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The probability distribution of a discrete random variable $X$ is given below. If $E\left(X^2\right)=\Sigma x^2 P(X=x)$, then $6 E\left(X^2\right)-\operatorname{Var}(X)=$
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Verified Answer
The correct answer is:
$\frac{113}{12}$
We have,
$$
\begin{aligned}
6 E\left(X^2\right)-\operatorname{Var}(X) & =6 E\left(X^2\right)-\left(E\left(X^2\right)-\left(E(X)^2\right)\right) \\
& =5 E\left(X^2\right)+(E(X))^2 \\
& =5\left(\frac{1}{3}+\frac{1}{6}+\frac{4}{3}\right)+\left(-\frac{1}{3}+\frac{1}{6}+\frac{2}{3}\right)^2 \\
& =\frac{55}{6}+\frac{1}{4}=\frac{110+3}{12}=\frac{113}{12}
\end{aligned}
$$
$$
\begin{aligned}
6 E\left(X^2\right)-\operatorname{Var}(X) & =6 E\left(X^2\right)-\left(E\left(X^2\right)-\left(E(X)^2\right)\right) \\
& =5 E\left(X^2\right)+(E(X))^2 \\
& =5\left(\frac{1}{3}+\frac{1}{6}+\frac{4}{3}\right)+\left(-\frac{1}{3}+\frac{1}{6}+\frac{2}{3}\right)^2 \\
& =\frac{55}{6}+\frac{1}{4}=\frac{110+3}{12}=\frac{113}{12}
\end{aligned}
$$
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