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, $npq = 4$. Since $npq = 4$ and $np = 8$, we have $8q = 4$, which imp

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

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(C) Given, mean of binomial variable, $np = 8$ and variance of binomial variable, $npq = 4$. Since $npq = 4$ and $np = 8$, we have $8q = 4$, which implies $q = \frac{1}{2}$. Then $p = 1 - q = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ into $np = 8$, we get $n(\frac{1}{2}) = 8$, so $n = 16$. We need to find $P(X Using the binomial probability 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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