If the mean and variance of a binomial distri | 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 me

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 $q = \frac{npq}{np} = \frac{2}{4} = \frac{1}{2}$. 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 getting $x$ successes is given by $P(X=x) = { }^n C_x p^x q^{n-x}$. For $x = 2$,$P(X=2) = { }^8 C_2 \left(\frac{1}{2}ight)^2 \left(\frac{1}{2}ight)^6 = \frac{8 	imes 7}{2 	imes 1} 	imes \left(\frac{1}{2}ight)^8$. $P(X=2) = 28 	imes \frac{1}{256} = \frac{28}{256}$.

If the mean and variance of a binomial distribution are $4$ and $2$ respectively,then the probability of getting $2$ successes is:

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