Using variance

The formulas can also be run backwards: to find missing values, to fix mistakes, and to see what happens when every value is changed.

Changing every value.

• Add a number a to every value: the mean goes up by a, but every deviation stays the same, so the variance and SD do not change. (Moving the whole class up the number line does not change how spread out it is.)
• Multiply every value by k: the mean is multiplied by k, the variance by k², and the SD by |k|.
• Example: 15 values have variance 4. Multiply each by 3: the new variance is 9 × 4 = 36 and the new SD is 3 × 2 = 6.

Two facts to set up equations. If n values have mean x̄ and variance σ²:

• Σx = n x̄.
• Σx² = n(σ² + x̄²), from σ² = Σx²/n – x̄².

Missing values. Five values have mean 6 and variance 2.8; three of them are 4, 6, 6. Call the other two x and y.

• Σx = 5 × 6 = 30, so x + y = 30 – 16 = 14.
• Σx² = 5(2.8 + 36) = 194, so x² + y² = 194 – (16 + 36 + 36) = 106.
• (x + y)² = 196 gives 2xy = 90; then (x – y)² = 106 – 90 = 16, so x – y = ±4.
• The values are 9 and 5.

A value copied wrongly. 20 values have mean 10 and SD 2, but one value 12 was copied as 8.

• Wrong Σx = 20 × 10 = 200. Correct Σx = 200 – 8 + 12 = 204, so the correct mean is 10.2.
• Wrong Σx² = 20(4 + 100) = 2080. Correct Σx² = 2080 – 64 + 144 = 2160.
• Correct variance = 2160/20 – 10.2² = 108 – 104.04 = 3.96, so the correct SD ≈ 1.99.

The method is always the same: rebuild Σx and Σx² from the given mean and SD, correct them, then recompute. Fix the mean first, because the variance formula needs the corrected mean.

Removing a value. 10 values have mean 15, so Σx = 150. Remove the value 24: the other 9 add to 126 and have mean 14.