The mean of a binomial variate X sim B(n, p) | vedclass

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Question

(A) For a binomial distribution $X \sim B(n, p)$, the mean is given by $E(X) = np = 1$. Thus, $p = \frac{1}{n}$ and $q = 1 - p = 1 - \frac{1}{n} = \fr

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(A) For a binomial distribution $X \sim B(n, p)$, the mean is given by $E(X) = np = 1$. Thus, $p = \frac{1}{n}$ and $q = 1 - p = 1 - \frac{1}{n} = \frac{n-1}{n}$. The probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k}$. Given $P(X=2) = \binom{n}{2} p^2 q^{n-2} = \frac{27}{128}$. Substituting $p$ and $q$: $\frac{n(n-1)}{2} 	imes (\frac{1}{n})^2 	imes (\frac{n-1}{n})^{n-2} = \frac{27}{128}$ $\frac{n-1}{2n} 	imes \frac{(n-1)^{n-2}}{n^{n-2}} = \frac{27}{128}$ $\frac{(n-1)^{n-1}}{2n^{n-1}} = \frac{27}{128}$ For $n=4$: $\frac{(4-1)^{4-1}}{2(4)^{4-1}} = \frac{3^3}{2(4^3)} = \frac{27}{2(64)} = \frac{27}{128}$. This satisfies the given condition. The variance is $Var(X) = npq = 1 	imes q = q$. Since $p = \frac{1}{4}$, $q = 1 - \frac{1}{4} = \frac{3}{4}$. Therefore, the variance is $\frac{3}{4}$.

The mean of a binomial variate $X \sim B(n, p)$ is $1$. If $n > 2$ and $P(X=2)=\frac{27}{128}$, then the variance of the distribution is

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