# Unbiased estimator and variance

A random sample of n people are asked whether they are against smoking or not. Suppose x are against smoking. What is the distribution of the random variable X (number of those against smoking). State an unbiased estimate of p, the true proportion of those against smoking. Show it is unbiased and derive the variance. Show the point estimate is consistent.

So p=x/n => E(p) = E(x/n) = 1/n(E(x)) = p.

I don't know what the distribution of this is though. If also know the variance is sigma^2 / n but how do i apply this to the question?

As stated the problem is meaningless because the random nature of $N$ is left unexplained in its statement.

For instance, it may be that $N$ is fixed [an extreme case of a random variable] and the questioned persons are independent and all with the same probability $p$ to smoke, in which case$$X|N=n\sim\mathcal{B}(n,p)$$which leads easily to $X/N$ being an unbiased estimator of $p$.

On the opposite extreme, it may be that $N$ is random so that $X$ is fixed, namely $N$ may be the number of attempts to reach $X$ smokers out of $N$ persons, in which case$$N|X=x\sim\mathcal{NB}(x,p)$$a Negative Binomial variate. Then $X/N$ is not unbiased for $p$, since $(X-1)/(N-1)$ is unbiased.