The mean and the variance of a binomial distr | vedclass

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(A) For a binomial distribution,the mean is given by $\mu = np = 4$ and the variance is given by $\sigma^2 = npq = 2$.Dividing the variance by the mea

Vedclass · Accepted Answer

$\frac{28}{256}$

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

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

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