The average of six squares, and the side of one.

Variance looks like an arbitrary formula until you draw it. It is the average area of six squares — and the standard deviation is the side of the square that average makes.

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Six values: 2, 4, 4, 5, 7, 8. The mean is the balance point, $\bar x = \tfrac{30}{6} = 5$. The median is the middle, $\tfrac{4+5}{2} = 4.5$, and the mode is 4.

Three answers to one question, and none of them says how spread out the data is.

Measure each value's distance from the mean and build a square on it. Squaring kills the signs — otherwise the distances would cancel to zero every time — and it punishes the far ones hardest.

One value sits exactly on the mean, so its square has no area at all.

Add the areas and divide by how many there are:

The square underneath has that area, and its side is the standard deviation $\sigma = 2$.

Now 2, 2, 2, 8, 8, 8. The mean is still 5 — but every value is now 3 from it, so every square is the same and bigger:

A mean on its own describes almost nothing. Always quote a measure of location and a measure of spread.

Put the first six back and add one value of 12. The mean is dragged a whole unit, from 5 to 6, while the median shuffles only half of one, from 4.5 to 5.

That is why the median and the interquartile range are preferred when the data has extreme values: they barely notice.

One number for the middle, one for the spread.

Quartiles and percentiles, linear interpolation for grouped data, the $\sum x^2$ shortcut, coding, comparing distributions in context, and twelve graded exam questions with full worked solutions are waiting.

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