The mean and variance of a binomial distribut | vedclass

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(A) For a binomial distribution,the mean is given by $\mu = np$ and the variance is given by $\sigma^2 = npq$.Given that the mean is $6$,we have $np =

Vedclass · Accepted Answer

$18$

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(A) For a binomial distribution,the mean is given by $\mu = np$ and the variance is given by $\sigma^2 = npq$.Given that the mean is $6$,we have $np = 6$.Given that the variance is $4$,we have $npq = 4$.Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{4}{6}$,which simplifies to $q = \frac{2}{3}$.Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.Substituting the value of $p$ into the mean equation: $n 	imes \frac{1}{3} = 6$.Therefore,$n = 6 	imes 3 = 18$.

The mean and variance of a binomial distribution are $6$ and $4$. The parameter $n$ is

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