Binomial exercise with different variances? Is it possible?

Question

Good evening,

I honestly can't understand why in this excercise the variance is computed normally but then...

In this one the variance is computed by dividing by n instead if multiplying by it as in the previous exercise?

Answer 1

In the first exercise, your random variable is a count of things, represented by a binomial, $Y \sim B(n, p)$ and hence $Var(Y) = np(1-p)$.

In the second exercise, the random variable is a proportion of things, which is equal to the count (a binomial) divided by the number of trials, i.e. $\hat{P} = \frac{Y}{n}$. And since $n$ is a constant, we can write $Var(\hat{P}) = Var(\frac{Y}{n}) = \frac{1}{n^2}Var(Y) = \frac{1}{n^2}np(1-p) = \frac{p(1-p)}{n}$.