Standard Deviation

Close to the mean, or far from it

The values 4, 5 and 6 and the values 1, 5 and 9 have the same mean, 5. In the first set every value is within 1 of the mean. In the second, two of the values are 4 away from it.

The standard deviation is the most widely used measure of how far values sit from their mean. It is small when the values are close to the mean and larger when they are spread out, so it is smaller for 4, 5 and 6 than for 1, 5 and 9.

The values 4, 5 and 6 sit close together around their mean, 5.

On the same scale, 1, 5 and 9 have the same mean, 5, and spread much further from it.

Square, average, then take the square root

The differences from the mean cannot simply be averaged, because they always add up to 0. The mean absolute deviation drops their signs. The standard deviation squares them instead: a square is never negative, so the squares cannot cancel.

For 1, 5 and 9, the differences from 5 are −4, 0 and +4, and their squares are 16, 0 and 16. The mean of the squares is $\left(16+0+16\right)÷3=32÷3\approx 10.67$(16 + 0 + 16) ÷ 3 = 32 ÷ 3 ≈ 10.67. This mean of the squared differences is called the variance.

Finally, take the square root of the variance: $\sqrt{32÷3}\approx 3.27$√(32 ÷ 3) ≈ 3.27. That is the standard deviation. For 4, 5 and 6, the differences are −1, 0 and +1, the squares are 1, 0 and 1, the variance is $2÷3\approx 0.67$2 ÷ 3 ≈ 0.67, and the standard deviation is $\sqrt{2÷3}\approx 0.82$√(2 ÷ 3) ≈ 0.82, much smaller, as the dot plots suggest.

The standard deviation is written $\sigma$σ, the Greek small letter sigma. In symbols, $\sigma =\sqrt{\frac{\sum {\left(x-\bar{x}\right)}^2}{n}}$σ = √(Σ(x − x̄)²/n): square each difference from the mean, add up the squares, divide by the number of values, and take the square root.

Why take the square root?

Squaring changes the units. If the values are lengths in centimeters, each squared difference is in square centimeters, and so is the variance. The square root brings the answer back to centimeters, the units of the data, so the standard deviation can be set beside the mean and the values themselves.

A standard deviation is 0 only when every value is the same, because then every difference from the mean is 0. It is never negative, and it is never more than the range.

The standard deviation and the MAD on one data set

Five seedlings grow 2, 6, 8, 10 and 14 millimeters in a week. The total is 40 mm, so the mean is 40 ÷ 5 = 8 mm. The differences from 8 are −6, −2, 0, +2 and +6.

For the mean absolute deviation, drop the signs: the distances 6, 2, 0, 2 and 6 total 16, and 16 ÷ 5 = 3.2 mm. For the standard deviation, square the differences instead: 36, 4, 0, 4 and 36 total 80, the variance is 80 ÷ 5 = 16 square millimeters, and $\sigma =\sqrt{16}=4$σ = √16 = 4 mm.

Both measure how far the growths typically are from the mean, and they give different answers, 3.2 mm and 4 mm. The standard deviation is the larger. It always is at least as large as the mean absolute deviation, because squaring gives more weight to the values furthest from the mean.

The distances total 16, so the mean absolute deviation is 16 ÷ 5 = 3.2 mm. The squares total 80, so the variance is 80 ÷ 5 = 16 and the standard deviation is $\sqrt{16}=4$√16 = 4 mm.

Why square?

The first reason is that squaring weights values far from the mean more heavily. The values 3, 3, 7 and 7 have mean 5, and every value is 2 from it, so the mean absolute deviation is 2 and the standard deviation is $\sqrt{\left(4+4+4+4\right)÷4}=\sqrt{4}=2$√((4 + 4 + 4 + 4) ÷ 4) = √4 = 2. The values 1, 5, 5 and 9 also have mean 5, and their distances 4, 0, 0 and 4 also average 2. But their squares are 16, 0, 0 and 16, so the standard deviation is $\sqrt{32÷4}=\sqrt{8}\approx 2.83$√(32 ÷ 4) = √8 ≈ 2.83.

The two sets have the same mean absolute deviation, but the second has its spread in two values far from the mean, and the standard deviation is larger for it. A value twice as far from the mean adds four times as much to the variance: $4^2=16$4² = 16 against $2^2=4$2² = 4.

The second reason is that squares are easier to work with in algebra. The distance |x − x̄| has to be worked out one way for a value above the mean and another way for a value below it, while the square ${\left(x-\bar{x}\right)}^2$(x − x̄)² is one expression that holds for every value and can be expanded like any bracket. Expanding it gives a shorter way to calculate the standard deviation, which is the subject of the next lesson.

The values 3, 3, 7 and 7: every value is 2 from the mean of 5, so the standard deviation is 2.

The values 1, 5, 5 and 9 have the same mean and the same mean absolute deviation, 2, but the two values 4 from the mean raise the standard deviation to $\sqrt{8}\approx 2.83$√8 ≈ 2.83.

The values 3, 5, 7, 9 and one more, each with a square drawn on its distance from the mean: the area of each square is that value’s squared difference. The figure writes the mean as $\mu$μ, another common symbol for it. Drag the last value away from the others. Doubling its distance from the mean doubles the bar under the line and makes its square four times as large, so one far value soon outweighs the other four together.