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(D) To find the variance of the probability distribution,we use the formula: $	ext{Variance} = E(X^2) - [E(X)]^2$,where $E(X) = \sum p_i x_i$ and $E(

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$\frac{3}{16}$

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(D) To find the variance of the probability distribution,we use the formula: $	ext{Variance} = E(X^2) - [E(X)]^2$,where $E(X) = \sum p_i x_i$ and $E(X^2) = \sum p_i x_i^2$. First,calculate $E(X)$: $E(X) = (0 	imes \frac{9}{16}) + (1 	imes \frac{3}{8}) + (2 	imes \frac{1}{16}) = 0 + \frac{3}{8} + \frac{2}{16} = \frac{6}{16} + \frac{2}{16} = \frac{8}{16} = \frac{1}{2}$. Next,calculate $E(X^2)$: $E(X^2) = (0^2 	imes \frac{9}{16}) + (1^2 	imes \frac{3}{8}) + (2^2 	imes \frac{1}{16}) = 0 + \frac{3}{8} + \frac{4}{16} = \frac{6}{16} + \frac{4}{16} = \frac{10}{16} = \frac{5}{8}$. Now,calculate the variance: $	ext{Variance} = E(X^2) - [E(X)]^2 = \frac{5}{8} - (\frac{1}{2})^2 = \frac{5}{8} - \frac{1}{4} = \frac{5}{8} - \frac{2}{8} = \frac{3}{8}$.

$Z$	$3$	$2$	$1$	$0$	$-1$
$P(Z)$	$0.3$	$0.2$	$0.4$	$0.1$	$0.05$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X = x)$	$k$	$\frac{k}{3}$	$\frac{k}{4}$	$\frac{k}{2}$	$\frac{k}{2}$

$X = x_i$	$0$	$1$	$2$	$3$
$P(X = x_i)$	$k$	$3k$	$3k$	$k$

The variance of the following probability distribution is:
$x$ $0$ $1$ $2$
$P(X)$ $\frac{9}{16}$ $\frac{3}{8}$ $\frac{1}{16}$

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State which of the following is not the probability distribution of a random variable. Give reasons for your answer.

$Z$ $3$ $2$ $1$ $0$ $-1$

$P(Z)$ $0.3$ $0.2$ $0.4$ $0.1$ $0.05$

If the probability mass function (p.m.f.) of a random variable $X$ is given by the table below,then find the value of $k$.
$X$ $1$ $2$ $3$ $4$ $5$
$P(X = x)$ $k$ $\frac{k}{3}$ $\frac{k}{4}$ $\frac{k}{2}$ $\frac{k}{2}$

If $X$ is a random variable with the distribution given below:
$X = x_i$ $0$ $1$ $2$ $3$
$P(X = x_i)$ $k$ $3k$ $3k$ $k$

Then the value of $k$ and its variance are respectively given by:

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The variance of the following probability distribution is: $x$$0$$1$$2$$P(X)$$\frac{9}{16}$$\frac{3}{8}$$\frac{1}{16}$