Q.Find the mean deviation about the mean of the following observations: $4, 7, 8, 9, 10, 12, 13, 17$. OR Find the mean deviation about the median for the following data: $3, 9, 5, 3, 12, 10, 18, 4, 7, 9, 21$.

Mean Deviation About Mean – The Intuition First

Imagine you have a small set of numbers: the marks of five students in a test: 4, 6, 8, 10, 12. The average (mean) is 8. Now, each student is some distance away from this average. The student who scored 4 is 4 marks below the mean; the one who scored 12 is 4 marks above. The student who scored 8 is exactly at the mean.

If you simply add these distances, the positives and negatives cancel out — you get zero. That's not useful. So instead, we ask: on average, how far is each data point from the mean? That's the mean deviation about mean.

The Precise Definition

For a set of $n$ observations $x_1, x_2, \dots, x_n$ with mean $\bar{x}$, the mean deviation about mean (often written as MD or M.D.) is:

That vertical bars mean absolute value — we take the distance without caring about direction. So every deviation is positive.

Step-by-Step Calculation

Let's use the marks example: 4, 6, 8, 10, 12.

Step 1: Find the mean.

Step 2: Find each absolute deviation $|x_i - \bar{x}|$.

| $x_i$ | $x_i - \bar{x}$ | $|x_i - \bar{x}|$ |

|------|----------------|-------------------|

| 4 | -4 | 4 |

| 6 | -2 | 2 |

| 8 | 0 | 0 |

| 10 | 2 | 2 |

| 12 | 4 | 4 |

Step 3: Sum the absolute deviations.

Step 4: Divide by $n = 5$.

So, on average, each student's mark is 2.4 marks away from the mean of 8.

Why Use Absolute Values?

You might wonder: why not just average the plain deviations (without absolute value)? Because the sum of $(x_i - \bar{x})$ is always zero — that's a property of the mean. The absolute value is the simplest way to make all deviations positive so they don't cancel.

When Is This Used?

Mean deviation is intuitive and easy to explain. It's used in:

• Quality control (checking how consistent a manufacturing process is) …