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(B) Given that the mean $np = 2$ and the variance $npq = 1$. Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{1}{2}$,which implies $q

Vedclass · Accepted Answer

$\frac{11}{16}$

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(B) Given that the mean $np = 2$ and the variance $npq = 1$. Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{1}{2}$,which implies $q = \frac{1}{2}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n \left( \frac{1}{2} ight) = 2$,so $n = 4$. We need to find $P(X > 1) = P(X = 2) + P(X = 3) + P(X = 4)$. Alternatively,$P(X > 1) = 1 - [P(X = 0) + P(X = 1)]$. $P(X = 0) = {}^4 C_0 \left( \frac{1}{2} ight)^0 \left( \frac{1}{2} ight)^4 = 1 	imes 1 	imes \frac{1}{16} = \frac{1}{16}$. $P(X = 1) = {}^4 C_1 \left( \frac{1}{2} ight)^1 \left( \frac{1}{2} ight)^3 = 4 	imes \frac{1}{2} 	imes \frac{1}{8} = \frac{4}{16}$. Therefore,$P(X > 1) = 1 - \left( \frac{1}{16} + \frac{4}{16} ight) = 1 - \frac{5}{16} = \frac{11}{16}$.

If the mean and the variance of a Binomial variate $X$ are $2$ and $1$ respectively,then the probability that $X$ takes a value greater than $1$ is equal to

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