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(D) Let $X \sim B(n, p)$.Given that the mean $np = 8$ and variance $npq = 4$.Since $q = 1 - p$,we have $8q = 4$,which implies $q = \frac{1}{2}$ and $p

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$137$

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(D) Let $X \sim B(n, p)$.Given that the mean $np = 8$ and variance $npq = 4$.Since $q = 1 - p$,we have $8q = 4$,which implies $q = \frac{1}{2}$ and $p = \frac{1}{2}$.Substituting $p = \frac{1}{2}$ into $np = 8$,we get $n = 16$.We need to find $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.Using the formula $P(X=r) = {}^{n}C_{r} p^{r} q^{n-r}$,we have:$P(X \leq 2) = {}^{16}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{16} + {}^{16}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{15} + {}^{16}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{14}$.$P(X \leq 2) = \frac{{}^{16}C_{0} + {}^{16}C_{1} + {}^{16}C_{2}}{2^{16}}$.Calculating the combinations: ${}^{16}C_{0} = 1$,${}^{16}C_{1} = 16$,and ${}^{16}C_{2} = \frac{16 	imes 15}{2} = 120$.Thus,$P(X \leq 2) = \frac{1 + 16 + 120}{2^{16}} = \frac{137}{2^{16}}$.Comparing this with $\frac{K}{2^{16}}$,we get $K = 137$.

Let a random variable $X$ have a Binomial distribution with mean $8$ and variance $4$. If $P(X \leq 2) = \frac{K}{2^{16}}$,then $K$ is

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