Suppose that $X$ is a $Binomial(n,p)$. Then the variance of $X/n$ is:

Question

In my attempts, I find that variance of $x=npq$, therefore $x/n$ must be $npq/n$, which becomes, ($npq/(variance/pq)$). And variance is $npq$. So I am left with $npq/npq/pq$. And I know this is totally wrong, so I just need some help to figure this out.

Answer 1

For a random variable $X$ and constant $a,V[aX]=E[(aX-E(aX))^2]=a^2V[X]$. So $V[\frac Xn]=\left(\frac1n\right)^2\times V[X]=\frac{npq}{n^2}=\frac{p(1-p)}n$, since $q=1-p$.