Variance

Why ordinary deviations do not work

For data values $x_i$ with mean $\bar x$, the deviations $x_i-\bar x$ always sum to zero. Positive and negative distances cancel, so their ordinary average cannot measure spread.

Squaring the deviations makes them nonnegative and gives greater weight to values farther from the mean.

Worked example

If the same values were a sample, the sample variance would be $8/3$.

Units

Variance has squared units. If observations are measured in centimetres, variance is in square centimetres. This makes variance algebraically convenient but less intuitive for direct interpretation.

Standard deviation takes the square root and returns to the original units.

Effect of transformations

If $Y=aX+b$, then

Adding $b$ shifts every value and the mean equally, so spread does not change. Multiplying by $a$ scales deviations by $a$ and squared deviations by $a^2$.

Common mistakes

Averaging unsquared deviations. They cancel to zero.

Using $n$ instead of $n-1$ for an estimating sample. Identify population versus sample.

Forgetting frequency weights. Repeated values contribute repeatedly.

Reporting variance in original units. Its units are squared.

Assuming large variance identifies why values differ. It measures spread, not cause.

Quick self-check

• Is the dataset a full population or a sample estimate?
• Is the correct mean used in every deviation?
• Have deviations been squared and frequencies included?
• Is the denominator $N$ or $n-1$ for the intended purpose?
• Are squared units reported?
• Could outliers or scale make another spread measure useful too?