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(C) Step $1$: Find the value of $k$. The sum of probabilities must be $1$.$3k + k + k + 2k + k = 1 \implies 8k = 1 \implies k = \frac{1}{8}$.Step $2$:

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$\sqrt{11}$

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(C) Step $1$: Find the value of $k$. The sum of probabilities must be $1$.$3k + k + k + 2k + k = 1 \implies 8k = 1 \implies k = \frac{1}{8}$.Step $2$: Calculate the mean $E(X) = \sum x_i P(X=x_i)$.$E(X) = 2(3k) + 3(k) + 5(k) + 7(2k) + 12(k) = 6k + 3k + 5k + 14k + 12k = 40k$.Since $k = \frac{1}{8}$, $E(X) = 40 	imes \frac{1}{8} = 5$.Step $3$: Calculate $E(X^2) = \sum x_i^2 P(X=x_i)$.$E(X^2) = 2^2(3k) + 3^2(k) + 5^2(k) + 7^2(2k) + 12^2(k) = 12k + 9k + 25k + 98k + 144k = 288k$.Since $k = \frac{1}{8}$, $E(X^2) = 288 	imes \frac{1}{8} = 36$.Step $4$: Calculate the variance $Var(X) = E(X^2) - [E(X)]^2$.$Var(X) = 36 - (5)^2 = 36 - 25 = 11$.Step $5$: Calculate the standard deviation $\sigma = \sqrt{Var(X)} = \sqrt{11}$.

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$	$0$

The probability distribution of a random variable $X$ is given below. Then, the standard deviation of $X$ is
$X=x_i$ $2$ $3$ $5$ $7$ $12$
$P(X=x_i)$ $3k$ $k$ $k$ $2k$ $k$

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The probability distribution of a random variable $X$ is given below. Then, the standard deviation of $X$ is$X=x_i$$2$$3$$5$$7$$12$$P(X=x_i)$$3k$$k$$k$$2k$$k$