A random variable X has the following probabi | vedclass

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(B) The mean $E(X)$ is calculated as $\sum x_i P(x_i) = (1 	imes 0.1) + (2 	imes 0.2) + (3 	imes 0.3) + (4 	imes 0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.

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$3$ and $1$

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(B) The mean $E(X)$ is calculated as $\sum x_i P(x_i) = (1 	imes 0.1) + (2 	imes 0.2) + (3 	imes 0.3) + (4 	imes 0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0$.The variance $Var(X)$ is calculated as $E(X^2) - [E(X)]^2$.$E(X^2) = \sum x_i^2 P(x_i) = (1^2 	imes 0.1) + (2^2 	imes 0.2) + (3^2 	imes 0.3) + (4^2 	imes 0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10.0$.$Var(X) = 10.0 - (3.0)^2 = 10.0 - 9.0 = 1.0$.The standard deviation $\sigma = \sqrt{Var(X)} = \sqrt{1.0} = 1.0$.Thus,the mean is $3$ and the standard deviation is $1$.

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.1$	$0.2$	$0.3$	$0.4$

Weather	Clear	Fair	Dubious	Rainy
Probability	$0.50$	$0.30$	$0.15$	$0.05$

$A$ random variable $X$ has the following probability distribution:
$X=x$ $1$ $2$ $3$ $4$
$P(X=x)$ $0.1$ $0.2$ $0.3$ $0.4$

The mean and standard deviation of $X$ are respectively:

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$A$ random variable $X$ has the following probability distribution:$X=x$$1$$2$$3$$4$$P(X=x)$$0.1$$0.2$$0.3$$0.4$The mean and standard deviation of $X$ are respectively: