Standard Deviation: Ungrouped

The differences from the mean

Take the three values 2, 4 and 6. Their mean is x̄ = (2 + 4 + 6) ÷ 3 = 12 ÷ 3 = 4. The differences from the mean, x − x̄, are 2 − 4 = −2, 4 − 4 = 0 and 6 − 4 = +2.

The bar for 2 ends 2 below the mean line at 4, the bar for 4 ends on it, and the bar for 6 ends 2 above it.

Square them so they cannot cancel

Added as they are, the differences cancel: −2 + 0 + 2 = 0. They always do, because the mean is the balance point of the values.

Square each difference first: ${\left(-2\right)}^2=4$(−2)² = 4, $0^2=0$0² = 0 and $2^2=4$2² = 4. A square is never negative, so the squares cannot cancel. Their total, 4 + 0 + 4 = 8, could be 0 only if every value were equal to the mean.

The squared differences of 2, 4 and 6 from the mean: 4, 0 and 4. None of them is below 0.

Average the squares, then take the square root

Average the squares: $8÷3=\frac{8}{3}\approx 2.67$8 ÷ 3 = 8/3 ≈ 2.67. That is the variance. Then take its square root: $\sqrt{\frac{8}{3}}\approx 1.63$√(8/3) ≈ 1.63. That is the standard deviation. The square root undoes the squaring and brings the answer back to the units of the data.

In symbols, $\sigma =\sqrt{\frac{\sum {\left(x-\bar{x}\right)}^2}{n}}$σ = √(Σ(x − x̄)²/n). Laid out as a table, there is one row for each value, a column for x − x̄ and a column for ${\left(x-\bar{x}\right)}^2$(x − x̄)², and the total of the last column is divided by n.

The squared differences total 8, so the variance is $\frac{8}{3}$8/3 and $\sigma =\sqrt{\frac{8}{3}}\approx 1.63$σ = √(8/3) ≈ 1.63. The last column, the squares of the values themselves, is used next.

A shorter formula

The definition needs the mean before any square can be worked out, and when the mean is not a whole number every difference is a decimal. Expanding the square gives a formula that avoids this.

Each squared difference expands like any bracket: ${\left(x-\bar{x}\right)}^2=x^2-2\bar{x}x+{\bar{x}}^2$(x − x̄)² = x² − 2x̄x + x̄². Add this up over all n values. The $x^2$x² terms give $\sum x^2$Σx². The middle terms give $2\bar{x}\sum x$2x̄Σx, and since $\sum x=n\bar{x}$Σx = n x̄, that is $2n{\bar{x}}^2$2n x̄². The last term, ${\bar{x}}^2$x̄², is added n times, which gives $n{\bar{x}}^2$n x̄². So $\sum {\left(x-\bar{x}\right)}^2=\sum x^2-2n{\bar{x}}^2+n{\bar{x}}^2=\sum x^2-n{\bar{x}}^2$Σ(x − x̄)² = Σx² − 2n x̄² + n x̄² = Σx² − n x̄².

Dividing by n gives the variance as $\frac{\sum x^2}{n}-{\bar{x}}^2$Σx²/n − x̄², the mean of the squares minus the square of the mean. So $\sigma =\sqrt{\frac{\sum x^2}{n}-{\bar{x}}^2}$σ = √(Σx²/n − x̄²).

Check it on 2, 4 and 6. The squares of the values are 4, 16 and 36, so $\sum x^2=56$Σx² = 56. Then $\frac{\sum x^2}{n}-{\bar{x}}^2=\frac{56}{3}-4^2=\frac{56}{3}-\frac{48}{3}=\frac{8}{3}$Σx²/n − x̄² = 56/3 − 4² = 56/3 − 48/3 = 8/3, the same variance as before, and $\sigma =\sqrt{\frac{8}{3}}\approx 1.63$σ = √(8/3) ≈ 1.63.

When the mean is not a whole number

A team scores 1, 2, 4 and 7 goals in four matches. The mean is 14 ÷ 4 = 3.5 goals. By the definition, the differences from 3.5 are −2.5, −1.5, 0.5 and 3.5, and their squares are 6.25, 2.25, 0.25 and 12.25, which total 21. The variance is 21 ÷ 4 = 5.25.

By the shorter formula, the squares of the values are 1, 4, 16 and 49, whole numbers that total 70. Then $\frac{\sum x^2}{n}-{\bar{x}}^2=70/4-3.5^2=17.5-12.25=5.25$Σx²/n − x̄² = 70 / 4 − 3.5² = 17.5 − 12.25 = 5.25, the same variance. Either way, $\sigma =\sqrt{5.25}\approx 2.29$σ = √5.25 ≈ 2.29 goals.

The two methods always agree. The definition shows what the standard deviation measures; the shorter formula is usually quicker by hand, because it works with the values themselves until the last step.

By the definition, 21 ÷ 4 = 5.25. By the shorter formula, $70/4-3.5^2=17.5-12.25=5.25$70 / 4 − 3.5² = 17.5 − 12.25 = 5.25. The variance is the same.

Mistakes with the shorter formula

$\sum x^2$Σx² is the sum of the squares, not the square of the sum. For 1, 2, 4 and 7, $\sum x^2=1+4+16+49=70$Σx² = 1 + 4 + 16 + 49 = 70, while the square of the sum is $14^2=196$14² = 196.

Divide $\sum x^2$Σx² by n first, and then subtract the square of the mean: 70 ÷ 4 − 12.25, not (70 − 12.25) ÷ 4.

Use the exact mean. The two numbers being subtracted are often close together, so a rounded mean can change the answer a great deal. Rounding the mean 3.5 up to 4 gives 17.5 − 16 = 1.5 in place of 5.25.