If the mean and variance of a binomial distri | vedclass

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(A) We know that for a binomial distribution: Mean $= np = 4$ Variance $= npq = \frac{4}{3}$ Dividing the variance by the mean, we get: $\frac{npq}{np

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$\frac{20}{243}$

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(A) We know that for a binomial distribution: Mean $= np = 4$ Variance $= npq = \frac{4}{3}$ Dividing the variance by the mean, we get: $\frac{npq}{np} = \frac{4/3}{4} \Rightarrow q = \frac{1}{3}$ Since $p + q = 1$, we have $p = 1 - \frac{1}{3} = \frac{2}{3}$ Substituting $p$ into the mean equation: $n 	imes \frac{2}{3} = 4 \Rightarrow n = 6$ The probability mass function is $P(X=k) = {}^nC_k p^k q^{n-k}$ For $X=2$: $P(X=2) = {}^6C_2 	imes (\frac{2}{3})^2 	imes (\frac{1}{3})^4$ $P(X=2) = \frac{6 	imes 5}{2 	imes 1} 	imes \frac{4}{9} 	imes \frac{1}{81}$ $P(X=2) = 15 	imes \frac{4}{729} = \frac{60}{729} = \frac{20}{243}$

If the mean and variance of a binomial distribution are $4$ and $\frac{4}{3}$ respectively, then $P(X=2)=$

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