For a binomial distribution with mean 6 and v | vedclass

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(B) Given Mean = $np = 6$ ... $(i)$ Variance = $npq = 2$ where $q = 1 - p$ ... $(ii)$ Substituting $(i)$ into $(ii)$: $6q = 2 \implies q = \frac{1}{3}

Vedclass · Accepted Answer

$\frac{2^8}{3^7}$

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(B) Given Mean = $np = 6$ ... $(i)$ Variance = $npq = 2$ where $q = 1 - p$ ... $(ii)$ Substituting $(i)$ into $(ii)$: $6q = 2 \implies q = \frac{1}{3}$ Since $p = 1 - q$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$ From $(i)$,$n = \frac{6}{p} = \frac{6}{2/3} = 9$ The probability mass function is $P(X = r) = {}^nC_r p^r q^{n-r}$ For $X = 8$: $P(X = 8) = {}^9C_8 \left(\frac{2}{3}ight)^8 \left(\frac{1}{3}ight)^{9-8}$ $P(X = 8) = 9 	imes \frac{2^8}{3^8} 	imes \frac{1}{3}$ $P(X = 8) = 9 	imes \frac{2^8}{3^9} = \frac{3^2 	imes 2^8}{3^9} = \frac{2^8}{3^7}$

For a binomial distribution with mean $6$ and variance $2$,the value of $P(X = 8)$ is

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