For which binomial distribution are the mean | vedclass

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(D) For a binomial distribution,the mean is given by $\mu = np = 6$ and the variance is given by $\sigma^2 = npq = 2$. Dividing the variance by the me

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$(\frac{1}{3} + \frac{2}{3})^9$

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(D) For a binomial distribution,the mean is given by $\mu = np = 6$ and the variance is given by $\sigma^2 = npq = 2$. Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{2}{6}$,which simplifies to $q = \frac{1}{3}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$. Substituting $p = \frac{2}{3}$ into $np = 6$,we get $n(\frac{2}{3}) = 6$,which implies $n = 6 	imes \frac{3}{2} = 9$. The binomial distribution is given by $(q + p)^n$,which is $(\frac{1}{3} + \frac{2}{3})^9$.

For which binomial distribution are the mean $= 6$ and variance $= 2$?

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