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(A) For a binomial distribution,the mean is given by $\mu = np = 3$ and the standard deviation is $\sigma = \sqrt{npq} = \frac{3}{2}$.Squaring the sta

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$(\frac{3}{4} + \frac{1}{4})^{12}$

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(A) For a binomial distribution,the mean is given by $\mu = np = 3$ and the standard deviation is $\sigma = \sqrt{npq} = \frac{3}{2}$.Squaring the standard deviation,we get $npq = \frac{9}{4}$.Dividing $npq$ by $np$,we find $q = \frac{npq}{np} = \frac{9/4}{3} = \frac{3}{4}$.Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{3}{4} = \frac{1}{4}$.Now,substituting $p = \frac{1}{4}$ into $np = 3$,we get $n(\frac{1}{4}) = 3$,which implies $n = 12$.The binomial distribution is given by $(q + p)^n = (\frac{3}{4} + \frac{1}{4})^{12}$.

In a binomial probability distribution,the mean is $3$ and the standard deviation is $\frac{3}{2}$. Then the probability distribution is:

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