Let a random variable X have a binomial distr | vedclass

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(B) Given that the mean of the binomial distribution is $np = 8$ and the variance is $npq = 4$.Dividing the variance by the mean,we get $q = \frac{npq

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$137$

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(B) Given that the mean of the binomial distribution is $np = 8$ and the variance is $npq = 4$.Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{4}{8} = \frac{1}{2}$.Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.Substituting $p = \frac{1}{2}$ into $np = 8$,we get $n 	imes \frac{1}{2} = 8$,which implies $n = 16$.The probability mass function is $P(X = r) = ^{16}C_r (\frac{1}{2})^r (\frac{1}{2})^{16-r} = ^{16}C_r (\frac{1}{2})^{16}$.We need to find $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.$P(X \le 2) = \frac{^{16}C_0 + ^{16}C_1 + ^{16}C_2}{2^{16}}$.Calculating the combinations: $^{16}C_0 = 1$,$^{16}C_1 = 16$,and $^{16}C_2 = \frac{16 	imes 15}{2} = 120$.Summing these values: $1 + 16 + 120 = 137$.Thus,$P(X \le 2) = \frac{137}{2^{16}}$.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 $P(X \le 2) = \frac{k}{2^{16}}$,then $k$ is equal to

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