A center is only half the story: two classes can have the same mean score while one is tightly bunched and the other spread from 40 to 100. In Grade 6 you measured spread with the mean absolute deviation, the average distance from the mean. The standard deviation is its more powerful cousin, and it needs square roots.
Key idea
The sample standard deviation is a typical distance of the values from their mean:
$$s = \sqrt{\dfrac{\sum (x - \bar{x})^2}{n - 1}}.$$
Its square, $s^2$, is the variance. A larger $s$ means the data are more spread out; $s = 0$ only when every value is the same.
Why it works
Deviations $x - \bar{x}$ always add up to 0 (the mean is the balance point), so we cannot just average them. Squaring makes them all positive, and squares have algebra that absolute values lack, which makes the standard deviation the measure used throughout statistics. Squaring also changes the units (minutes become square minutes), so the last step takes a square root to return to the original units. We divide by $n - 1$ rather than $n$ because the deviations are measured from $\bar{x}$, which was fit to this same data; that makes them a little too small on average, and $n - 1$ corrects for it. (For a whole population you divide by $N$ and write $\sigma$.)
Worked example: Step by step
Find $s$ for the data $13, 21, 21, 22, 23$.
- Mean: $\bar{x} = 100 \div 5 = 20$.
- Deviations: $-7, 1, 1, 2, 3$ (check: they add to 0).
- Squares: $49, 1, 1, 4, 9$; their sum is $64$.
- Divide by $n - 1 = 4$: $s^2 = 16$.
- Square root: $s = \sqrt{16} = 4$.
Answer: $s = 4$
Worked example: Variance to standard deviation
A data set has variance $s^2 = 2.25$ square inches. What is its standard deviation, and in what units?
- Standard deviation is the square root of the variance: $\sqrt{2.25} = 1.5$.
- Taking the square root also undoes the squared units.
Answer: $s = 1.5$ inches.
Common mistake
Two common slips. (1) Stopping at the variance: $s^2 = 16$ is not the standard deviation; take the square root to get $s = 4$. (2) Confusing $s$ with the range. The range (max − min $= 10$ here) uses only two values; $s$ uses all of them and is usually much smaller than the range. Check: $s$ should be a "typical distance" from the mean, so it should look like the size of a typical deviation.
In the first example the mean absolute deviation is $(7 + 1 + 1 + 2 + 3)/5 = 2.8$, smaller than $s = 4$. Squaring gives extra weight to the one far value (the 13). For bell-shaped data, about two thirds of the values lie within one standard deviation of the mean.