I know that the exact probability that I get $n+j$ heads is $(}\genfrac{}{}{0ex}{}{2n}{n+j}{\textstyle )}\cdot {\displaystyle \frac{1}{{2}^{n}}$ and I tried to use Stirling's formula on the binomial but it just made it more complicated.

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 $\varphi (x)={\displaystyle \frac{1}{\sqrt{2\pi}}}{e}^{-\frac{{x}^{2}}{2}}$, 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 $(}\genfrac{}{}{0ex}{}{2n}{n+j}{\textstyle )}\cdot {\displaystyle \frac{1}{{2}^{n}}$ and I tried to use Stirling's formula on the binomial but it just made it more complicated.

I know that the exact probability that I get $n+j$ heads is $(}\genfrac{}{}{0ex}{}{2n}{n+j}{\textstyle )}\cdot {\displaystyle \frac{1}{{2}^{n}}$ and I tried to use Stirling's formula on the binomial but it just made it more complicated.

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$\begin{array}{|cc|}\hline k& P(X=k)\\ 0& \phantom{\rule{0ex}{0ex}}\\ 1& \phantom{\rule{0ex}{0ex}}\\ 2& \phantom{\rule{0ex}{0ex}}\\ 3& \phantom{\rule{0ex}{0ex}}\\ 4& \phantom{\rule{0ex}{0ex}}\\ 5& \phantom{\rule{0ex}{0ex}}\\ \hline\end{array}$

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