The probability distribution of a random vari | vedclass

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(A) The mean $E(X)$ is calculated as:$E(X) = \sum x_i p_i = 0 	imes \frac{1}{10} + 1 	imes \frac{2}{10} + 2 	imes \frac{3}{10} + 3 	imes \frac{4}{

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(A) The mean $E(X)$ is calculated as:$E(X) = \sum x_i p_i = 0 	imes \frac{1}{10} + 1 	imes \frac{2}{10} + 2 	imes \frac{3}{10} + 3 	imes \frac{4}{10}$$E(X) = 0 + \frac{2}{10} + \frac{6}{10} + \frac{12}{10} = \frac{20}{10} = 2$The variance $Var(X)$ is calculated as:$Var(X) = E(X^2) - [E(X)]^2$First, calculate $E(X^2) = \sum x_i^2 p_i$:$E(X^2) = 0^2 	imes \frac{1}{10} + 1^2 	imes \frac{2}{10} + 2^2 	imes \frac{3}{10} + 3^2 	imes \frac{4}{10}$$E(X^2) = 0 + \frac{2}{10} + \frac{12}{10} + \frac{36}{10} = \frac{50}{10} = 5$Now, $Var(X) = 5 - (2)^2 = 5 - 4 = 1$Thus, the variance is $1$.

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

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

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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