The binomial distribution for which mean = 6 | vedclass

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(D) For a binomial distribution,the mean is given by $np = 6$ and the variance is given by $npq = 2$.Dividing the variance by the mean,we get $\frac{n

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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 $np = 6$ and the variance is given by $npq = 2$.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 $p = 1 - q = 1 - \frac{1}{3} = \frac{2}{3}$.Substituting $p = \frac{2}{3}$ into the mean equation $np = 6$,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 $(\frac{1}{3} + \frac{2}{3})^9$.

The binomial distribution for which mean $= 6$ and variance $= 2$,is

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