Let a random variable X have a Binomial distr | vedclass

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(C) For a Binomial distribution,the mean is given by $np = 8$ and the variance is given by $npq = 4$. Dividing the variance by the mean,we get $q = \f

Vedclass · Accepted Answer

$137$

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(C) For a Binomial distribution,the mean is given by $np = 8$ and the variance is given by $npq = 4$. Dividing the variance by the mean,we get $q = \frac{4}{8} = \frac{1}{2}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ into $np = 8$,we get $n 	imes \frac{1}{2} = 8$,so $n = 16$. Now,$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$. Using the formula $P(X=r) = {}^{n}C_{r} p^{r} q^{n-r}$,we have: $P(X \leq 2) = {}^{16}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{16} + {}^{16}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{15} + {}^{16}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{14}$ $P(X \leq 2) = ({}^{16}C_{0} + {}^{16}C_{1} + {}^{16}C_{2}) (\frac{1}{2})^{16}$ Calculating the combinations: ${}^{16}C_{0} = 1$ ${}^{16}C_{1} = 16$ ${}^{16}C_{2} = \frac{16 	imes 15}{2} = 120$ Sum $= 1 + 16 + 120 = 137$. Thus,$P(X \leq 2) = \frac{137}{2^{16}}$. 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 $P(X \leq 2) = \frac{k}{2^{16}}$,then $k$ is equal to:

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