For a binomial variate X with n=6,if P(X=4)=f | vedclass

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(D) Given,$P(X=4) = \frac{135}{2^{12}}$.Using the binomial probability formula $P(X=k) = {}^nC_k p^k q^{n-k}$,we have:${}^6C_4 p^4 q^2 = \frac{135}{2^

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$\frac{9}{8}$

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(D) Given,$P(X=4) = \frac{135}{2^{12}}$.Using the binomial probability formula $P(X=k) = {}^nC_k p^k q^{n-k}$,we have:${}^6C_4 p^4 q^2 = \frac{135}{2^{12}}$.Since ${}^6C_4 = 15$,we get $15 p^4 q^2 = \frac{135}{2^{12}}$.Dividing by $15$,$p^4 q^2 = \frac{9}{2^{12}} = \frac{3^2}{(2^6)^2} = \left(\frac{3}{64}ight)^2$.Taking the square root,$p^2 q = \frac{3}{64}$.Substituting $q = 1-p$,we have $p^2(1-p) = \frac{3}{64}$.By inspection,if $p = \frac{1}{4}$,then $p^2(1-p) = (\frac{1}{16})(\frac{3}{4}) = \frac{3}{64}$.Thus,$p = \frac{1}{4}$ and $q = \frac{3}{4}$.The variance of a binomial distribution is given by $npq$.Variance $= 6 	imes \frac{1}{4} 	imes \frac{3}{4} = \frac{18}{16} = \frac{9}{8}$.

For a binomial variate $X$ with $n=6$,if $P(X=4)=\frac{135}{2^{12}}$,then its variance is

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