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 – 1st PUC 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

Karnataka PUCShort· 5mImportance★★★★★est

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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.

COMEDK 2025Set 2025-A1 markMCQ
$4,7,2,8,6$
›Reveal solutionSolution
The problem gives the mean of six numbers as 7, which lets us solve for the missing value $k$ . Once we have all six numbers, we compute the mean deviation (average absolute deviation from the mean). The result is 2, which corresponds to option (B).

Concept & Intuition

Mean deviation measures the average “spread” of data points around the mean. It’s the arithmetic mean of the absolute differences between each observation and the mean. Here, we’re told the mean itself is 7, so we first find the missing number $k$ that makes the mean 7, then compute the deviations.

Find $k$ using the given mean The mean of $4, 7, 2, 8, 6, k$ is 7.

$\frac{4 + 7 + 2 + 8 + 6 + k}{6} = 7$
Sum the known numbers: $4+7+2+8+6 = 27$ .

So
$\frac{27 + k}{6} = 7 \quad \Rightarrow \quad 27 + k = 42 \quad \Rightarrow \quad k = 15.$

List all observations

The full data set: $4, 7, 2, 8, 6, 15$ .

The mean is already known to be 7.

Compute absolute deviations from the mean

For each observation $x_i$ , find $|x_i - 7|$ :

$|4 - 7| = 3$

$|7 - 7| = 0$

$|2 - 7| = 5$

$|8 - 7| = 1$

$|6 - 7| = 1$

$|15 - 7| = 8$

Calculate the mean deviation

Mean deviation = $\frac{\text{sum of absolute deviations}}{\text{number of observations}}$

$\frac{3 + 0 + 5 + 1 + 1 + 8}{6} = \frac{18}{6} = 3.$
Watch out

A common mistake is to forget the absolute value and use signed deviations, which would sum to zero. Always take the absolute value for mean deviation.

Tip

Notice that the outlier 15 (far from the mean) contributes a large deviation of 8, but the mean deviation still comes out to a modest 3 because the other values are close to the mean.

✓Final answer
$The correct option is (B) . ANSWER: B$
COMEDK 2024Set 2024-A1 markMCQ
$Q-P=$
›Reveal solutionSolution
The mean deviation about the mean is the average absolute distance from the mean. For the first five odd natural numbers (1,3,5,7,9) and the first five prime numbers (2,3,5,7,11), we compute each mean deviation and subtract; the result is $Q-P = 0.32$ , so option (C) is correct.

Concept and intuition

Mean deviation about the mean measures how spread out a set of numbers is, on average, from their central value. It’s calculated as:
$\text{MD} = \frac{1}{n}\sum_{i=1}^{n} |x_i - \bar{x}|$
where $\bar{x}$ is the arithmetic mean.

We are comparing two small, well-known sequences: the first five odd natural numbers and the first five prime numbers. Both are symmetric-ish but not perfectly so; the primes are slightly more spread because 11 is farther from the mean than 9 is from its mean. The difference $Q-P$ will be a small decimal.

Step-by-step solution

First five odd natural numbers These are: $1, 3, 5, 7, 9$ . Their mean:

$\bar{x}_P = \frac{1+3+5+7+9}{5} = \frac{25}{5} = 5$
Deviations from the mean:

$|1-5|=4,\; |3-5|=2,\; |5-5|=0,\; |7-5|=2,\; |9-5|=4$

Sum of deviations: $4+2+0+2+4 = 12$

Mean deviation:
$P = \frac{12}{5} = 2.4$

First five prime numbers These are: $2, 3, 5, 7, 11$ . Their mean:

$\bar{x}_Q = \frac{2+3+5+7+11}{5} = \frac{28}{5} = 5.6$
Deviations from the mean:

$|2-5.6| = 3.6,\; |3-5.6| = 2.6,\; |5-5.6| = 0.6,\; |7-5.6| = 1.4,\; |11-5.6| = 5.4$

Sum of deviations: $3.6 + 2.6 + 0.6 + 1.4 + 5.4 = 13.6$

Mean deviation:
$Q = \frac{13.6}{5} = 2.72$

Difference $Q - P$

$Q - P = 2.72 - 2.4 = 0.32$
Tip

Notice that the mean of the odd numbers is exactly the middle value (5), making the deviations symmetric and easy to sum. For the primes, the mean is not an element of the set, so the deviations are not symmetric — that’s why the sum is slightly larger.

Watch out

A common mistake is to forget the absolute values or to compute the mean incorrectly (e.g., using 1 as the first odd number but forgetting 9 is the fifth). Always list the sequence explicitly.

✓Final answer
$The correct option is (C) . ANSWER: C$

←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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