Mean deviation for $n$ observations $x_1, x_2, \ldots, x_n$ from their mean $\overline{x}$ is given by (A) $\sum_{i=1}^{n}(x_i - \overline{x})$ (B) $\dfrac{1}{n}\sum_{i=1}^{n}\left|x_i - \overline{x}\right|$ (C) $\sum_{i=1}^{n}(x_i - \overline{x})^2$ (D) $\dfrac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2$

> [!TLDR] > Mean deviation about the mean measures the average absolute distance of observations from their mean; it is the arithmetic mean of the absolute deviations, so **option (B)**. ## Why mean deviation uses absolute values When we want to measure how spread out a dataset is, the natural first thought is to look at how far each observation sits from the center (the mean). If we compute $x_i - \overline{x}$ for each observation, we get the *deviation* of that point from the mean. But here's the problem: some deviations are positive (observations above the mean) and some are negative (observations below the mean). If we simply add them up, the positives and negatives cancel out perfectly — in fact, $\sum_{i=1}^{n}(x_i - \overline{x}) = 0$ always, by the very definition of the mean. That tells us nothing about spread. To capture the *magnitude* of deviation without letting signs cancel, we have two main strategies: 1. **Square the deviations** (leading to variance and standard deviation) 2. **Take absolute values** (leading to mean deviation) Mean deviation takes the second route: it measures the average of the absolute distances from the mean. --- ## Step-by-step reasoning **1. Start with deviations** For each observation $x_i$, compute its deviation from the mean: $$ x_i - \overline{x} $$ **2. Remove the sign by taking absolute value** To prevent cancellation, we take: $$ |x_i - \overline{x}| $$ This gives the *distance* of $x_i$ from $\overline{x}$, always non-negative. **3. Sum all absolute deviations** Add these distances across all $n$ observations: $$ \sum_{i=1}^{n} |x_i - \overline{x}| $$ **4. Average them** To get the *mean* deviation, divide by $n$: $$ \text{Mean Deviation} = \frac{1}{n}\sum_{i=1}^{n} |x_i - \overline{x}| $$ This is the average absolute deviation from the mean — a direct, intuitive measure of spread. --- ## Eliminating the other options …

Mean deviation for n observations x₁, x₂, …, xₙ fr – Class 11 Mathematics

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Concept understanding — Mean Deviation About Mean

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.

Note

Mean deviation is a measure of spread or dispersion. It tells you how scattered the data is around the central value. A small mean deviation means most data points are close to the mean; a large one means they are spread out.

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:

\text{MD}(\bar{x}) = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|

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

\text{Mean Deviation about Mean} = \frac{\sum |x_i - \bar{x}|}{n}

Step-by-Step Calculation

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

Step 1: Find the mean.

\bar{x} = \frac{4+6+8+10+12}{5} = \frac{40}{5} = 8

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.

4 + 2 + 0 + 2 + 4 = 12

Step 4: Divide by $n = 5$ .

\text{MD} = \frac{12}{5} = 2.4

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

Tip

Notice that the mean deviation is always less than or equal to the standard deviation (another measure of spread). For this data, standard deviation is about 2.83, which is larger than 2.4. This is because standard deviation squares deviations, giving more weight to extreme values.

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.

Watch out

A common mistake: forgetting to take absolute values and getting zero. Always check: if your sum of deviations is zero, you forgot the absolute value.

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) …

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Mean deviation about the mean measures the average absolute distance of observations from their mean; it is the arithmetic mean of the absolute deviations, so option (B).

Why mean deviation uses absolute values

When we want to measure how spread out a dataset is, the natural first thought is to look at how far each observation sits from the center (the mean). If we compute $x_i - \overline{x}$ for each observation, we get the deviation of that point from the mean.

But here's the problem: some deviations are positive (observations above the mean) and some are negative (observations below the mean). If we simply add them up, the positives and negatives cancel out perfectly — in fact, $\sum_{i=1}^{n}(x_i - \overline{x}) = 0$ always, by the very definition of the mean. That tells us nothing about spread.

To capture the magnitude of deviation without letting signs cancel, we have two main strategies:

Square the deviations (leading to variance and standard deviation)
Take absolute values (leading to mean deviation)

Mean deviation takes the second route: it measures the average of the absolute distances from the mean.

Step-by-step reasoning

1. Start with deviations

For each observation $x_i$ , compute its deviation from the mean:

x_i - \overline{x}

2. Remove the sign by taking absolute value

To prevent cancellation, we take:

|x_i - \overline{x}|

This gives the distance of $x_i$ from $\overline{x}$ , always non-negative.

3. Sum all absolute deviations

Add these distances across all $n$ observations:

\sum_{i=1}^{n} |x_i - \overline{x}|

4. Average them

To get the mean deviation, divide by $n$ :

\text{Mean Deviation} = \frac{1}{n}\sum_{i=1}^{n} |x_i - \overline{x}|

This is the average absolute deviation from the mean — a direct, intuitive measure of spread.

Eliminating the other options …

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Why mean deviation uses absolute values

Step-by-step reasoning

Eliminating the other options …

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