Q.The sum of the squares of the deviations of the values of the variable is _______ when taken about their arithmetic mean.

The key idea is that the arithmetic mean minimises the sum of squared deviations — this is a fundamental property of the mean as a measure of central tendency.

Reasoning:

1. For any set of values $x_1, x_2, \dots, x_n$, consider the function $f(a) = \sum (x_i - a)^2$.
2. Differentiating with respect to $a$ and setting to zero gives $-2\sum (x_i - a) = 0 \implies a = \frac{\sum x_i}{n} = \bar{x}$. …

Why This Property Matters

The arithmetic mean isn't just an average — it's the unique point that minimises the total squared distance from all data points. This is why the mean is used in regression, variance calculations, and optimisation: it gives the "centre" in a least-squares sense.

Think of it physically: if you have weights placed on a number line at each data value, the mean is the balance point where the sum of squared distances is smallest. Any other point — median, mode, or a random guess — will always give a larger total.

Step-by-Step Proof

1. Set up the problem

Let $x_1, x_2, \dots, x_n$ be $n$ observations. Their arithmetic mean is $\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i$.

We want to show that for any other value $a \neq \bar{x}$, the sum of squared deviations is larger:

2. Express the sum for a general $a$

Write the sum of squares about any point $a$:

Expand this:

3. Complete the square in $a$

Notice this is a quadratic in $a$:

Complete the square:

But $\frac{\sum x_i}{n} = \bar{x}$, so:

4. Identify the minimum

The term $n(a - \bar{x})^2$ is always non-negative, and equals zero only when $a = \bar{x}$. The remaining terms $-n\bar{x}^2 + \sum x_i^2$ are constant (they don't depend on $a$).

Therefore:

$S(a) = \underbrace{n(a - \bar{x})^2}_{\geq 0} + \underbrace{\left(\sum x_i^2 - n\bar{x}^2\right)}_{\text{constant}}$ …