Prove standard deviation $=\sqrt{npq}$

Question

In class we learnt that for a binomial experiment, the standard deviation is equal to $\sqrt{npq}$. Is there a way to derive/prove this? I have a basic understanding of standard deviation and binomial but have no idea where to start in this proof.

Answer 1

Hint: Derive with direct computation the variance $σ^2$ of Bernoulli(p). You will find that it is equal to $p(1-p)$. Then, write the Binomial$(n,p)$ denoted with $X_n$ as the sum of $n$ independent Bernoulli$(p)$, $X_i$ for $i=1,\dots,n$ $$X_n=\sum_{i=1}^n X_i$$ and use the formula $$σ^2_n=\sum_{i=1}^nσ^2_i=\sum_{i=1}^np(1-p)=np(1-p)$$ Now, to find the standard deviation $σ_n$, just take the root of the previous expression.