The probability distribution of a random vari | vedclass

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(A) To find the variance of the random variable $X$,we use the formula $Var(X) = E(X^2) - [E(X)]^2$.$x_{i}$$p(x_{i})$$p_{i} x_{i}$$p_{i} x_{i}^{2}$$0$

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$\frac{14}{25}$

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(A) To find the variance of the random variable $X$,we use the formula $Var(X) = E(X^2) - [E(X)]^2$.$x_{i}$$p(x_{i})$$p_{i} x_{i}$$p_{i} x_{i}^{2}$$0$$\frac{1}{5}$$0$$0$$1$$\frac{2}{5}$$\frac{2}{5}$$\frac{2}{5}$$2$$\frac{2}{5}$$\frac{4}{5}$$\frac{8}{5}$First,calculate the mean $E(X) = \sum p_{i} x_{i} = 0 + \frac{2}{5} + \frac{4}{5} = \frac{6}{5}$.Next,calculate $E(X^2) = \sum p_{i} x_{i}^{2} = 0 + \frac{2}{5} + \frac{8}{5} = \frac{10}{5} = 2$.Now,calculate the variance: $Var(X) = E(X^2) - [E(X)]^2 = 2 - (\frac{6}{5})^2 = 2 - \frac{36}{25} = \frac{50 - 36}{25} = \frac{14}{25}$.

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

The probability distribution of a random variable $X$ is given by
$X = x$ $0$ $1$ $2$
$P(X = x)$ $\frac{1}{5}$ $\frac{2}{5}$ $\frac{2}{5}$

Then the variance of $X$ is

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$X$ $0$ $1$ $2$ $3$

$P(X=x)$ $\frac{1+p}{5}$ $\frac{2-2p}{5}$ $\frac{2-p}{5}$ $\frac{2p}{5}$

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If $X$ is a random variable with the following probability distribution:
$X=x$ $-3$ $6$ $9$
$P(X=x)$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{1}{3}$

Then the variance of $X$ is:

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$X = x$	$0$	$1$	$2$
$P(X = x)$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{2}{5}$

The probability distribution of a random variable $X$ is given by$X = x$$0$$1$$2$$P(X = x)$$\frac{1}{5}$$\frac{2}{5}$$\frac{2}{5}$Then the variance of $X$ is