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$

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

$X$	$30$	$10$	$-10$
$P(X)$	$\frac{1}{5}$	$A$	$B$

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 random variable $X$ take values $\{0, 1, 2, 3\}$ with $P(X=0) = P(X=1) = p$,$P(X=2) = P(X=3) = q$,and $E(X^2) = 2E(X)$. Then the value of $8p - 1$ is:

Let $X$ be a discrete random variable. The probability distribution of $X$ is given below:
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For a minimum value of $p$,the value of $5 E(X)$ is:

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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: