Q.If $\overline{x}$ is the mean of $n$ values of $x$, then $\sum_{i=1}^{n}(x_i - \overline{x})$ is always equal to _______. If $a$ has any value other than $\overline{x}$, then $\sum_{i=1}^{n}(x_i - \overline{x})^2$ is _________ than $\sum(x_i - a)^2$

The key idea is that the sum of deviations from the mean is always zero, and the sum of squared deviations is minimized when taken about the mean (this is the Effect of Scaling Variance — the mean is the point of least squares).

Step 1: For the first blank, use the definition of the mean: $\overline{x} = \frac{1}{n}\sum x_i$.

Then $\sum (x_i - \overline{x}) = \sum x_i - n\overline{x} = n\overline{x} - n\overline{x} = 0$.

Step 2: For the second blank, consider $\sum (x_i - a)^2$. Expand and compare with $\sum (x_i - \overline{x})^2$: …

Why this works — the idea of a "balancing point"

The mean $\overline{x}$ is not just an average — it is the unique point where the positive and negative deviations from it exactly cancel out. That is why $\sum (x_i - \overline{x}) = 0$. For squared deviations, the mean acts as the point of minimum total squared distance — a property that makes it the centre of the "least squares" idea. Any shift away from $\overline{x}$ adds a positive penalty term, so the sum of squares increases.

Step-by-step reasoning

1. First blank: the sum of deviations from the mean

By definition, the mean is

Multiply both sides by $n$:

Now consider the sum of deviations:

Since $\overline{x}$ is a constant, $\sum_{i=1}^{n} \overline{x} = n\overline{x}$. So:

But from the definition, $\sum x_i = n\overline{x}$. Hence:

2. Second blank: comparing sums of squared deviations

We need to compare $S_{\overline{x}} = \sum (x_i - \overline{x})^2$ with $S_a = \sum (x_i - a)^2$ for any $a \neq \overline{x}$.

Start by rewriting $S_a$:

Expand the square:

$S_a = \sum_{i=1}^{n} (x_i - \overline{x})^2 + 2(\overline{x} - a)\sum_{i=1}^{n}(x_i - \overline{x}) + \sum_{i=1}^{n} (\overline{x} - a)^2.$ …