TheoremProved

The Deviations from the Mean Sum to Zero

Statement

For any data set $x_1, x_2, \ldots, x_n$ with mean $\bar x$ , $\sum_{i=1}^{n}(x_i-\bar x) = 0$ .

Why is it true?

The mean is defined precisely as the balance point of the data — think of a see-saw with weights at each $x_i$ — so the values above the mean must exactly offset the values below it, or else $\bar x$ would not be the balance point.

Proof sketch

Expand the sum by distributing: $\sum_{i=1}^{n}(x_i-\bar x) = \sum_{i=1}^{n}x_i - \sum_{i=1}^{n}\bar x$ . The second sum adds the constant $\bar x$ to itself $n$ times, so $\sum_{i=1}^{n}\bar x = n\bar x$ .

By definition, $\bar x = \frac{1}{n}\sum_{i=1}^{n}x_i$ , so multiplying both sides by $n$ gives $n\bar x = \sum_{i=1}^{n}x_i$ . This is exactly the same quantity as the first sum in the expansion above.

Substituting back, $\sum_{i=1}^{n}(x_i-\bar x) = \sum_{i=1}^{n}x_i - n\bar x = \sum_{i=1}^{n}x_i - \sum_{i=1}^{n}x_i = 0$ , which proves the claim.

Topics that use this theorem

Descriptive statistics

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

David Freedman, Robert Pisani, Roger Purves (2007). Statistics
David S. Moore, William I. Notz (2020). The Basic Practice of Statistics