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(D) Since the sum of probabilities is $1$:$\frac{1}{6} + k + \frac{1}{4} + k + \frac{1}{6} = 1$$2k + \frac{7}{12} = 1 \implies 2k = \frac{5}{12} \impl

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$\frac{7}{4}$

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(D) Since the sum of probabilities is $1$:$\frac{1}{6} + k + \frac{1}{4} + k + \frac{1}{6} = 1$$2k + \frac{7}{12} = 1 \implies 2k = \frac{5}{12} \implies k = \frac{5}{24}$Calculation for Variance:$\begin{array}{|c|c|c|c|} \hline x_i & P(X=x_i) & x_i P(x_i) & x_i^2 P(x_i) \ \hline -2 & 1/6 & -1/3 & 2/3 \ \hline -1 & 5/24 & -5/24 & 5/24 \ \hline 0 & 1/4 & 0 & 0 \ \hline 1 & 5/24 & 5/24 & 5/24 \ \hline 2 & 1/6 & 1/3 & 2/3 \ \hline 	ext{Total} & 1 & 0 & 21/12 \ \hline \end{array}$$	ext{Variance} = E(X^2) - [E(X)]^2$$= \frac{21}{12} - (0)^2 = \frac{7}{4}$

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$1/6$	$k$	$1/4$	$k$	$1/6$

$X = x$	$1$	$2$	$3$	$\dots$	$n$
$P(X = x)$	$\frac{1}{n}$	$\frac{1}{n}$	$\frac{1}{n}$	$\dots$	$\frac{1}{n}$

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

$A$ random variable $X$ has the following probability distribution:

$X=x_i$ $-2$ $-1$ $0$ $1$ $2$

$P(X=x_i)$ $1/6$ $k$ $1/4$ $k$ $1/6$

The variance of this random variable is

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The probability distribution of a random variable $X$ is given by the following table:
$X = x$ $1$ $2$ $3$ $\dots$ $n$
$P(X = x)$ $\frac{1}{n}$ $\frac{1}{n}$ $\frac{1}{n}$ $\dots$ $\frac{1}{n}$

Then $\operatorname{Var}(X) = $

The probability distribution of a random variable $X$ is given below.

$X = x$ $0$ $1$ $2$ $3$

$P(X = x)$ $\frac{1}{10}$ $\frac{2}{10}$ $\frac{3}{10}$ $\frac{4}{10}$

Then the variance of $X$ is

$A$ man draws a card from a pack of $52$ playing cards,replaces it,and shuffles the pack. He continues this process until he gets a spade card. The probability that he will fail the first two times is:

If the following function is a probability density function of a random variable $X$, $f(x) = kx^2(1 - x)$ for $0 < x < 1$ and $f(x) = 0$ otherwise, then the value of $k$ is:

In a Poisson distribution with unit mean,calculate the value of $\sum_{x=0}^{\infty} |x-\bar{x}| P(X=x)$,where $\bar{x}$ is the mean of the distribution.

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