Variance

Two archers each fire five arrows. Both average dead-centre — same mean. But one groups every arrow in a tight cluster, while the other sprays them all over the target. The averages can't tell them apart; only their spread can. The range is one way to measure that spread — but the range looks only at the two most extreme arrows, so a single wild shot wrecks it and the other eight arrows have no say at all.

Variance does something far more honest: it asks how far every data point sits from the mean, and averages those distances. Every point votes. It is the workhorse measure of spread — the one nearly all of statistics is quietly built on.

For data x_1, x_2, \dots, x_n with mean \bar{x}:

\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}\left(x_i - \bar{x}\right)^2.

Each term (x_i - \bar{x}) is a deviation — how far that one point strays from the centre. Variance is the average squared deviation.

The recipe, in four steps

Every variance is cooked exactly the same way:

Find the mean \bar{x} of the data.
For each value, take its deviation x_i - \bar{x}.
Square every deviation.
Average the squares.

That average of squares is the variance. The two ideas that make it work — why we square, and how big deviations get punished — are worth pausing on.

Why square? Because raw deviations cancel

It is tempting to just average the deviations themselves. But that always gives zero — the mean sits exactly at the balance point, so the points below it cancel the points above:

\sum_{i=1}^{n}\left(x_i - \bar{x}\right) = 0 \quad\text{always.}

Squaring fixes the sign: a deviation of -3 and one of +3 both contribute 9 instead of cancelling. Squaring also punishes big misses — a point twice as far away contributes four times as much — so variance is especially sensitive to far-flung values.

Worked example: five test scores

Five students scored 2,\ 4,\ 4,\ 5,\ 5 out of ten. Let's find the variance step by step.

Step 1 — the mean.

\bar{x} = \frac{2+4+4+5+5}{5} = \frac{20}{5} = 4.

Step 2 & 3 — deviations and their squares.

\begin{array}{c|c|c} x_i & x_i-\bar{x} & (x_i-\bar{x})^2 \\ \hline 2 & -2 & 4 \\ 4 & 0 & 0 \\ 4 & 0 & 0 \\ 5 & 1 & 1 \\ 5 & 1 & 1 \end{array}

Step 4 — average the squares.

\sigma^2 = \frac{4+0+0+1+1}{5} = \frac{6}{5} = 1.2.

Notice the two scores that are the mean contribute nothing — a zero deviation squares to zero. Only distance from the centre counts.

A second example: spot the wider set

Compare \{4, 5, 6\} with \{1, 5, 9\}. Both have mean 5, so the mean can't separate them — but the variance can.

\sigma^2_{\{4,5,6\}} = \frac{(-1)^2+0^2+1^2}{3} = \frac{2}{3} \approx 0.67,

\sigma^2_{\{1,5,9\}} = \frac{(-4)^2+0^2+4^2}{3} = \frac{32}{3} \approx 10.7.

The second set's variance is sixteen times larger, even though its points are only four times farther out — that is the squaring, magnifying the gap and flagging the wider set loud and clear.

Why square deviations rather than take their absolute value |x_i - \bar{x}|, which also kills the sign? Two quiet reasons. First, x^2 is smooth — you can differentiate it, and |x| has a sharp corner at zero that calculus hates. Second, it turns out the mean is exactly the value that minimises the total squared deviation: no other centre makes the sum of squares smaller. That "minimise the sum of squares" idea is the seed of least-squares fitting, which runs straight lines through scattered data all across science. Variance isn't an arbitrary choice of formula — it is the natural companion of the mean.

See it: variance is the average square

The five points share a mean of 7 (the dashed line). Above each point sits a square whose side is that point's distance from the mean, so its area is the squared deviation (x_i - \bar{x})^2. Drag the slider to spread the points out: the squares grow, and the reported variance — their average area — climbs with them.

Notice the point sitting on the mean has a square of side zero: it adds nothing. Only distance from the centre counts.

Population variance vs sample variance

There are two variances, and they differ only in what you divide by. When your data are the whole story — every pupil in the class, every planet in the system — you have a population, and you divide by n:

\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}\left(x_i - \bar{x}\right)^2 \quad\text{(population)}.

But usually your data are just a sample — a handful drawn from a much larger group you can't measure in full — and you want to estimate the spread of that larger group. Then you divide by n-1 instead:

s^2 = \frac{1}{n-1}\sum_{i=1}^{n}\left(x_i - \bar{x}\right)^2 \quad\text{(sample)}.

The one-line reason: a sample's own mean is nudged toward the sample's own points, so measuring spread from it under-counts the true spread — dividing by the smaller n-1 nudges the estimate back up to compensate.

The units are squared too

Because we squared the deviations, the variance carries squared units. If the data are heights in centimetres, the variance is in \text{cm}^2 — a slightly awkward quantity to interpret. That single inconvenience is exactly what the standard deviation is built to repair.

Variance is the average squared deviation from the mean: \sigma^2 = \dfrac{1}{n}\sum (x_i - \bar{x})^2.
Raw deviations always sum to zero; squaring removes the sign so they no longer cancel, and makes large deviations count disproportionately.
Its units are the square of the data's units — the reason the standard deviation takes a square root.
A larger variance means the data are more spread out around the mean; a variance of 0 means every value equals the mean.

The commonest trap with variance is forgetting its units are squared. Measure heights in centimetres and the variance comes out in \text{cm}^2 — square centimetres — which is nonsense to picture: nobody is "1.2 square centimetres tall". A variance of 36\ \text{cm}^2 does not mean the data span 36 cm. That's why variance is rarely reported on its own; instead people take its square root, the standard deviation, which lands back in plain centimetres and can sit right next to the original numbers.

It looks like a typo. To average five squared deviations, surely you divide by five? For a sample, you divide by four — and this quiet n-1, called Bessel's correction, has puzzled students for two centuries. Here's the intuition. To measure spread you need a centre to measure from — but you don't know the true centre of the big population, so you use the sample's own mean. That mean is computed from the very points you're measuring, so it sits slightly closer to them than the true mean would. The deviations come out a touch too small, and the naïve variance under-estimates the real spread every single time. Shrinking the divisor from n to n-1 inflates the answer by just the right amount to cancel that bias. The "-1" is the price you pay for having spent one piece of your data estimating the mean.