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As written in the title, is the distribution of sample mean of Bernoulli random variable a Binomial distribution? And I was taught that we can approximate Binomial distribution to normal distribution due to CLT. Is the title the reason we can approximate it?

I think, if $X \sim Bern(p)$, and say $X_n$ is the sum of the elements of the sample of scale $n$, i.e. $X_n=n\bar{X_n}$ then $$Pr(X_n=k)={n \choose k}p^kq^{n-k}$$so $X_n \sim Bin(n,p)$.

And due to CLT, which ststes $$\sqrt{n}( \bar{X_n} -\mu)\overset {d}{\to} N(0,\sigma^2),$$ the distribution of sample mean $\bar{X_n}$ tents to follow $N(\mu,\frac{\sigma^2}{n})$ as $n$ gets bigger, and therefore $X_n=n\bar{X_n}$ tends to follow $N(n\mu, n\sigma^2)$.

And if we apply this consequence to random variable $X \sim Bern(p), $ then since $E(X)=p$ and $V(x)=pq$,
$X_n \sim Bin(n,p)$ tends to follow $N(n\mu ,n \sigma^2 )=N(np, npq).$

Is this the right way to suggest we can approximate binomial distribution to normal distribution by CLT? Or is it wrong and does there exist other way to suggest it?

#Since I'm new to statistic and I'm Korean, I would appreciate if you forgive some improper expressions.

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  • $\begingroup$ I think you should write $P(X_n = k) = \binom{n}{k}p^kq^{\color{red}{n-k}}$. $\endgroup$ Commented Dec 26, 2023 at 11:30
  • $\begingroup$ @Riemann Oh that was right. I will edit $\endgroup$ Commented Dec 26, 2023 at 11:49

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No, as you said, sum of n independent bernulli random variables is binomial but if $Y\sim Bin(n,p)$ then $aY$ is not binomial (a≠1). You can compare the support of $aY$ to binomial to see why $aY$ is not binomial. The support of binomial is like $\{0,1,...,n\}$ but support of $aY$ is $\{0,a,2a,...,an\}$ which is different. Also there are other ways to show that $aY$ is not binomial but if you plot the probability distributions of $aY$ and $Y$, you see that they are similar but it doesn't mean that $aY$ is binomial.

Also, take a look at this question.

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