n/2}}$. Approximate this area by a rectangle. The rectangle will have height $\varphi \left(z\right)$, where $z=\frac{j}{\sqrt{n/2}}$ is the midpoint of L and R. The width of the rectangle will be the difference between the two endpoints L and R.

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