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(D) We know that the sum of probabilities in a probability distribution is $1$.$\Sigma P(X = x_i) = 1$$\Rightarrow \frac{1}{6} + k + \frac{1}{4} + k +

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$\frac{7}{4}$

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(D) We know that the sum of probabilities in a probability distribution is $1$.$\Sigma P(X = x_i) = 1$$\Rightarrow \frac{1}{6} + k + \frac{1}{4} + k + \frac{1}{6} = 1$$\Rightarrow 2k + \frac{2}{6} + \frac{1}{4} = 1$$\Rightarrow 2k + \frac{1}{3} + \frac{1}{4} = 1$$\Rightarrow 2k + \frac{7}{12} = 1$$\Rightarrow 2k = 1 - \frac{7}{12} = \frac{5}{12}$$\Rightarrow k = \frac{5}{24}$Now,the mean $E(X) = \Sigma x_i P(x_i)$:$E(X) = (-2)(\frac{1}{6}) + (-1)(\frac{5}{24}) + (0)(\frac{1}{4}) + (1)(\frac{5}{24}) + (2)(\frac{1}{6})$$E(X) = -\frac{2}{6} - \frac{5}{24} + 0 + \frac{5}{24} + \frac{2}{6} = 0$Now,$E(X^2) = \Sigma x_i^2 P(x_i)$:$E(X^2) = (-2)^2(\frac{1}{6}) + (-1)^2(\frac{5}{24}) + (0)^2(\frac{1}{4}) + (1)^2(\frac{5}{24}) + (2)^2(\frac{1}{6})$$E(X^2) = 4(\frac{1}{6}) + 1(\frac{5}{24}) + 0 + 1(\frac{5}{24}) + 4(\frac{1}{6})$$E(X^2) = \frac{4}{6} + \frac{5}{24} + \frac{5}{24} + \frac{4}{6} = \frac{8}{6} + \frac{10}{24} = \frac{4}{3} + \frac{5}{12} = \frac{16+5}{12} = \frac{21}{12} = \frac{7}{4}$Variance $	ext{Var}(X) = E(X^2) - [E(X)]^2$$	ext{Var}(X) = \frac{7}{4} - (0)^2 = \frac{7}{4}$

$X = x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x_i)$	$\frac{1}{6}$	$k$	$\frac{1}{4}$	$k$	$\frac{1}{6}$

$A$ random variable $X$ has the following probability distribution:
$X = x_i$ $-2$ $-1$ $0$ $1$ $2$
$P(X = x_i)$ $\frac{1}{6}$ $k$ $\frac{1}{4}$ $k$ $\frac{1}{6}$

The variance of this random variable is:

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$A$ random variable $X$ has the following probability distribution:$X = x_i$$-2$$-1$$0$$1$$2$$P(X = x_i)$$\frac{1}{6}$$k$$\frac{1}{4}$$k$$\frac{1}{6}$The variance of this random variable is: