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(D) For a binomial distribution,the mean is given by $np = 4$ and the variance is given by $npq = 2$. Dividing the variance by the mean,we get $q = \f

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$28$

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(D) For a binomial distribution,the mean is given by $np = 4$ and the variance is given by $npq = 2$. Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{2}{4} = \frac{1}{2}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ into $np = 4$,we get $n 	imes \frac{1}{2} = 4$,which implies $n = 8$. The probability mass function for a binomial distribution is $P(X = k) = \binom{n}{k} p^k q^{n-k}$. For $X = 2$,$P(X = 2) = \binom{8}{2} (\frac{1}{2})^2 (\frac{1}{2})^{8-2} = \binom{8}{2} (\frac{1}{2})^8$. Calculating the combination,$\binom{8}{2} = \frac{8 	imes 7}{2 	imes 1} = 28$. Thus,$P(X = 2) = 28 	imes \frac{1}{256} = \frac{28}{256}$.

For a binomial distribution,the mean and variance are $4$ and $2$ respectively. What is the probability of $X = 2$ (in $/256$)?

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