In a binomial distribution,the probability of | vedclass

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(C) Given that the probability of success $p = \frac{1}{4}$.Since $q = 1 - p$,we have $q = 1 - \frac{1}{4} = \frac{3}{4}$.The standard deviation $\sig

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$12$

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(C) Given that the probability of success $p = \frac{1}{4}$.Since $q = 1 - p$,we have $q = 1 - \frac{1}{4} = \frac{3}{4}$.The standard deviation $\sigma = 3$,so the variance $\sigma^2 = 9$.For a binomial distribution,the variance is given by $npq = 9$.Substituting the values of $p$ and $q$: $n 	imes \frac{1}{4} 	imes \frac{3}{4} = 9$.$n 	imes \frac{3}{16} = 9 \Rightarrow n = 9 	imes \frac{16}{3} = 48$.The mean of the binomial distribution is $\mu = np$.Therefore,$\mu = 48 	imes \frac{1}{4} = 12$.

In a binomial distribution,the probability of success is $\frac{1}{4}$ and the standard deviation is $3$. Find its mean.

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