If the mean and variance of a binomial variat | vedclass

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(C) Given,mean of binomial variable,$np = 8$ and variance of binomial variable,$npq = 4$. $	herefore q = \frac{npq}{np} = \frac{4}{8} = \frac{1}{2}$.

Vedclass · Accepted Answer

$\frac{137}{2^{16}}$

Vedclass · Answer

(C) Given,mean of binomial variable,$np = 8$ and variance of binomial variable,$npq = 4$. $	herefore q = \frac{npq}{np} = \frac{4}{8} = \frac{1}{2}$. Since $p = 1 - q$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ in $np = 8$,we get $n 	imes \frac{1}{2} = 8$,so $n = 16$. We need to find $P(X Using the formula $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$: $P(X=0) = {}^{16}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{16} = 1 	imes \frac{1}{2^{16}} = \frac{1}{2^{16}}$. $P(X=1) = {}^{16}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{15} = 16 	imes \frac{1}{2^{16}} = \frac{16}{2^{16}}$. $P(X=2) = {}^{16}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{14} = \frac{16 	imes 15}{2} 	imes \frac{1}{2^{16}} = 120 	imes \frac{1}{2^{16}} = \frac{120}{2^{16}}$. Therefore,$P(X < 3) = \frac{1 + 16 + 120}{2^{16}} = \frac{137}{2^{16}}$.

If the mean and variance of a binomial variate $X$ are $8$ and $4$ respectively,then $P(X < 3)$ equals to

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