A random variable X has the following distrib | vedclass

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(C) For a probability distribution,the sum of all probabilities must be $1$.$\sum P(X = x_{i}) = 0.1 + k + 0.2 + 2k + 3k + k = 1$$7k + 0.3 = 1 \Righta

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$2.16$

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(C) For a probability distribution,the sum of all probabilities must be $1$.$\sum P(X = x_{i}) = 0.1 + k + 0.2 + 2k + 3k + k = 1$$7k + 0.3 = 1 \Rightarrow 7k = 0.7 \Rightarrow k = 0.1$.Now,we calculate the mean $\mu = E(X) = \sum x_{i} P(x_{i})$:$\mu = (-2)(0.1) + (-1)(0.1) + (0)(0.2) + (1)(0.2) + (2)(0.3) + (3)(0.1)$$\mu = -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3 = 0.8$.Next,we calculate $E(X^2) = \sum x_{i}^2 P(x_{i})$:$E(X^2) = (-2)^2(0.1) + (-1)^2(0.1) + (0)^2(0.2) + (1)^2(0.2) + (2)^2(0.3) + (3)^2(0.1)$$E(X^2) = 4(0.1) + 1(0.1) + 0 + 1(0.2) + 4(0.3) + 9(0.1)$$E(X^2) = 0.4 + 0.1 + 0 + 0.2 + 1.2 + 0.9 = 2.8$.The variance is given by $Var(X) = E(X^2) - [E(X)]^2$.$Var(X) = 2.8 - (0.8)^2 = 2.8 - 0.64 = 2.16$.

$X = x_{i}$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x_{i})$	$0.1$	$k$	$0.2$	$2k$	$3k$	$k$

$X$	$0$	$1$	$2$
$P(X)$	$0.4$	$0.4$	$0.2$

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

$A$ random variable $X$ has the following distribution.
$X = x_{i}$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X = x_{i})$ $0.1$ $k$ $0.2$ $2k$ $3k$ $k$

Then the variance of this distribution is

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$A$ random variable $X$ has the following distribution.$X = x_{i}$$-2$$-1$$0$$1$$2$$3$$P(X = x_{i})$$0.1$$k$$0.2$$2k$$3k$$k$Then the variance of this distribution is