The probability distribution of a random vari | vedclass

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(A) The expected value $E(X)$ is calculated as:$E(X) = \sum x_i \cdot P(x_i)$$E(X) = 0(0.4) + 1(0.3) + 2(0.1) + 3(0.1) + 4(0.1)$$E(X) = 0 + 0.3 + 0.2

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$1.76$

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(A) The expected value $E(X)$ is calculated as:$E(X) = \sum x_i \cdot P(x_i)$$E(X) = 0(0.4) + 1(0.3) + 2(0.1) + 3(0.1) + 4(0.1)$$E(X) = 0 + 0.3 + 0.2 + 0.3 + 0.4 = 1.2$The expected value of the square of the random variable $E(X^2)$ is:$E(X^2) = \sum x_i^2 \cdot P(x_i)$$E(X^2) = 0^2(0.4) + 1^2(0.3) + 2^2(0.1) + 3^2(0.1) + 4^2(0.1)$$E(X^2) = 0(0.4) + 1(0.3) + 4(0.1) + 9(0.1) + 16(0.1)$$E(X^2) = 0 + 0.3 + 0.4 + 0.9 + 1.6 = 3.2$The variance of $X$ is given by:$	ext{Var}(X) = E(X^2) - [E(X)]^2$$	ext{Var}(X) = 3.2 - (1.2)^2$$	ext{Var}(X) = 3.2 - 1.44 = 1.76$

$X = x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

Outcome	Probability
$\omega_{1}$	$1/8$
$\omega_{2}$	$2/3$
$\omega_{3}$	$1/3$
$\omega_{4}$	$1/3$
$\omega_{5}$	$-1/4$
$\omega_{6}$	$-1/3$

$X = x_i$	$3$	$5$	$7$	$9$
$P(X = x_i)$	$k$	$2k$	$3k$	$4k$

The probability distribution of a random variable $X$ is given by:
$X = x_i$ $0$ $1$ $2$ $3$ $4$
$P(X = x_i)$ $0.4$ $0.3$ $0.1$ $0.1$ $0.1$

Then the variance of $X$ is:

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Let a sample space be $S = \{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?

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The probability distribution of a random variable $X$ is given by the following table:
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The probability distribution of a random variable $X$ is given by:$X = x_i$$0$$1$$2$$3$$4$$P(X = x_i)$$0.4$$0.3$$0.1$$0.1$$0.1$Then the variance of $X$ is: