Theoretical Distribution

Concept:For a binomial distribution, the mean is $np$ and the variance is $np(1-p)$, where $0 \le p \le 1$.Explanation:• Mean $\mu = np$.• Variance $\sigma^2 = np(1-p) = npq$.• Since $0 \le p \le 1$, we have $q = 1-p \le 1$, so $npq \le np$.• Therefore, mean is always greater than or equal to variance. Equality occurs only when $p = 0$ or $p = 1$ (degenerate cases).• Thus, for any non-degenerate binomial distribution, mean is strictly more than variance.Answer:The mean is always more than its variance.