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(D) The expected value $E(X^2)$ is calculated using the formula $E(X^2) = \sum x_i^2 P(x_i)$.Given the values:$x_1 = 1, P(x_1) = \frac{1}{10} \implies

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$10$

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(D) The expected value $E(X^2)$ is calculated using the formula $E(X^2) = \sum x_i^2 P(x_i)$.Given the values:$x_1 = 1, P(x_1) = \frac{1}{10} \implies x_1^2 P(x_1) = 1^2 	imes \frac{1}{10} = \frac{1}{10}$$x_2 = 2, P(x_2) = \frac{1}{5} \implies x_2^2 P(x_2) = 2^2 	imes \frac{1}{5} = 4 	imes \frac{1}{5} = \frac{4}{5} = \frac{8}{10}$$x_3 = 3, P(x_3) = \frac{3}{10} \implies x_3^2 P(x_3) = 3^2 	imes \frac{3}{10} = 9 	imes \frac{3}{10} = \frac{27}{10}$$x_4 = 4, P(x_4) = \frac{2}{5} \implies x_4^2 P(x_4) = 4^2 	imes \frac{2}{5} = 16 	imes \frac{2}{5} = \frac{32}{5} = \frac{64}{10}$Summing these values:$E(X^2) = \frac{1}{10} + \frac{8}{10} + \frac{27}{10} + \frac{64}{10} = \frac{1 + 8 + 27 + 64}{10} = \frac{100}{10} = 10$.Therefore,the correct option is $D$.

$X$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$

For the given probability distribution,find $E(X^2)$.
$X$ $1$ $2$ $3$ $4$
$P(X)$ $\frac{1}{10}$ $\frac{1}{5}$ $\frac{3}{10}$ $\frac{2}{5}$

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For the given probability distribution,find $E(X^2)$.$X$$1$$2$$3$$4$$P(X)$$\frac{1}{10}$$\frac{1}{5}$$\frac{3}{10}$$\frac{2}{5}$