Q.Find the mean deviation from the median, and its coefficient, for the data: $4, 6, 8, 10, 12, 14, 16$.

Mean Deviation: The Intuition

You have a set of numbers — say, the marks of five students in a test: 40, 50, 60, 70, 80. The average (mean) is 60. Now, not every student scored 60. Some are above, some below. The question is: on average, how far away from the centre are these marks?

That is exactly what Mean Deviation measures. It answers: "If I pick any one mark at random, how many units away from the mean should I expect it to be?"

The Problem with Signs

If you simply add up the differences (40 − 60 = −20, 50 − 60 = −10, 60 − 60 = 0, 70 − 60 = +10, 80 − 60 = +20), the positives and negatives cancel: (−20) + (−10) + 0 + 10 + 20 = 0. That tells you nothing about spread — it always gives zero for any symmetric data.

So we need to get rid of the sign. The simplest way: take the absolute value of each deviation. That is the core idea.

The Definition

For a set of $n$ observations $x_1, x_2, \dots, x_n$ with mean $\bar{x}$, the Mean Deviation about the mean is:

You can also compute it about the median instead of the mean — the formula is the same, just replace $\bar{x}$ with the median $M$.

Worked Example

Take the marks: 40, 50, 60, 70, 80.

1. Mean $\bar{x} = 60$.
2. Deviations: |40−60|=20, |50−60|=10, |60−60|=0, |70−60|=10, |80−60|=20.
3. Sum of absolute deviations = 20 + 10 + 0 + 10 + 20 = 60.
4. Mean Deviation = 60 / 5 = 12.

Interpretation: On average, a student's mark is 12 points away from the class average of 60.

Why Not Just Use Standard Deviation?

Mean Deviation is simpler to explain and compute — no squaring, no square roots. But squaring (as in standard deviation) gives more weight to extreme values, which is often desirable in statistics. Mean Deviation treats every deviation equally, which can be both a strength (robustness to outliers) and a weakness (less sensitive to large errors).

For Grouped Data

When data is grouped into classes with frequencies $f_i$ and midpoints $x_i$: …