The maximum value of the variance of a Binomi | vedclass

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(B) The variance of a Binomial distribution is given by $\sigma^2 = n p q$,where $q = 1 - p$. To find the maximum value of the variance,we express it

Vedclass · Accepted Answer

$\frac{n}{4}$

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(B) The variance of a Binomial distribution is given by $\sigma^2 = n p q$,where $q = 1 - p$. To find the maximum value of the variance,we express it as a function of $p$: $f(p) = n p(1 - p) = n(p - p^2)$. Taking the derivative with respect to $p$ and setting it to zero: $f'(p) = n(1 - 2p) = 0$. This gives $1 - 2p = 0$,or $p = \frac{1}{2}$. Since $q = 1 - p$,we have $q = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting these values into the variance formula: $\sigma^2_{\max} = n 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{n}{4}$.

The maximum value of the variance of a Binomial distribution with parameters $n$ and $p$ is

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