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(B) Let $\mu = np$ be the mean and $\sigma^2 = npq$ be the variance of the binomial distribution $X \sim B(n, p)$.Given $\mu + \sigma^2 = 24$ and $\mu

Vedclass · Accepted Answer

$529$

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(B) Let $\mu = np$ be the mean and $\sigma^2 = npq$ be the variance of the binomial distribution $X \sim B(n, p)$.Given $\mu + \sigma^2 = 24$ and $\mu \sigma^2 = 128$.Let $x = \mu$ and $y = \sigma^2$. Then $x + y = 24$ and $xy = 128$.The quadratic equation $t^2 - 24t + 128 = 0$ has roots $t = 16$ and $t = 8$.Since $\mu > \sigma^2$ (as $q Thus,$np = 16$ and $npq = 8$. Dividing these gives $q = \frac{8}{16} = \frac{1}{2}$.Since $p = 1 - q$,we have $p = \frac{1}{2}$.Then $n 	imes \frac{1}{2} = 16$,so $n = 32$.We need to find $P(X > n - 3) = P(X > 29) = P(X = 30) + P(X = 31) + P(X = 32)$.$P(X = r) = {}^{n}C_r p^r q^{n-r} = {}^{32}C_r (\frac{1}{2})^r (\frac{1}{2})^{32-r} = \frac{{}^{32}C_r}{2^{32}}$.So,$P(X > 29) = \frac{{}^{32}C_{30} + {}^{32}C_{31} + {}^{32}C_{32}}{2^{32}} = \frac{k}{2^{32}}$.Thus,$k = {}^{32}C_{30} + {}^{32}C_{31} + {}^{32}C_{32} = {}^{32}C_2 + {}^{32}C_1 + {}^{32}C_0$.$k = \frac{32 	imes 31}{2} + 32 + 1 = 496 + 32 + 1 = 529$.

Let $X$ have a binomial distribution $B(n, p)$ such that the sum and the product of the mean and variance of $X$ are $24$ and $128$ respectively. If $P(X > n - 3) = \frac{k}{2^n}$,then $k$ is equal to.

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