Q.Let $x_1, x_2, \ldots, x_n$ be $n$ observations and $\overline{x}$ be their arithmetic mean. The formula for the standard deviation is given by
(A) $\sum(x_i - \overline{x})^2$
(B) $\dfrac{\sum(x_i - \overline{x})^2}{n}$
(C) $\sqrt{\dfrac{\sum(x_i - \overline{x})^2}{n}}$
(D) $\sqrt{\dfrac{\sum x_i^2}{n} + \overline{x}^2}$

Concept: Effect of Scaling Variance — Standard deviation measures spread in the original units, so it must be the square root of the average squared deviation.

Step 1: Variance is the mean of squared deviations from the mean:

$\sigma^2 = \frac{\sum (x_i - \overline{x})^2}{n}$.

Step 2: Standard deviation is the square root of variance, restoring the original unit of measurement:

$\sigma = \sqrt{\frac{\sum (x_i - \overline{x})^2}{n}}$. …

The key idea here is that standard deviation is a measure of dispersion — how far the individual observations are from the centre (the mean). If you just took the average of the deviations $(x_i - \overline{x})$, the positives and negatives would cancel out, giving zero every time. That’s useless. So instead, we square each deviation to make everything positive, average those squares, and then take the square root to bring the units back to the original scale.

Let’s walk through the options one by one.

1. Option (A): $\sum(x_i - \overline{x})^2$

   This is just the sum of squared deviations. It grows with the number of observations — a dataset with 100 points will almost always have a larger sum than one with 10 points, even if the spread is the same. So this can’t be a proper measure of spread; it’s not standardised for sample size.

2. Option (B): $\dfrac{\sum(x_i - \overline{x})^2}{n}$

   This is the variance — the average of the squared deviations. It fixes the sample-size problem from (A). But the units are now the square of the original units (e.g., if $x_i$ are in cm, variance is in cm²). That’s awkward to interpret directly.

3. Option (C): $\sqrt{\dfrac{\sum(x_i - \overline{x})^2}{n}}$

   This is the standard deviation. By taking the square root of the variance, we return to the original units. This is the standard formula for the population standard deviation (when we have all $n$ observations). It tells you, on average, how far a typical observation lies from the mean.

4. Option (D): $\sqrt{\dfrac{\sum x_i^2}{n} + \overline{x}^2}$

   This one is a trap. Let’s check it algebraically.

   The correct expression inside the square root for standard deviation is $\frac{\sum (x_i - \overline{x})^2}{n}$. Expand that:

Since $\frac{\sum x_i}{n} = \overline{x}$, this becomes: