Q.
For the following probability distribution, the standard deviation of the random variable $X$ is
(X)
$2$
$3$
$4$
P(X)
$0.2$
$0.5$
$0.3$
| (X) | $2$ | $3$ | $4$ |
| P(X) | $0.2$ | $0.5$ | $0.3$ |
Probability - Part 2
Solution:
We have,
$X$
$P(X)$
$XP(X)$
$X^2P(X)$
$2$
$0.2$
$0.4$
$0.8$
$3$
$0.5$
$1.5$
$4.5$
$4$
$0.3$
$1.2$
$4.8$
$\sum XP\left(X\right) = 3.1$
$\sum X^{2}P\left(X\right) = 10.1$
$\mu = \sum XP\left(X\right) = 3.1$
var $X$, $\sigma^{2} = \sum X^{2}P\left(X\right)-\mu^{2} = 10.1-\left(3.1\right)^{2}$
$= 10.1 - 9.61 = 0.49$
$\therefore $ Standard deviation $=\sqrt{\sigma^{2}} = \sqrt{0.49} = 0.7$
| $X$ | $P(X)$ | $XP(X)$ | $X^2P(X)$ |
|---|---|---|---|
| $2$ | $0.2$ | $0.4$ | $0.8$ |
| $3$ | $0.5$ | $1.5$ | $4.5$ |
| $4$ | $0.3$ | $1.2$ | $4.8$ |
| $\sum XP\left(X\right) = 3.1$ | $\sum X^{2}P\left(X\right) = 10.1$ |