# How do you compute the variance of the probability distribution in the table provided?

## Outcome | Probability 50 | 0.5 51 | 0.2 52 | 0.1 53 | 0.2?

Feb 27, 2017

$V a r \left(X\right) = 1.4$

#### Explanation:

Let $X$ be the Random Variable that represents a possible outcome: First we quickly check that $\sum P \left(x\right) = 1$ which is indeed the case.

The, we calculate ${x}^{2}$, $x P \left(x\right)$, and ${x}^{2} P \left(x\right)$: So then the Expectation is calculated using:

$E \left(X\right) = \sum x P \left(x\right)$
$\text{ } = 25 + 10.2 + 5.2 + 10.6$
$\text{ } = 51$

Next prior to calculating the Variance we calculate E(X^2):

$E \left({X}^{2}\right) = \sum {x}^{2} P \left(x\right)$
$\text{ } = 1250 + 520.2 + 270.4 + 561.8$
$\text{ } = 2602.4$

Then we can calculate the variance:

$V a r \left(X\right) = E \left({X}^{2}\right) - {E}^{2} \left(X\right)$
$\text{ } = 2602.4 - {\left(51\right)}^{2}$
$\text{ } = 2602.4 - 2601$
$\text{ } = 1.4$

We can also calculate the Standard Deviation (if required); as

${\sigma}^{2} = V a r \left(X\right) \implies \sigma = 1.18$ (3sf)