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Q. For the following probability distribution:

X $4$ $3$ $2$ $1$ $0$

P(X) $0.1$ $0.2$ $0.3$ $0.2$ $0.2$

$E(X)$ is equal to

X	$4$	$3$	$2$	$1$	$0$
P(X)	$0.1$	$0.2$	$0.3$	$0.2$	$0.2$

Probability - Part 2

$0$

14%

$-1$

29%

$-2$

14%

$-1.8$

43%

Solution:

$E\left(X\right) =\Sigma XP\left(X\right)$
$ = \left(-4\right)\times0.1+\left(-3\right)\times0.2 +\left(-2\right) \times0.3 +\left(-1\right) \times0.2+0\times0.2 $
$ =-0.4 - 0.6 - 0.6 - 0.2 = -1.8 $