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 $p$?

  • 1
    $\begingroup$ 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, $p(1-p) = \frac14-\left(p - \frac12\right)^2$ $\endgroup$ – Henry Feb 27 at 23:50
  • $\begingroup$ what do you mean by latter and former? $\endgroup$ – Elchavo18 Feb 28 at 0:06
  • $\begingroup$ You said originally you wanted to find "for what value of $p$ is the standard deviation of $X$ the greatest" $\endgroup$ – Henry Feb 28 at 0:11
  • $\begingroup$ yes correct @Henry $\endgroup$ – 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).

  • $\begingroup$ Still a bit confused $\endgroup$ – Elchavo18 Feb 27 at 23:50

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