MathsISC Class 11 · Statistics

Measures of Dispersion: Range and Mean Deviation about the Mean and Median

Updated 17 October 2026 · Fact-checked

Dispersion shows how spread out the data are. The range is the largest minus the smallest value. The mean deviation about a point a is MD = Σfᵢ|xᵢ − a|/N, taking a as the mean x̄ or the median M. The coefficient of mean deviation is MD/a.

Range uses only the two end values, while mean deviation averages every point's gap from the mean: 14 ÷ 5 = 2.8.

Illustration drawn for QuestionBuddy.

Understand Range and Mean Deviation about the Mean and Median

Range. Largest value minus smallest value. The coefficient of range is $\frac{L - S}{L + S}$ . It is simple but depends only on two values.

Mean deviation about the mean. For raw data $x_{1}, ..., x_{n}$ with mean $\overset{x}{ˉ}$ : $M D (\overset{x}{ˉ}) = \frac{\sum ∣ x _{i} - x ˉ ∣}{n}$ . For a discrete frequency distribution: $M D = \frac{\sum f _{i} ∣ x _{i} - x ˉ ∣}{\sum f _{i}}$ with $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{\sum f _{i}}$ . For a continuous distribution use the class marks as $x_{i}$ .

Mean deviation about the median. Arrange the data in order and find the median M (the middle value for odd n; the mean of the two middle values for even n), then $M D (M) = \frac{\sum ∣ x _{i} - M ∣}{n}$ . For frequency data, find M from the cumulative frequency, then use $\frac{\sum f _{i} ∣ x _{i} - M ∣}{N}$ . The mean deviation about the median is the smallest of all mean deviations.

Coefficient. The relative measure: $\frac{M D}{mean (or median)}$ , used to compare the variability of two series with different units or sizes.

Steps for a table. (1) Find $N = \sum f$ and the mean (or median). (2) Compute $∣ x_{i} - a ∣$ . (3) Multiply by f. (4) Sum and divide by N.

Range and Mean Deviation about the Mean and Median: formulas to remember

Range: $\text{Range} = L - S$; Coefficient of range $= \frac{L - S}{L + S}$ .
Mean deviation: $MD = \frac{\sum f_i |x_i - a|}{\sum f_i}$; a = mean or median.
Mean: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$; Use class marks for grouped data.
Coefficient of MD: $\frac{MD}{a}$; Compare different series.

Range and Mean Deviation about the Mean and Median 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.

Daily sales of a shop

Suppose a shop's sales (in thousands of rupees) over 8 days are 6, 7, 10, 12, 13, 4, 8, 12.

Find the mean and the mean deviation about the mean.

Try it first, then show the solutionSolution

Mean $= \frac{72}{8} = 9$ .
Deviations $∣ x - 9∣$ : 3, 2, 1, 3, 4, 5, 1, 3; sum $= 22$ .
$M D = \frac{22}{8} = 2.75$ thousand rupees.

Answer: Mean 9; MD 2.75 thousand rupees.

Memory hook: Take absolute values before adding.

Tricks to remember Range and Mean Deviation about the Mean and Median

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

Absolute first
Never add signed deviations: they sum to zero for the mean.
Use it when: Mean deviation.
Median minimises MD
MD about the median is never larger than about the mean.
Use it when: Comparison questions.

Range and Mean Deviation about the Mean and Median: solved examples

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

Example 1: Range · 1 mark · MCQBoard level

The range of the data 4, 7, 10, 15 is:
(a) 4
(b) 8
(c) 11
(d) 15

Try it first, then show the solutionSolution

$15 - 4 = 11$ .

Answer: (c) 11

Example 2: MD about the mean · 3 marks · Short answerBoard level

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

Try it first, then show the solutionSolution

Mean $= \frac{72}{8} = 9$ . (1 mark)
$∣ x - 9∣$ : 3, 2, 1, 3, 4, 5, 1, 3; sum 22. (1 mark)
$M D = \frac{22}{8} = 2.75$ . (1 mark)

Answer: 2.75

Example 3: MD about the median · 4 marks · Short answerHarder board level

Find the mean deviation about the median for 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.

Try it first, then show the solutionSolution

Arranged: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21; n = 11, median is the 6th value $= 9$ . (1.5 marks)
Deviations $∣ x - 9∣$ : 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12; sum $= 58$ . (1.5 marks)
$M D = \frac{58}{11} \approx 5.27$ . (1 mark)

Answer: ≈ 5.27

Example 4: Grouped data · 4 marks · Short answerHarder board level

The waiting times (minutes) of 40 patients at a clinic are 0–10: 4, 10–20: 6, 20–30: 12, 30–40: 14, 40–50: 4. Find the mean deviation about the mean.

Try it first, then show the solutionSolution

Class marks 5, 15, 25, 35, 45; $N = 40$ ; $\sum f x = 20 + 90 + 300 + 490 + 180 = 1080$ ; mean $= 27$ . (1.5 marks)
$∣ x - 27∣$ : 22, 12, 2, 8, 18. (0.5 mark)
$f ∣ x - 27∣$ : 88, 72, 24, 112, 72; sum $= 368$ . (1 mark)
$M D = \frac{368}{40} = 9.2$ minutes. (1 mark)

Answer: 9.2 minutes

Example 5: Discrete frequency distribution · 5 marks · Long answerHOTS / competency

For x = 3, 5, 7, 9, 11 with frequencies 2, 4, 6, 5, 3, find the mean, the mean deviation about the mean and its coefficient. What does the coefficient allow you to do?

Try it first, then show the solutionSolution

$N = 20$ ; $\sum f x = 6 + 20 + 42 + 45 + 33 = 146$ ; mean $= 7.3$ . (1.5 marks)
$∣ x - 7.3∣$ : 4.3, 2.3, 0.3, 1.7, 3.7; $f ∣ x - 7.3∣$ : 8.6, 9.2, 1.8, 8.5, 11.1; sum $= 39.2$ . (1.5 marks)
$M D = \frac{39.2}{20} = 1.96$ . (1 mark)
Coefficient $= \frac{1.96}{7.3} = 0.268$ ; it lets us compare the variability of series with different means or units. (1 mark)

Answer: Mean 7.3; MD 1.96; coefficient 0.268.

Where students lose marks in Range and Mean Deviation about the Mean and Median

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

Mistake: Adding signed deviations
Getting zero for the deviations about the mean.
Do this: Use absolute values.
Usually costs: 2 marks.
Mistake: Not arranging data for the median
Taking the middle of the unordered list.
Do this: Sort the data first.
Usually costs: 1 to 2 marks.
Mistake: Using class limits instead of class marks
Taking x as 0–10 etc.
Do this: Use the mid-point of each class.
Usually costs: 1 mark.

Exam tips

Present the working in a table.
Show N, the mean and the sum separately.
Check that Σf(x − x̄) = 0 as a verification.
State the units of MD.

Practice questions on Range and Mean Deviation about the Mean and Median

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 is least when measured about the:
(a) mean
(b) median
(c) mode
(d) range
Hint
A known property.
Answer
(b) median
Q2 · 2 marks · Short answerBoard level
Find the mean deviation about the mean for 2, 4, 6, 8.
Hint
Mean 5.
Answer
Deviations 3, 1, 1, 3: $M D = \frac{8}{4} = 2$ .
Q3 · 3 marks · Short answerHarder board level
Find the mean deviation about the median for 6, 8, 10, 12, 14, 16.
Hint
Median is the average of the middle values.
Answer
Median $= 11$ ; $∣ x - 11∣$ : 5, 3, 1, 1, 3, 5; sum 18; $M D = 3$ .
Q4 · 3 marks · Short answerHarder board level
Find the range and its coefficient for 35, 42, 28, 50, 41, 60.
Hint
L = 60, S = 28.
Answer
Range $= 32$ ; coefficient $= \frac{32}{88} = 0.364$ .
Q5 · 3 marks · Short answerHarder board level
Two series have means 20 and 50 and mean deviations 4 and 8. Which is more variable?
Hint
Compare the coefficients.
Answer
Coefficients: $\frac{4}{20} = 0.2$ and $\frac{8}{50} = 0.16$ . The first series is more variable.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a teacher records the number of books read by 10 pupils in a month: 2, 4, 4, 5, 6, 6, 7, 8, 9, 9.
(i) Find the mean. (1 mark)
(ii) Find the median. (1 mark)
(iii) Find the mean deviation about the mean. (1 mark)
(iv) Find the mean deviation about the median. (1 mark)
Hint
Sum is 60.
Answer
(i) Mean $= 6$ . (ii) Median $= \frac{6 + 6}{2} = 6$ . (iii) $∣ x - 6∣$ : 4, 2, 2, 1, 0, 0, 1, 2, 3, 3 = 18; $M D = 1.8$ . (iv) The same as the median equals the mean: 1.8.

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.Range?Show answer
Largest − smallest.
2.MD about the mean?Show answer
Σf|x − x̄|/N.
3.Class mark?Show answer
Mid-point of the class.
4.Coefficient of MD?Show answer
MD/mean.

Range and Mean Deviation about the Mean and Median: frequently asked questions

Why use absolute values?

Signed deviations from the mean add to zero, hiding the spread.

Is the range a good measure of dispersion?

It uses only two values, so it is sensitive to extreme values.

Which is smaller: MD about the mean or the median?

The MD about the median is the smaller (or equal).