The mean and variance of a binomial distribut | vedclass

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(A) 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 =

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$\frac{7}{64}$

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(A) 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 of exactly $x$ successes is given by $P(X=x) = {}^{n}C_{x} p^{x} q^{n-x}$. For $x = 2$, $P(X=2) = {}^{8}C_{2} \left(\frac{1}{2}ight)^{2} \left(\frac{1}{2}ight)^{8-2} = {}^{8}C_{2} \left(\frac{1}{2}ight)^{8}$. Calculating the value, ${}^{8}C_{2} = \frac{8 	imes 7}{2 	imes 1} = 28$. Thus, $P(X=2) = 28 	imes \frac{1}{256} = \frac{28}{256} = \frac{7}{64}$.

The mean and variance of a binomial distribution are $4$ and $2$ respectively. Then the probability of exactly two successes is

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