I am looking at Central Limit Theorem. It says that for a series of random sample $W_1,\cdots,W_2$ under the same mean $\mu$ and standard deviation $\sigma$, then the random variable \begin{equation} \dfrac{\overline{W}-\mu}{\sigma/\sqrt{n}} \end{equation} follows normal distribution. However, it seems like for Binomial samples, $$\dfrac{\overline{W}-\mu}{\sigma}=\dfrac{\overline{W}-np}{\sqrt{np(1-p)}}$$ is normal. I want to know where the $\sqrt{n}$ term in the denominator went?
Z-score for binomial distribution does not have $\sqrt{n}$ term
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probability
central-limit-theorem
1 Answers
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Since the binomial random variable $W$ is the sum of $n$ independent Bernoulli variables, the expression from CLT which is close to the standard normal distribution can be formulated for just one $W$: $$ \frac{W-np}{\sqrt{np(1-p)}}=\frac{W/n-p}{\sqrt{p(1-p)/n}}. $$ Here you can interpret $W=X_1+\dots+X_n$ where $X_i$ - i.i.d. Bernoulli variables. For the series of $X_i$ the expression becomes pretty standard: $$ \frac{X_1+\dots+X_n-np}{\sqrt{np(1-p)}}=\frac{\overline X-p}{\sqrt{p(1-p)/n}}. $$