If the mean and variance of a binomial distri | vedclass

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(D) For a binomial distribution,the mean is given by $\mu = np = 4$ ... $(i)$ The variance is given by $\sigma^2 = npq = 2$ ... (ii) Dividing equation

Vedclass · Accepted Answer

$\frac{7}{64}$

Vedclass · Answer

(D) For a binomial distribution,the mean is given by $\mu = np = 4$ ... $(i)$ The variance is given by $\sigma^2 = npq = 2$ ... (ii) Dividing equation (ii) by equation $(i)$,we get: $\frac{npq}{np} = \frac{2}{4} \implies q = \frac{1}{2}$ Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$ Substituting $p = \frac{1}{2}$ in equation $(i)$: $n 	imes \frac{1}{2} = 4 \implies n = 8$ The probability of $X$ successes in a binomial distribution is given by $P(X=k) = { }^n C_k p^k q^{n-k}$ For $k = 2$: $P(X=2) = { }^8 C_2 \left(\frac{1}{2}ight)^2 \left(\frac{1}{2}ight)^6 = 28 	imes \left(\frac{1}{2}ight)^8 = \frac{28}{256} = \frac{7}{64}$

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

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