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(C) The sum of probabilities in a probability distribution is $1$.$0.07 + 0.2 + 0.3 + k + 0.07 + 0.04 + 0.02 = 1$$0.7 + k = 1 \implies k = 0.3$Now,we

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(C) The sum of probabilities in a probability distribution is $1$.$0.07 + 0.2 + 0.3 + k + 0.07 + 0.04 + 0.02 = 1$$0.7 + k = 1 \implies k = 0.3$Now,we calculate the mean $E(X) = \sum x_i p_i$ and $E(X^2) = \sum x_i^2 p_i$.$x_i$$p_i$$x_i p_i$$x_i^2 p_i$$5$$0.07$$0.35$$1.75$$6$$0.2$$1.2$$7.2$$7$$0.3$$2.1$$14.7$$8$$0.3$$2.4$$19.2$$9$$0.07$$0.63$$5.67$$10$$0.04$$0.4$$4$$11$$0.02$$0.22$$2.42$Total$1$$7.3$$55.04$Mean $E(X) = \sum x_i p_i = 7.3$$E(X^2) = \sum x_i^2 p_i = 55.04$Variance $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$$\operatorname{Var}(X) = 55.04 - (7.3)^2 = 55.04 - 53.29 = 1.75$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{2}{20}$	$\frac{4}{20}$	$\frac{6}{20}$	$\frac{8}{20}$

$X=x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.1$	$k$	$0.2$	$2k$	$0.3$	$k$

$X = x_i$	$3$	$5$	$7$	$9$
$P(X = x_i)$	$k$	$2k$	$3k$	$4k$

For the probability distribution given below,find $\operatorname{Var}(X)$.
$X$ $5$ $6$ $7$ $8$ $9$ $10$ $11$
$P(X=x)$ $0.07$ $0.2$ $0.3$ $k$ $0.07$ $0.04$ $0.02$

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For the probability distribution given below,find $\operatorname{Var}(X)$.$X$$5$$6$$7$$8$$9$$10$$11$$P(X=x)$$0.07$$0.2$$0.3$$k$$0.07$$0.04$$0.02$