Let X be a binomially distributed random vari | vedclass

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(B) For a binomial distribution,the mean is $np = 4$ and the variance is $npq = \frac{4}{3}$.Dividing the variance by the mean,we get $q = \frac{4/3}{

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$\frac{146}{27}$

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(B) For a binomial distribution,the mean is $np = 4$ and the variance is $npq = \frac{4}{3}$.Dividing the variance by the mean,we get $q = \frac{4/3}{4} = \frac{1}{3}$.Since $p + q = 1$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$.Substituting $p$ into $np = 4$,we get $n 	imes \frac{2}{3} = 4$,which gives $n = 6$.The probability mass function is $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k} = {}^{6}C_{k} (\frac{2}{3})^{k} (\frac{1}{3})^{6-k}$.We need to calculate $54 P(X \leq 2) = 54 [P(X=0) + P(X=1) + P(X=2)]$.$P(X=0) = {}^{6}C_{0} (\frac{2}{3})^{0} (\frac{1}{3})^{6} = 1 	imes 1 	imes \frac{1}{729} = \frac{1}{729}$.$P(X=1) = {}^{6}C_{1} (\frac{2}{3})^{1} (\frac{1}{3})^{5} = 6 	imes \frac{2}{3} 	imes \frac{1}{243} = \frac{12}{729}$.$P(X=2) = {}^{6}C_{2} (\frac{2}{3})^{2} (\frac{1}{3})^{4} = 15 	imes \frac{4}{9} 	imes \frac{1}{81} = \frac{60}{729}$.Summing these,$P(X \leq 2) = \frac{1 + 12 + 60}{729} = \frac{73}{729}$.Finally,$54 P(X \leq 2) = 54 	imes \frac{73}{729} = \frac{2 	imes 73}{27} = \frac{146}{27}$.

Let $X$ be a binomially distributed random variable with mean $4$ and variance $\frac{4}{3}$. Then $54 P(X \leq 2)$ is equal to.

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