Variance and standard deviation

${\sigma }^2=\frac{\sum (x-\bar{x}{)}^2}{n}\sigma =\sqrt{{\sigma }^2}$

Example. $12;15;17;18;20;22;25;31$

• Mean: $\bar{x}=\frac{160}{8}=20$
• Deviations $x-\bar{x}$: $-8;-5;-3;-2;0;2;5;11$
• Squared deviations: $64;25;9;4;0;4;25;121$, with sum 252

${\sigma }^2=\frac{252}{8}=31,5\sigma \approx 5,61$

The deviations always add up to 0, which is why they are squared.

In statistics mode:

1. Clear old data.
2. Enter each value (and its frequency if needed).
3. Read $\bar{x}$ and ${\sigma }_x$ (the population standard deviation).

Use ${\sigma }_x$, not $s_x$. The variance is ${\sigma }_x^2$.

Write down what you entered and what you read off. If the calculator answer is wrong, the method shown can still earn marks.

Find the interval $\bar{x}-\sigma \le x\le \bar{x}+\sigma$ and count the values in it.

Example. $\bar{x}-\sigma \approx 14,39$ and $\bar{x}+\sigma \approx 25,61$. The values 15, 17, 18, 20, 22 and 25 lie inside: 6 of the 8, or 75%.

Interpreting. A small standard deviation means the data is close to the mean (consistent). A large one means it is spread out.

• Adding a constant to every value changes the mean but not the standard deviation.
• Removing an extreme value usually lowers the standard deviation.