When a question summarises x − a, first treat those shifted values as a sample. Find their mean and add a to recover the original sample mean. For the unbiased estimate of the population variance, subtract the squared-sum correction and divide by n − 1. A constant shift leaves the variance unchanged.
The 2027 A-Level H2 Mathematics 9758 syllabus includes unbiased estimates from shifted summary data. This guide shows why the mean changes when you shift the data back and why the variance does not.
- Define the shifted variable and identify the sample size.
- Add the shift back to the mean.
- Calculate the corrected sum of squares before dividing by n − 1.
- Give variance in squared units.
CHAPTER 1 OF 4 · Back to contents
Read the two sums carefully
Write u = x − a. The given summaries Σu and Σu² are different quantities. The first adds the shifted values, which can be positive or negative. The second adds their individual squares.
The corrected sum of squares is Σu² − (Σu)²/n. This measures squared distances from the sample mean. Simply adding the squares of u measures squared distances from the chosen reference a on the original scale.
Shifting every value by the same constant shifts its mean by that constant as well. Each distance from the mean stays the same, so the corrected sum of squares is unchanged.
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Worked example | Five measurements centred around 50
Five lengths in centimetres have Σ(x − 50) = 10 and Σ(x − 50)² = 36. Find unbiased estimates of the population mean and variance.
| Quantity | Working | Result |
|---|---|---|
| Mean of x − 50 | 10 ÷ 5 | 2 cm |
| Mean of x | 50 + 2 | 52 cm |
| Corrected sum of squares | 36 − 10²/5 | 16 cm² |
| Unbiased variance estimate | 16 ÷ 4 | 4 cm² |
The shifted values average 2 cm above the reference, so the estimate of the mean is 52 cm. For variance, the mean correction removes 20 cm² from the supplied 36 cm², leaving 16 cm². Divide by 5 − 1 to obtain 4 cm².
To check with raw data, one sample matching the summaries is 50, 50, 52, 54 and 54. Its deviations from 52 are −2, −2, 0, 2 and 2. Their squares add to 16, which confirms the corrected sum of squares.
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CHAPTER 3 OF 4 · Back to contents
Practise with a negative shifted sum
For six values with Σ(x − 20) = −6 and Σ(x − 20)² = 24, the shifted mean is −1, so the original mean is 19.
The corrected sum of squares is 24 − (−6)²/6 = 18. Divide by 5 to get an unbiased variance estimate of 3.6 in squared units.
The negative sign matters for the mean, while the correction uses the square of the shifted sum. Write the brackets before entering a negative value on a calculator.
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Frequently asked questions
When can the correction term vanish?
It vanishes when the shifted sum is zero. Then the reference value equals the sample mean.
Should I add 50 to the variance?
A constant shift changes the mean and leaves all distances from the mean unchanged. The variance therefore stays the same.
Is the square root of the variance estimate also unbiased?
The square root estimates the population standard deviation. The unbiasedness claim here applies to the variance estimate.
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