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(D) The variance of a random variable $X$ is given by $	ext{Var}(X) = E(X^2) - (E(X))^2$.Since the sum of probabilities in a probability distribution

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$\frac{3}{4}$

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(D) The variance of a random variable $X$ is given by $	ext{Var}(X) = E(X^2) - (E(X))^2$.Since the sum of probabilities in a probability distribution is $1$,we have:$\frac{1}{8} + \frac{3}{8} + 3K + K = 1$$\frac{4}{8} + 4K = 1$$\frac{1}{2} + 4K = 1 \Rightarrow 4K = \frac{1}{2} \Rightarrow K = \frac{1}{8}$.Now,we calculate $E(X)$ and $E(X^2)$:$E(X) = \sum x_i P(x_i) = (0 	imes \frac{1}{8}) + (1 	imes \frac{3}{8}) + (2 	imes \frac{3}{8}) + (3 	imes \frac{1}{8}) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2}$.$E(X^2) = \sum x_i^2 P(x_i) = (0^2 	imes \frac{1}{8}) + (1^2 	imes \frac{3}{8}) + (2^2 	imes \frac{3}{8}) + (3^2 	imes \frac{1}{8}) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$.Therefore,$	ext{Var}(X) = E(X^2) - (E(X))^2 = 3 - (\frac{3}{2})^2 = 3 - \frac{9}{4} = \frac{12 - 9}{4} = \frac{3}{4}$.

$X = x_i$	$0$	$1$	$2$	$3$
$P(X = x_i)$	$\frac{1}{8}$	$\frac{3}{8}$	$3K$	$K$

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

$X = x$	$0$	$1$	$2$
$P(X = x)$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{2}{5}$

If the probability distribution of a random variable $X$ is given by:
$X = x_i$ $0$ $1$ $2$ $3$
$P(X = x_i)$ $\frac{1}{8}$ $\frac{3}{8}$ $3K$ $K$

Find the variance of $X$.

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If the probability distribution of a random variable $X$ is given by:$X = x_i$$0$$1$$2$$3$$P(X = x_i)$$\frac{1}{8}$$\frac{3}{8}$$3K$$K$Find the variance of $X$.