Q.The standard deviation is _______ to the mean deviation taken from the arithmetic mean.

The question asks for the relationship between standard deviation and mean deviation about the arithmetic mean.

Both measures quantify dispersion, but they treat deviations differently. Mean deviation uses absolute deviations $|x_i - \bar{x}|$, while standard deviation uses squared deviations $(x_i - \bar{x})^2$ before taking the square root.

For any dataset, squaring amplifies larger deviations more than smaller ones. This makes the root-mean-square (standard deviation) systematically larger than the mean of absolute values (mean deviation).

The theoretical relationship for most distributions: …

Understanding the Relationship

Both standard deviation and mean deviation measure dispersion—how spread out data points are from a central value. But they penalize deviations differently, and that difference creates a fundamental inequality.

Mean deviation uses absolute deviations $|x_i - \bar{x}|$, treating a deviation of 2 units the same as two deviations of 1 unit each. Standard deviation uses squared deviations $(x_i - \bar{x})^2$, which amplifies larger deviations. Squaring before averaging, then taking the square root, always produces a value at least as large as simply averaging the absolute values.

The mathematical heart of this relationship lies in the power mean inequality: for non-negative numbers, the root mean square is always greater than or equal to the mean of absolute values.

Step-by-Step Proof

1. Define the two measures

For a dataset $x_1, x_2, \ldots, x_n$ with mean $\bar{x}$:

2. Apply the Cauchy-Schwarz inequality

Let $d_i = x_i - \bar{x}$ be the deviation of each observation. By Cauchy-Schwarz:

Dividing both sides by $n^2$:

3. Take square roots

Since both sides are non-negative:

This is exactly:

4. Identify when equality holds …