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(B) For a probability distribution,the sum of probabilities is $1$.$\Sigma P(X=x) = k + 2k + 3k + 2k + k = 9k = 1 \implies k = \frac{1}{9}$.The mean $

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$\frac{4}{3}$

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(B) For a probability distribution,the sum of probabilities is $1$.$\Sigma P(X=x) = k + 2k + 3k + 2k + k = 9k = 1 \implies k = \frac{1}{9}$.The mean $E(X) = \Sigma x_i P(x_i) = (1 	imes k) + (2 	imes 2k) + (3 	imes 3k) + (4 	imes 2k) + (5 	imes k) = k + 4k + 9k + 8k + 5k = 27k$.Substituting $k = \frac{1}{9}$,$E(X) = 27 	imes \frac{1}{9} = 3$.The expected value of $X^2$ is $E(X^2) = \Sigma x_i^2 P(x_i) = (1^2 	imes k) + (2^2 	imes 2k) + (3^2 	imes 3k) + (4^2 	imes 2k) + (5^2 	imes k) = k + 8k + 27k + 32k + 25k = 93k$.Substituting $k = \frac{1}{9}$,$E(X^2) = 93 	imes \frac{1}{9} = \frac{93}{9} = \frac{31}{3}$.Variance $Var(X) = E(X^2) - [E(X)]^2 = \frac{31}{3} - (3)^2 = \frac{31}{3} - 9 = \frac{31-27}{3} = \frac{4}{3}$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$k$	$2k$	$3k$	$2k$	$k$

$A$ random variable $X$ has the probability distribution given below. Its variance is
$X$ $1$ $2$ $3$ $4$ $5$
$P(X=x)$ $k$ $2k$ $3k$ $2k$ $k$

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$A$ random variable $X$ has the probability distribution given below. Its variance is$X$$1$$2$$3$$4$$5$$P(X=x)$$k$$2k$$3k$$2k$$k$