MathsCBSE Class 11 · Statistics

Mean Deviation About the Mean and the Median (Class 11 Statistics)

Q: What is the difference between mean deviation and standard deviation?

Mean deviation averages the absolute deviations. Standard deviation is the square root of the average of the squared deviations; it is more widely used.

Q: Can the mean deviation be negative?

No. It is an average of absolute values, so it is zero or positive.

Q: When is the mean deviation zero?

Only when all observations are equal.

Q: Is the range a measure of dispersion?

Yes, but it uses only the largest and smallest observations, which makes it less informative.

Updated 17 October 2026 · Fact-checked

Mean deviation about a point a is the average of the absolute deviations from a: MD = Σ f|x − a| / Σ f. Take a = mean or median. For raw data use Σ|x − a|/n; for a discrete table use Σ f|x − a|/N; for a continuous distribution use the class mid-points. Mean deviation is least about the median, and multiplying every observation by k multiplies the mean deviation by |k|.

Each arrow is an absolute deviation from the median 5; add the lengths (20) and divide by n = 5 to get MD = 4.

Illustration drawn for QuestionBuddy.

Understand Mean Deviation: Ungrouped, Discrete and Continuous Data

Dispersion measures how widely data are spread around a central value. The range uses only two values; the mean deviation uses every observation.

Definition. The mean deviation about a value $a$ is
$ $MD (a) = \frac{\sum f _{i} ∣ x _{i} - a ∣}{\sum f _{i}},$ $
the average of the absolute deviations. Use $a = \overset{x}{ˉ}$ (mean) or $a = M$ (median). The absolute value is needed because $\sum (x_{i} - \overset{x}{ˉ}) = 0$ : positive and negative deviations cancel.

Raw (ungrouped) data. $MD = \frac{1}{n} \sum ∣ x_{i} - a ∣$ . Steps: find the mean (or median), write each absolute deviation, sum, divide by $n$ . For the median of $n$ ordered values: the middle one if $n$ is odd, the average of the two middle ones if $n$ is even.

Discrete frequency distribution. Compute $N = \sum f$ and the mean $\overset{x}{ˉ} = \frac{\sum f x}{N}$ (or the median from cumulative frequencies: the observation at position $\frac{N + 1}{2}$ , or the average of positions $\frac{N}{2}$ and $\frac{N}{2} + 1$ ). Then $MD = \frac{\sum f ∣ x - a ∣}{N}$ . Build a table with columns $x$ , $f$ , $f x$ , $∣ x - a ∣$ , $f ∣ x - a ∣$ .

Continuous distribution. Use the mid-point $x_{i}$ of each class as the representative value, then proceed as for a discrete table. (For the median of a continuous distribution, locate the median class from cumulative frequencies.)

Properties.

Mean deviation about the median is the smallest possible: for any value $a$ , $\sum ∣ x_{i} - a ∣$ is least at $a =$ median. (For $2, 3, 5, 10, 15$ : sum about the median $5$ is $20$ ; about the mean $7$ it is $22$ ; about $4$ it is $21$ .)
Change of scale: if every observation is changed to $p x + q$ , then every deviation (from the new mean or median) is multiplied by $∣ p ∣$ , so MD becomes $∣ p ∣ \times MD$ . Adding a constant changes nothing.
It has the same units as the data, so it can be compared only among data in the same units.

Limitations. Mean deviation is easy to understand, but it is not convenient for further mathematics because of the absolute value, which is why variance and standard deviation (next page) are preferred.

Mean Deviation: Ungrouped, Discrete and Continuous Data: formulas to remember

Mean deviation about a: $M D = Σ f ∣ x - a ∣/Σ f$; a is the mean or the median.
Raw data: $M D = Σ∣ x - a ∣/ n$; All frequencies equal to 1.
Mean of a table: $\overset{x}{ˉ} = Σ f x / N$; N = Σ f.
Median position (discrete): $I t e mn u mb er (N + 1) /2 in t h eor d er e d l i s t$; Use cumulative frequency.
Change of scale: $N e w M D = ∣ p ∣ \times M D f or o b ser v a t i o n s p x + q$; The shift q does not matter.

How to solve Mean Deviation: Ungrouped, Discrete and Continuous Data questions

Use these steps for mean deviation about the mean or median.

1Decide whether the data are raw, discrete or continuous; for continuous classes use mid-points.
2Find the mean $\overset{x}{ˉ} = \frac{\sum f x}{N}$ or the median $M$ (from the ordered list or the cumulative frequencies).
3For each $x$ write $∣ x - a ∣$ and, for tables, $f ∣ x - a ∣$ .
4Add the column $f ∣ x - a ∣$ and divide by $N$ .
5Check that the sum of signed deviations from the mean is zero as a test of the mean.
6State the answer with units, to the required decimal places.

Mean Deviation: Ungrouped, Discrete and Continuous Data in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Mean deviation about the median of delivery times

Suppose seven delivery times (minutes) are $20, 13, 17, 25, 21, 15, 9$ .

Find the mean deviation about the median.

Try it first, then show the solutionSolution

Ordered: $9, 13, 15, 17, 20, 21, 25$ ; median $= 17$ (4th value).
Absolute deviations: $8, 4, 2, 0, 3, 4, 8$ ; sum $= 29$ .
$MD = \frac{29}{7} \approx 4.14$ minutes.

Answer: About 4.14 minutes.

Memory hook: Order first for the median; then add the absolute gaps.

Comparing consistency

Suppose two batters have three-match scores: A: 20, 40, 60 and B: 38, 40, 42. Both average 40.

Which is more consistent? Compare the mean deviations about the mean.

Try it first, then show the solutionSolution

A: deviations $20, 0, 20$ ; $MD = \frac{40}{3} \approx 13.3$ .
B: deviations $2, 0, 2$ ; $MD = \frac{4}{3} \approx 1.33$ .
B has the smaller mean deviation, so B is more consistent.

Answer: B (MD 1.33 against 13.3).

Memory hook: Same mean, smaller spread means more consistent.

Tricks to remember Mean Deviation: Ungrouped, Discrete and Continuous Data

Short memory aids. Say each one aloud once, then use it on a question the same day.

Order before median
Sort the data first; the median is the middle item.
Use it when: Median-based questions.
Absolute values only
Never subtract the mean and add signed values; they cancel to zero.
Use it when: Every MD question.
Mid-points for classes
Replace each class by its mid-point, then treat like a discrete table.
Use it when: Continuous data.
Median beats mean
The sum of absolute deviations is smallest about the median.
Use it when: Checking answers and property questions.

Mean Deviation: Ungrouped, Discrete and Continuous Data: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Ungrouped data · 2 marks · Very short answerBoard level

Find the mean deviation about the mean of $6, 7, 10, 12, 13, 4, 8, 12$ .

Try it first, then show the solutionSolution

$\overset{x}{ˉ} = \frac{72}{8} = 9$ . (½ mark)
Absolute deviations: $3, 2, 1, 3, 4, 5, 1, 3$ ; sum $= 22$ . (1 mark)
$MD = \frac{22}{8} = 2.75$ . (½ mark)

Answer: 2.75

Example 2: About the median · 3 marks · Short answerBoard level

Find the mean deviation about the median of $20, 13, 17, 25, 21, 15, 9$ .

Try it first, then show the solutionSolution

Ordered: $9, 13, 15, 17, 20, 21, 25$ , so the median is $17$ ( $n = 7$ is odd, 4th value). (1 mark)
Absolute deviations: $8, 4, 2, 0, 3, 4, 8$ ; sum $= 29$ . (1 mark)
$MD = \frac{29}{7} \approx 4.14$ . (1 mark)

Answer: $\frac{29}{7} \approx 4.14$

Example 3: Discrete frequency table · 3 marks · Short answerHarder board level

Find the mean deviation about the mean for $x = 2, 4, 6, 8, 10$ with frequencies $3, 5, 8, 3, 1$ .

Try it first, then show the solutionSolution

$N = 20$ , $\sum f x = 6 + 20 + 48 + 24 + 10 = 108$ , $\overset{x}{ˉ} = 5.4$ . (1 mark)
$∣ x - 5.4∣ = 3.4, 1.4, 0.6, 2.6, 4.6$ ; $f ∣ x - 5.4∣ = 10.2, 7, 4.8, 7.8, 4.6$ ; sum $= 34.4$ . (1 mark)
$MD = \frac{34.4}{20} = 1.72$ . (1 mark)

Answer: 1.72

Example 4: Continuous distribution · 4 marks · Short answerHarder board level

Find the mean deviation about the mean for the classes $0$ – $10$ , $10$ – $20$ , $20$ – $30$ , $30$ – $40$ , $40$ – $50$ with frequencies $4, 7, 12, 9, 3$ .

Try it first, then show the solutionSolution

Mid-points $5, 15, 25, 35, 45$ ; $N = 35$ . $\sum f x = 20 + 105 + 300 + 315 + 135 = 875$ , $\overset{x}{ˉ} = 25$ . (1½ marks)
$∣ x - 25∣ = 20, 10, 0, 10, 20$ ; $f ∣ x - 25∣ = 80, 70, 0, 90, 60$ ; sum $= 300$ . (1½ marks)
$MD = \frac{300}{35} = \frac{60}{7} \approx 8.57$ . (1 mark)

Answer: $\frac{60}{7} \approx 8.57$

Example 5: Median or mean? · 5 marks · Long answerHOTS / competency

For the frequency table $x = 5, 10, 15, 20, 25$ with $f = 7, 4, 6, 3, 5$ , find the mean deviation about (i) the median and (ii) the mean, and comment.

Try it first, then show the solutionSolution

$N = 25$ ; cumulative frequencies $7, 11, 17, 20, 25$ . The median is the $13$ th item, which lies in the class of $x = 15$ , so $M = 15$ . (1 mark)
(i) $∣ x - 15∣ = 10, 5, 0, 5, 10$ ; $f ∣ x - 15∣ = 70, 20, 0, 15, 50$ ; sum $= 155$ ; $MD = \frac{155}{25} = 6.2$ . (1 mark)
(ii) $\sum f x = 35 + 40 + 90 + 60 + 125 = 350$ , $\overset{x}{ˉ} = 14$ . (1 mark)
$f ∣ x - 14∣ = 63, 16, 6, 18, 55$ ; sum $= 158$ ; $MD = \frac{158}{25} = 6.32$ . (1 mark)
$6.2 < 6.32$ : the mean deviation is smaller about the median, as the property predicts. (1 mark)

Answer: (i) 6.2 (ii) 6.32; smaller about the median

Where students lose marks in Mean Deviation: Ungrouped, Discrete and Continuous Data

The slips examiners see most often, and what to write instead.

Mistake: Using signed deviations
They sum to zero about the mean.
Wrong working
$(6 - 9) + (7 - 9) + \dots = 0$ , so MD $= 0$ .
Right working
Use $∣ x - 9∣$ for each term.
Do this: Write the vertical bars in every column heading.
Usually costs: The whole answer.
Mistake: Using unordered data for the median
The median is defined on ordered data.
Wrong working
Median of $20, 13, 17, 25, 21, 15, 9$ taken as 25 (4th listed).
Right working
Order first: median $= 17$ .
Do this: Rewrite the data in ascending order.
Usually costs: The answer.
Mistake: Dividing by the number of classes
The divisor is N, the total frequency.
Wrong working
$\frac{300}{5}$ .
Right working
$\frac{300}{35}$ .
Do this: Compute N first.
Usually costs: The answer.
Mistake: Using class limits instead of mid-points
Each class is represented by its mid-point.
Wrong working
Using 0, 10, 20, ... as x.
Right working
Use $5, 15, 25, \dots$
Do this: Add a mid-point column to the table.
Usually costs: 1 to 2 marks.

Exam tips

Always build a table with columns $x$ , $f$ , $f x$ , $∣ x - a ∣$ , $f ∣ x - a ∣$ .
Order the data before locating the median.
State the divisor ( $n$ or $N$ ) explicitly.
For continuous data write the mid-point column first.
Round only at the end; give the fraction and the decimal.

Practice questions on Mean Deviation: Ungrouped, Discrete and Continuous Data

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The mean deviation about the mean of $2, 4, 6, 8$ is:
(a) 2
(b) 1
(c) 4
(d) 5
Hint
The mean is 5.
Answer
(a). Deviations $3, 1, 1, 3$ ; $\frac{8}{4} = 2$ .
Q2 · 1 mark · MCQHarder board level
The mean deviation of a data set is 4. If every observation is multiplied by 3 and then increased by 5, the new mean deviation is:
(a) 4
(b) 12
(c) 17
(d) 9
Hint
The added constant does not affect the spread.
Answer
(b). Multiplying by 3 multiplies the deviations by 3: $3 \times 4 = 12$ .
Q3 · 3 marks · Short answerHarder board level
The mean of the five observations $4, 6, x, 12, 18$ is 10. Find $x$ and the mean deviation about the mean.
Hint
Use $\sum x = 50$ .
Answer
$4 + 6 + x + 12 + 18 = 50$ gives $x = 10$ . Deviations $6, 4, 0, 2, 8$ ; sum $= 20$ ; $MD = \frac{20}{5} = 4$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The mean deviation is calculated with absolute values of the deviations.
Reason (R): The signed deviations from the mean add up to zero.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Think about why the sign must be removed.
Answer
(a). Because $\sum (x - \overset{x}{ˉ}) = 0$ , absolute values are used to prevent cancelling.
Q5 · 4 marks · Short answerHOTS / competency
For the data $2, 3, 5, 10, 15$ find the sum of the absolute deviations about (i) the median, (ii) the mean, (iii) the value 4, and state what you notice.
Hint
Median $= 5$ , mean $= 7$ .
Answer
(i) About $5$ : $3 + 2 + 0 + 5 + 10 = 20$ . (ii) About $7$ : $5 + 4 + 2 + 3 + 8 = 22$ . (iii) About $4$ : $2 + 1 + 1 + 6 + 11 = 21$ . The sum is least about the median.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the mean deviation about a?Show answer
Σ f|x − a| / Σ f.
2.Why use absolute values?Show answer
Signed deviations from the mean add to zero.
3.About which value is the sum of absolute deviations smallest?Show answer
The median.
4.How does MD change if every observation is multiplied by k?Show answer
It is multiplied by |k|.
5.What represents a class in grouped data?Show answer
Its mid-point.

Mean Deviation: Ungrouped, Discrete and Continuous Data: frequently asked questions

What is the difference between mean deviation and standard deviation?

Mean deviation averages the absolute deviations. Standard deviation is the square root of the average of the squared deviations; it is more widely used.

Can the mean deviation be negative?

No. It is an average of absolute values, so it is zero or positive.

When is the mean deviation zero?

Only when all observations are equal.

Is the range a measure of dispersion?

Yes, but it uses only the largest and smallest observations, which makes it less informative.