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(C) The expected value $E(X)$ is calculated as follows:$E(X) = \sum x_i P(x_i)$$E(X) = (-2)(0.1) + (-1)(0.2) + (0)(0.2) + (1)(0.3) + (2)(0.15) + (3)(0

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$1.8275$

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(C) The expected value $E(X)$ is calculated as follows:$E(X) = \sum x_i P(x_i)$$E(X) = (-2)(0.1) + (-1)(0.2) + (0)(0.2) + (1)(0.3) + (2)(0.15) + (3)(0.05)$$E(X) = -0.2 - 0.2 + 0 + 0.3 + 0.3 + 0.15 = 0.35$The expected value of the square $E(X^2)$ is:$E(X^2) = \sum x_i^2 P(x_i)$$E(X^2) = (-2)^2(0.1) + (-1)^2(0.2) + (0)^2(0.2) + (1)^2(0.3) + (2)^2(0.15) + (3)^2(0.05)$$E(X^2) = (4)(0.1) + (1)(0.2) + 0 + (1)(0.3) + (4)(0.15) + (9)(0.05)$$E(X^2) = 0.4 + 0.2 + 0 + 0.3 + 0.6 + 0.45 = 1.95$The variance $Var(X)$ is given by:$Var(X) = E(X^2) - [E(X)]^2$$Var(X) = 1.95 - (0.35)^2$$Var(X) = 1.95 - 0.1225 = 1.8275$

$X$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X)$	$0.1$	$0.2$	$0.2$	$0.3$	$0.15$	$0.05$

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

Outcome	Probability
$\omega_{1}$	$\frac{1}{12}$
$\omega_{2}$	$\frac{1}{12}$
$\omega_{3}$	$\frac{1}{6}$
$\omega_{4}$	$\frac{1}{6}$
$\omega_{5}$	$\frac{1}{6}$
$\omega_{6}$	$\frac{3}{2}$

For the following probability distribution,find the $Var(X)$.
$X$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X)$ $0.1$ $0.2$ $0.2$ $0.3$ $0.15$ $0.05$

(Given : $(0.25)^2 = 0.0625$,$(0.35)^2 = 0.1225$,$(0.45)^2 = 0.2025$)

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For the following probability distribution,find the $Var(X)$.$X$$-2$$-1$$0$$1$$2$$3$$P(X)$$0.1$$0.2$$0.2$$0.3$$0.15$$0.05$(Given : $(0.25)^2 = 0.0625$,$(0.35)^2 = 0.1225$,$(0.45)^2 = 0.2025$)