Find the mean deviation about the mean of the distribution: | Size | 20 | 21 | 22 | 23 | 24 | |---|---|---|---|---|---| | Frequency | 6 | 4 | 5 | 1 | 4 |

Find the mean deviation about the mean of the dist – Class 11 Mathematics

NCERT Exemplar · Q1

Find the mean deviation about the mean of the distribution:

Size	20	21	22	23	24
Frequency	6	4	5	1	4

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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)
Economics (measuring income inequality)
Early statistics courses (before introducing variance and standard deviation)

However, it has a limitation: absolute values are mathematically tricky to work with in advanced statistics (they aren't differentiable at zero). That's why standard deviation (which squares the deviations) is more common in higher-level work.

Quick Summary

Concept	Meaning
Mean deviation about mean	Average absolute distance from the mean
Formula	$\frac{1}{n} \sum
Tells you	How spread out the data is, in the same units as the data
Key property	Always non-negative; zero only if all values are identical

Final answer: Mean deviation about mean is the average of the absolute differences between each data point and the arithmetic mean. For the set $\{4,6,8,10,12\}$ , it equals 2.4.

Mean Deviation about Mean is one of the measures of dispersion covered in the NCERT Class 11 Mathematics chapter on Statistics, matching searches like "mean deviation: formula and examples" or "statistics important questions class 11 maths". It's a regularly tested, calculation-based topic in CBSE boards, and understanding it also builds the intuition needed for standard deviation and variance questions in JEE Main and CET exams.

For this distribution the mean is $\bar{x}=21.65$ and the mean deviation about the mean is $1.25$ .

What we are finding

Mean deviation about the mean measures, on average, how far each value sits from the mean. For a frequency distribution it is

\text{M.D.}(\bar{x}) = \frac{\sum f_i\,\lvert x_i - \bar{x}\rvert}{\sum f_i}.

Step 1 — Total frequency and mean

N = \sum f_i = 6+4+5+1+4 = 20

\sum f_i x_i = 20(6)+21(4)+22(5)+23(1)+24(4) = 120+84+110+23+96 = 433

\bar{x} = \frac{433}{20} = 21.65

Step 2 — Absolute deviations, weighted by frequency

$x_i$	$f_i$	$\lvert x_i-\bar{x}\rvert$	$f_i\lvert x_i-\bar{x}\rvert$
20	6	1.65	9.90
21	4	0.65	2.60
22	5	0.35	1.75
23	1	1.35	1.35
24	4	2.35	9.40
Total	20		25.00

Step 3 — Divide by $N$

\text{M.D.}(\bar{x}) = \frac{\sum f_i\lvert x_i-\bar{x}\rvert}{N} = \frac{25.00}{20} = 1.25

✓Final answer

=21.65

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQ
$3, 10, 10, 4, 7, 10, 5$
›Reveal solutionSolution

Mean $=7$ ; $\frac{1}{7}\sum|x_i-7| = \frac{18}{7} \approx 2.57$ .

Data: $3, 10, 10, 4, 7, 10, 5$ ( $n = 7$ ).

Mean $\bar{x} = \dfrac{3+10+10+4+7+10+5}{7} = \dfrac{49}{7} = 7$ .

Absolute deviations $|x_i - 7|$ : $4, 3, 3, 3, 0, 3, 2$ .

Sum $= 4+3+3+3+0+3+2 = 18$ .

Mean deviation $= \dfrac{18}{7} \approx 2.57$ .

✓Final answer
$2.57$
AHSEC Higher Secondary (HS) 1st Year Examination 2018Set ANNUAL1 mark
$Q. Define mean deviation.$
›Reveal solutionSolution

Mean deviation is the average of the absolute deviations of the data from a central value (mean or median).

For a set of observations $x_1,x_2,\ldots,x_n$ with a chosen central value $a$ (usually the mean $\bar x$ or the median $M$ ), the mean deviation about $a$ is defined as

$\text{MD}(a) = \dfrac{1}{n}\sum_{i=1}^n |x_i-a|.$

We take the absolute value of each deviation $(x_i-a)$ because the ordinary (signed) deviations always sum to zero about the mean, so a signed average would be uninformative; taking absolute values captures the true average spread of the data. For grouped/frequency data,

$\text{MD}(a) = \dfrac{\sum_i f_i|x_i-a|}{\sum_i f_i}.$

✓Final answer
$\text{MD}=\dfrac1n\sum|x_i-\bar x|$

←Previous section · summarySummary →Next questionQ2 — Find the mean deviation about the median of the following distribution…

What we are finding

Step 1 — Total frequency and mean

Step 2 — Absolute deviations, weighted by frequency

Step 3 — Divide by NNN

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Step 3 — Divide by $N$

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