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(D) We know that the sum of probabilities in a distribution is $1$.$\Sigma P(X=x) = K + 2K + 3K^2 + K^2 = 1$$4K^2 + 3K - 1 = 0$$(4K - 1)(K + 1) = 0$Si

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$3$

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(D) We know that the sum of probabilities in a distribution is $1$.$\Sigma P(X=x) = K + 2K + 3K^2 + K^2 = 1$$4K^2 + 3K - 1 = 0$$(4K - 1)(K + 1) = 0$Since $P(X=x) \geq 0$, we reject $K = -1$. Thus, $K = \frac{1}{4}$.The distribution is:$\begin{array}{|c|c|c|c|c|}\hline X=x & 2 & 3 & 5 & 9 \\hline P(X=x) & \frac{1}{4} & \frac{1}{2} & \frac{3}{16} & \frac{1}{16} \\hline\end{array}$$E(X) = \Sigma x P(x) = (2 	imes \frac{1}{4}) + (3 	imes \frac{1}{2}) + (5 	imes \frac{3}{16}) + (9 	imes \frac{1}{16}) = \frac{8}{16} + \frac{24}{16} + \frac{15}{16} + \frac{9}{16} = \frac{56}{16} = \frac{7}{2}$$E(X^2) = \Sigma x^2 P(x) = (4 	imes \frac{1}{4}) + (9 	imes \frac{1}{2}) + (25 	imes \frac{3}{16}) + (81 	imes \frac{1}{16}) = 1 + \frac{9}{2} + \frac{75}{16} + \frac{81}{16} = \frac{16 + 72 + 75 + 81}{16} = \frac{244}{16} = \frac{61}{4}$$	ext{Variance} = E(X^2) - [E(X)]^2 = \frac{61}{4} - (\frac{7}{2})^2 = \frac{61}{4} - \frac{49}{4} = \frac{12}{4} = 3$.

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(a)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$0.2$	$0.3$	$0.15$	$0.25$	$0.1$

If the probability distribution of a random variable $X$ is as follows, then the variance of $X$ is
$\begin{array}{|c|c|c|c|c|}\hline X=x & 2 & 3 & 5 & 9 \\\hline P(X=x) & K & 2 K & 3 K^2 & K^2 \\\hline\end{array}$

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If the probability distribution of a random variable $X$ is as follows, then the variance of $X$ is$\begin{array}{|c|c|c|c|c|}\hline X=x & 2 & 3 & 5 & 9 \\\hline P(X=x) & K & 2 K & 3 K^2 & K^2 \\\hline\end{array}$