If X sim B(n, p) with n = 10, p = 0.4,then E( | vedclass

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(D) Given $X \sim B(n, p)$ with $n = 10$ and $p = 0.4$. Since $q = 1 - p$,we have $q = 1 - 0.4 = 0.6$. The mean of a binomial distribution is $E(X) =

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$18.4$

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(D) Given $X \sim B(n, p)$ with $n = 10$ and $p = 0.4$. Since $q = 1 - p$,we have $q = 1 - 0.4 = 0.6$. The mean of a binomial distribution is $E(X) = np = 10 	imes 0.4 = 4$. The variance of a binomial distribution is $V(X) = npq = 10 	imes 0.4 	imes 0.6 = 2.4$. We know the relationship $V(X) = E(X^2) - (E(X))^2$. Substituting the values,we get $2.4 = E(X^2) - (4)^2$. $2.4 = E(X^2) - 16$. Therefore,$E(X^2) = 16 + 2.4 = 18.4$.

If $X \sim B(n, p)$ with $n = 10, p = 0.4$,then $E(X^2) =$

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