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(D) For a binomial distribution,the mean is given by $E(X) = np = 2$ and the variance is given by $Var(X) = npq = 1$. Dividing the variance by the mea

Vedclass · Accepted Answer

$15/16$

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(D) For a binomial distribution,the mean is given by $E(X) = np = 2$ and the variance is given by $Var(X) = npq = 1$. Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{1}{2}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n(\frac{1}{2}) = 2$,which implies $n = 4$. The probability $P(X \geq 1)$ is given by $1 - P(X = 0)$. Using the binomial probability formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$,we have $P(X = 0) = \binom{4}{0} (\frac{1}{2})^0 (\frac{1}{2})^4 = 1 	imes 1 	imes \frac{1}{16} = \frac{1}{16}$. Therefore,$P(X \geq 1) = 1 - \frac{1}{16} = \frac{15}{16}$.

If the mean and variance of a binomial variable $X$ are $2$ and $1$ respectively,then what is the probability that $X \geq 1$?

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