A random variable X has the probability distr | vedclass

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(B) For a probability distribution, the sum of all probabilities must be $1$:$\Sigma P(X=x) = K + 2K + 3K + 2K + K = 9K = 1$$	herefore K = \frac{1}{9

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$\frac{4}{3}$

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(B) For a probability distribution, the sum of all probabilities must be $1$:$\Sigma P(X=x) = K + 2K + 3K + 2K + K = 9K = 1$$	herefore K = \frac{1}{9}$The mean $E(X) = \Sigma x_i P(x_i) = (1 	imes K) + (2 	imes 2K) + (3 	imes 3K) + (4 	imes 2K) + (5 	imes K)$$= K + 4K + 9K + 8K + 5K = 27K = 27 	imes \frac{1}{9} = 3$$E(X^2) = \Sigma x_i^2 P(x_i) = (1^2 	imes K) + (2^2 	imes 2K) + (3^2 	imes 3K) + (4^2 	imes 2K) + (5^2 	imes K)$$= K + 8K + 27K + 32K + 25K = 93K = 93 	imes \frac{1}{9} = \frac{93}{9} = \frac{31}{3}$Variance $Var(X) = E(X^2) - [E(X)]^2$$= \frac{31}{3} - (3)^2 = \frac{31}{3} - 9 = \frac{31 - 27}{3} = \frac{4}{3}$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

$X = x$	$0$	$1$	$2$	$3$	$4$
$P(X = x)$	$k$	$2k$	$4k$	$2k$	$k$

$A$ random variable $X$ has the probability distribution given below. Its variance is:
$X$ $1$ $2$ $3$ $4$ $5$
$P(X=x)$ $K$ $2K$ $3K$ $2K$ $K$

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$A$ random variable $X$ has the probability distribution given below. Its variance is:$X$$1$$2$$3$$4$$5$$P(X=x)$$K$$2K$$3K$$2K$$K$