Question

Prove that the mean of a binomial distribution is always greater than the variance.

Answer 1

Let X be a binomial variate with parameters n and p. Then,

Mean = np

And, Variance = npq

Mean - Variance = $np-npq=np(1-q)=n{p}^2$

$Mean-Variance>0$ $[n\in N,p>0,therefore,n{p}^2>0]$

$Mean>Variance$