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.

KEAM 2026Set eng-2026-04174 marksMCQ
$Q. The mean of the data : 4, 7, x, 13, 16 is 10. Then the mean deviation of the data is (A) 3 (B) 3.3 (C) 3.6 (D) 3.8 (E) 4$
›Reveal solutionSolution

Find $x$ from the mean, then compute $\dfrac{1}{n}\sum|x_i-\bar x|$ .

Mean $=10$ : $\dfrac{4+7+x+13+16}{5}=10\Rightarrow 40+x=50\Rightarrow x=10$ .

Data: $4,7,10,13,16$ . Absolute deviations from $10$ : $6,3,0,3,6$ , sum $=18$ .

Mean deviation $=\dfrac{18}{5}=3.6$ .

✓Final answer
$The correct option is (C) .$
KEAM 2026Set eng-2026-04184 marksMCQ
$5$
›Reveal solutionSolution

Median is $82$ ; the average absolute deviation from it is $60/10=6$ .

The ordered data $72,72,74,80,82,82,86,88,92,92$ has $10$ values, so the median is the mean of the 5th and 6th: $\tfrac{82+82}{2}=82$ .

Absolute deviations from $82$ : $10,10,8,2,0,0,4,6,10,10$ , summing to $60$ .

Mean deviation $=\dfrac{60}{10}=6$ .

✓Final answer
$The correct option is (B) .$
KEAM 2026Set eng-2026-04194 marksMCQ
$Q. The mean deviation about the mean for the data: 5, 6, 14, 15 is (A) 3.5 (B) 4.5 (C) 4.2 (D) 3.8 (E) 4.0$
›Reveal solutionSolution

Compute the mean, then average the absolute deviations.

Mean $=\dfrac{5+6+14+15}{4}=\dfrac{40}{4}=10$ .

Absolute deviations: $|5-10|=5,\ |6-10|=4,\ |14-10|=4,\ |15-10|=5$ ; sum $=18$ .
$\text{MD}=\frac{18}{4}=4.5.$
✓Final answer
$The correct option is (B) .$
KEAM 2026Set eng-2026-04204 marksMCQ
$x_1,x_2,\ldots,x_n$
›Reveal solutionSolution

Mean deviation $=\dfrac{\sum f_i|x_i-\bar{x}|}{\sum f_i}$ ; solve for $\sum f_i$ .

$10=\dfrac{400}{\sum f_i}$ , so $\sum_{i=1}^{n} f_i=\dfrac{400}{10}=40$ .

✓Final answer
$The correct option is (D) .$
KEAM 2025Set eng-2025-04264 marksMCQ
$400, 410, 420, 430, 440$
›Reveal solutionSolution

Mean of the data is $420$ ; average of absolute deviations $\{20,10,0,10,20\}$ is $60/5=12$ .

The values $400,410,420,430,440$ have mean $\dfrac{400+410+420+430+440}{5}=\dfrac{2100}{5}=420$ .

Absolute deviations from the mean: $|400-420|=20$ , $10$ , $0$ , $10$ , $20$ . Sum $=60$ .

Mean deviation $=\dfrac{60}{5}=12$ .

✓Final answer
$The correct option is (D) .$
KEAM 2025Set eng-2025-04274 marksMCQ
$x$
›Reveal solutionSolution

The mean is $5$ ; summing $f|x-5|=50$ over $N=20$ gives mean deviation $2.5$ .

Data: $x=2,4,6,10$ with $f=7,4,5,4$ , so $N=20$ .

Mean: $\sum fx=2(7)+4(4)+6(5)+10(4)=14+16+30+40=100$ , $\bar{x}=\dfrac{100}{20}=5$ .

Deviations $|x-5|=3,1,1,5$ , so $f|x-5|=21,4,5,20$ , summing to $50$ .

Mean deviation about mean $=\dfrac{\sum f|x-\bar{x}|}{N}=\dfrac{50}{20}=2.5$ .

✓Final answer
$The correct option is (B) .$
KEAM 2025Set eng-2025-04284 marksMCQ
$2, 4, 6, 8, 10$
›Reveal solutionSolution

With mean $6$ , the average absolute deviation of $2,4,6,8,10$ is $\tfrac{12}{5}=2.4$ .

The mean of $2,4,6,8,10$ is
$\bar x=\frac{2+4+6+8+10}{5}=6.$
The absolute deviations are $|2-6|,|4-6|,|6-6|,|8-6|,|10-6|=4,2,0,2,4$ , summing to $12$ .

Mean deviation $=\dfrac{12}{5}=2.4$ .

✓Final answer
$The correct option is (A) .$
KEAM 2024Set eng-2024-06094 marksMCQ
$2$
›Reveal solutionSolution

Compute the mean, then average the absolute deviations.

Data: $3,10,10,4,7,10,5$ . Sum $=49$ , so mean $=\dfrac{49}{7}=7$ .

Absolute deviations from $7$ : $|3-7|=4,\ |10-7|=3,\ 3,\ |4-7|=3,\ 0,\ 3,\ |5-7|=2$ . Sum $=18$ .
$\text{Mean deviation}=\frac{18}{7}\approx2.57.$
✓Final answer
$The correct option is (C) .$
KEAM 2023Set eng-2023-P2-B24 marksMCQ
$\dfrac{23}{7}$
›Reveal solutionSolution
For the data with median 7, the mean deviation about the median is 16/7.

Concept and Intuition

Sort the values, find the median (the middle of 7 values), take absolute deviations from it, and average them.

Step-by-Step Solution

Sorted: 3,3,5,7,8,9,10; median = 4th value = 7.

Absolute deviations: 4,4,2,0,1,2,3; sum = 16.

Mean deviation = 16/7.

Common Mistakes

Using the mean instead of the median.

Forgetting to sort before picking the middle value.

✓Final answer
$The correct option is (D) — 16/7.$

ANSWER: D

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

$x$	2	4	6	10
$f$	7	4	5	4
(A) $2$
(B) $2.5$
(C) $4$
(D) $6$
(E) $5$

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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