The variance of the following probability dis | vedclass

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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}$.

$X$	$\alpha$	$1$	$0$	$-3$
$P(X)$	$\frac{1}{3}$	$K$	$\frac{1}{6}$	$\frac{1}{4}$

$x = x_{i}$	$0$	$1$	$2$
$P_{i}$	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

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