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(B) Given that $X$ is a binomial variate with mean $np = 6$ and variance $npq = 2$. From these,we have $6q = 2$,which implies $q = \frac{1}{3}$. Thus,

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$\frac{4672}{6561}$

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(B) Given that $X$ is a binomial variate with mean $np = 6$ and variance $npq = 2$. From these,we have $6q = 2$,which implies $q = \frac{1}{3}$. Thus,$p = 1 - q = 1 - \frac{1}{3} = \frac{2}{3}$. Substituting $p$ in $np = 6$,we get $n 	imes \frac{2}{3} = 6$,so $n = 9$. We need to calculate $P(5 \leq X \leq 7) = P(X=5) + P(X=6) + P(X=7)$. Using the formula $P(X=k) = {}^nC_k p^k q^{n-k}$: $P(X=5) = {}^9C_5 (\frac{2}{3})^5 (\frac{1}{3})^4 = 126 	imes \frac{32}{3^9} = \frac{4032}{19683}$. $P(X=6) = {}^9C_6 (\frac{2}{3})^6 (\frac{1}{3})^3 = 84 	imes \frac{64}{3^9} = \frac{5376}{19683}$. $P(X=7) = {}^9C_7 (\frac{2}{3})^7 (\frac{1}{3})^2 = 36 	imes \frac{128}{3^9} = \frac{4608}{19683}$. Summing these values: $P(5 \leq X \leq 7) = \frac{4032 + 5376 + 4608}{19683} = \frac{14016}{19683}$. Dividing numerator and denominator by $3$: $\frac{4672}{6561}$.

If $X$ is a binomial variate with mean $6$ and variance $2$,then the value of $P(5 \leq X \leq 7)$ is

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