If the standard deviation of the random varia | vedclass

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(B) Key Idea: Use $p + q = 1$ and $Var(X) = E(X^2) - (E(X))^2$. Given that the standard deviation of $X$ is $\sqrt{3pq}$,the variance is $Var(X) = (\s

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$3p(1 + 2p)$

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(B) Key Idea: Use $p + q = 1$ and $Var(X) = E(X^2) - (E(X))^2$. Given that the standard deviation of $X$ is $\sqrt{3pq}$,the variance is $Var(X) = (\sqrt{3pq})^2 = 3pq$. Given that the mean $E(X) = 3p$. Using the formula $Var(X) = E(X^2) - (E(X))^2$,we have: $3pq = E(X^2) - (3p)^2$ $E(X^2) = 3pq + 9p^2$. Since $p + q = 1$,we can substitute $q = 1 - p$: $E(X^2) = 3p(1 - p) + 9p^2$ $E(X^2) = 3p - 3p^2 + 9p^2$ $E(X^2) = 3p + 6p^2$ $E(X^2) = 3p(1 + 2p)$.

If the standard deviation of the random variable $X$ is $\sqrt{3pq}$ and the mean is $3p$,then $E(X^2) = . . . . . . .$

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