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(D) For a binomial distribution,the mean is given by $np = 2$ and the variance is given by $npq = 1$.Dividing the variance by the mean,we get $q = \fr

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$\frac{11}{16}$

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(D) For a binomial distribution,the mean is given by $np = 2$ and the variance is given by $npq = 1$.Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{1}{2}$.Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n 	imes \frac{1}{2} = 2$,which implies $n = 4$.The probability mass function is $P(X = k) = {^nC_k} p^k q^{n-k} = {^4C_k} (\frac{1}{2})^k (\frac{1}{2})^{4-k} = {^4C_k} (\frac{1}{2})^4$.We need to find $P(X > 1) = 1 - P(X \le 1) = 1 - [P(X = 0) + P(X = 1)]$.$P(X = 0) = {^4C_0} (\frac{1}{2})^4 = 1 	imes \frac{1}{16} = \frac{1}{16}$.$P(X = 1) = {^4C_1} (\frac{1}{2})^4 = 4 	imes \frac{1}{16} = \frac{4}{16}$.Therefore,$P(X > 1) = 1 - (\frac{1}{16} + \frac{4}{16}) = 1 - \frac{5}{16} = \frac{11}{16}$.

If the mean and the variance of a binomial variate $X$ are $2$ and $1$ respectively,then the probability that $X$ takes a value greater than one is equal to

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