# The value of p in a Standard deviation of a binomial Distribution

The random variable $$X$$ has a binomial distribution $$b(n,p)$$. For what value of $$p$$ is the standard deviation of $$X$$ the greatest (note: the answer is independent of $$n$$).

Can someone help with this question! I know the formula of standard deviation, it's $$\sqrt{n p( 1 - p)}$$. My question is: Are we trying to find a value closest to zero? Can anyone explain and help to find the value of $$p$$?

• You have the expression $n p (1-p)$ which is in fact the variance. The standard deviation would be $\sqrt{np(1-p)}$. You want to find the value of $p$ which maximises the latter (and this will also maximise the former). As a hint, $p(1-p) = \frac14-\left(p - \frac12\right)^2$ – Henry Feb 27 at 23:50
• what do you mean by latter and former? – Elchavo18 Feb 28 at 0:06
• You said originally you wanted to find "for what value of $p$ is the standard deviation of $X$ the greatest" – Henry Feb 28 at 0:11
• yes correct @Henry – Elchavo18 Feb 28 at 0:12

Hint: What you want to do is find $$p\in[0,1]$$ such that the function $$f\colon[0,1]\to\mathbb R, \quad p \mapsto np(1-p),$$ attains its maximum in $$p$$. You can use straight-forward calculations or brute force (this function is twice differentiable).

• Still a bit confused – Elchavo18 Feb 27 at 23:50