Class 11 Statistics Notes · GSEB

Mean Deviation

Mean Deviation — average the absolute distances from the mean and see why absolute values are unavoidable. GSEB Class 11 Statistics notes with raw and grouped formulas, solved sums and the signed-to-absolute visual.

Last updated: 5 Oct 2026

Notes

Mean Deviation — the Mean of the Distances

Words marked with a dotted underline carry the question they usually appear in — tap them to reveal it.

“Uncontrolled variation is the enemy of quality.”— Edward Deming

Two archers, one average

Two archers shoot at a bullseye. Archer A misses by 3, 4, 5 and 4 metres; Archer B misses by 0, 9, 1 and 6 metres. Both average exactly 4 metres from the centre — the mean alone calls them equals. Their mean deviations (0.5 m vs 3.5 m) tell the real story: A is steady, B is wild. That is the whole job of a measure of dispersion.

Mean deviation (MD)

The mean of the absolute deviations of the observations of a data from their mean.

From Chapter 3 we already know that Σ(x − x̄) = 0— every positive deviation is cancelled by a negative one. Add the signed deviations of any data set and you always get zero, so a “mean of the deviations” would be useless. Two fixes exist:

Fix 1 — take absolute values

Ignore the sign and average the distances: this page's mean deviation.

Fix 2 — square the deviations

Squaring also removes negatives and keeps the algebra friendly: the next page's standard deviation.

Raw data

MD = \frac{\Sigma |x - \bar{x}|}{n}

Discrete & continuous

MD = \frac{\Sigma f|x - \bar{x}|}{n}

Coefficient of mean deviation (relative form)

\text{Coeff. of MD} = \frac{MD}{\bar{x}}

Key identification — when a question speaks of the absolute difference of the observation and its mean, it is asking for the mean deviation.

Signed → Absolute → Average

Illustration 8, one idea in three moves: the signed deviations around x̄ = 60 kg, the same distances with their signs removed, then their single average — the mean deviation.

Step 1 — each student’s weight minus the mean. Four bars go up, four go down, and the signed sum is zero.

Why the flip matters

If you skip the absolute values, the eight signed deviations of the weights (−14, −2, 0, −17, 15, 6, −9, 21) add up to 0 and every data set would report a mean deviation of zero. The flip is not cosmetic — it is the definition.

Worked Example — Weights of 8 Students

Solved Example

Problem

The weights (in kg) of 8 students of a class are 46, 58, 60, 43, 75, 66, 51, 81. Find the mean deviation and the coefficient of mean deviation.

Solution

MD = 10.5 kg, coefficient of mean deviation = 0.18

Median, Mode and Open-Ended Classes

Deviations may come from three averages

The deviations can be taken from the mean, the median or the mode, as per the need of the data. The mean is the standard choice (and the one the formulas on this page use), but for heavily skewed data — say monthly incomes of a locality with one very high earner — the median gives the more typical picture, and MD about the median is computed the same way: absolute distances, then divide by n.

Failure point — open-ended classes

A class like “150 and above” has no upper limit, hence no mid-value — and every grouped MD formula divides through mid-values. The mean deviation cannot be computed for open-ended frequency distributions, while the quartile deviation still can. That single contrast is a favourite of question setters.

Computing Skeletons

Two skeletons, one idea — add the absolute distances (weighted by f when data is grouped), then divide by n:

Raw data

MD = Σ|x − x̄| ÷ n

mean first, then the absolute distances, then divide by the count

Grouped data

MD = Σf|x − x̄| ÷ n

x is the mid-value for continuous classes — each distance carries its frequency

Two denominators, never one

Raw data divides by n (the count of observations). Grouped data also divides by n = Σf — never by the number of classes. Dividing by the class count is the fastest way to lose an entire sum.

Where Mean Deviation Is Used

The book's note: mean deviation is frequently used for problems in the social sciences; in economics it measures economic inequality, the distribution of personal wealth in a community or country, and helps in forecasting weather and business cycles.

Indian startup example — Zepto

Zepto promises groceries in 10 minutes. The meanof a store's delivery times can sit at a comfortable 10.4 minutes while the mean deviation is 5.5 minutes — meaning a large share of orders lands at 16 minutes. Startups watch the deviation, not just the average, because customers feel every late order.

Negative example — only the average was checked

A ₹299 tiffin service advertises “average delivery in 25 minutes” — but its mean deviation is 18 minutes: some tiffins arrive in 7 minutes, many after 40. They tracked the mean and never the spread, and customers who judge the service by their worst days churned anyway. An average without a dispersion measure is a half-told story.

Reading inequality with MD

Take per-capita income of five regions: ₹40k, ₹42k, ₹41k, ₹39k, ₹38k (MD = ₹1.2k) versus ₹20k, ₹25k, ₹40k, ₹55k, ₹80k (MD = ₹16k) — the same mean, wildly different inequality. This is exactly how mean deviation of incomes becomes a proxy for economic inequality.

Worked Example Gallery

Mean Deviation — solved practicals

5 problems

Raw data — weights of 8 students (Illustration 8)

Full question

The measurements of weight (in kg) of 8 students of a class of a school are given below :

46, 58, 60, 43, 75, 66, 51, 81

Find the mean deviation and coefficient of mean deviation of weights of the students.

x (kg)	x − x̄	\|x − x̄\|
46	−14	14
58	−2	2
60	0	0
43	−17	17
75	+15	15
66	+6	6
51	−9	9
81	+21	21
Total	480 → x̄ = 60	84

MD = 10.5 kg, coefficient = 0.18

x̄ = 480 / 8 = 60 kg.
Deviations: −14, −2, 0, −17, 15, 6, −9, 21 → signed sum = 0.
Absolute deviations: 14, 2, 0, 17, 15, 6, 9, 21 → Σ = 84.
MD = 84 / 8 = 10.5 kg · Coefficient = 10.5 / 60 = 0.175 ≈ 0.18.

Tiny raw set (exercise)

Full question

Find the mean deviation for the observations -1, 0, 4.

MD = 6/3 = 2

x̄ = (−1 + 0 + 4) / 3 = 1.
Absolute deviations: |−1 − 1| = 2, |0 − 1| = 1, |4 − 1| = 3 → Σ = 6.
MD = 6 / 3 = 2.

Discrete frequency — typing time (Illustration 9)

Full question

Calculate mean deviation and coefficient of mean deviation of typing time from the following information of time (in minutes) taken to type a report by 32 typists.

Typing time (minutes)	10	11	12	13	14
No. of typists	2	8	12	8	2

x (min)	f	fx	\|x − x̄\|	f\|x − x̄\|
10	2	20	2	4
11	8	88	1	8
12	12	144	0	0
13	8	104	1	8
14	2	28	2	4
Total	32	384	—	24

MD = 0.75 min, coefficient = 0.06

x̄ = Σfx / n = 384 / 32 = 12 min.
f|x − x̄|: 4, 8, 0, 8, 4 → Σ = 24.
MD = 24 / 32 = 0.75 min.
Coefficient = 0.75 / 12 = 0.0625 ≈ 0.06.

Continuous classes — spelling marks (Illustration 10)

Full question

20 children are selected for a district level spelling test. The distribution of their marks out of 50 marks is given below :

Marks	0 - 9	10 - 19	20 - 29	30 - 39	40 - 49
No. of Children	1	3	8	6	2

Find the mean deviation of the marks obtained by the children from this data.

Marks	f	x (mid)	fx	\|x − 27\|	f\|x − 27\|
0 – 9	1	4.5	4.5	22.5	22.5
10 – 19	3	14.5	43.5	12.5	37.5
20 – 29	8	24.5	196	2.5	20
30 – 39	6	34.5	207	7.5	45
40 – 49	2	44.5	89	17.5	35
Total	20	—	540	—	160

MD = 8 marks

Mid-values: 4.5, 14.5, 24.5, 34.5, 44.5 → Σfx = 540.
x̄ = 540 / 20 = 27 marks.
f|x − x̄|: 22.5, 37.5, 20, 45, 35 → Σ = 160.
MD = 160 / 20 = 8 marks.

Unequal classes — two-wheelers sold (Illustration 11)

Full question

From the following information of fortnight sale of two wheelers by 30 dealers of a city, find the mean deviation of 'number of two wheelers sold'.

No. of two wheelers	12 - 16	17 - 21	22 - 26	27 - 31	32 - 36
No. of dealers	2	3	14	8	3

No. of two-wheelers	f	x (mid)	fx	\|x − x̄\|	f\|x − x̄\|
12 – 16	2	14	28	11.17	22.34
17 – 21	3	19	57	6.17	18.51
22 – 26	14	24	336	1.17	16.38
27 – 31	8	29	232	3.83	30.64
32 – 36	3	34	102	8.83	26.49
Total	30	—	755	—	114.36

MD ≈ 3.81 two wheelers

Mid-values: 14, 19, 24, 29, 34 → Σfx = 755.
x̄ = 755 / 30 = 25.1667 ≈ 25.17 two wheelers.
f|x − x̄| = 22.34 + 18.51 + 16.38 + 30.64 + 26.49 = 114.36.
MD = 114.36 / 30 = 3.812 ≈ 3.81 two wheelers.

Inclusive limits (12 – 16, then 17 – 21) have a gap of 1 — boundaries (11.5 – 16.5 …) change the class width but not the mid-values, so the mean deviation is unaffected.

Advantages and Limitations

Mean deviation — merits and demerits

Advantages

It is a clearly defined measure of dispersion.
It is superior to the range and the quartile deviation, because all the observations are used in its computation.
Its value is less affected by extreme values compared with other measures.
The absolute difference is an appropriate measure of the distance between an observation and the mean.

Disadvantages

Its computation is more complicated than the range and the quartile deviation.
It is not suitable for algebraic operations.
It is less used in advanced study of statistics, because its definition rests on absolute values.
It cannot be computed when the frequency distribution has open-ended classes.

Quick Reference

Idea	Statement
Formulas	MD = Σ\|x − x̄\| / n (raw) · MD = Σf\|x − x̄\| / n (grouped, x = mid-value)
Definition	Mean deviation = the mean of the absolute deviations of the observations from their mean
Coefficient	Coefficient of MD = MD / x̄
Why absolute values	Σ(x − x̄) = 0 — signed deviations cancel, so each distance is taken as absolute
Key identification	Mean deviation uses the absolute difference of the observation and its mean
Coverage	Uses every observation — but cannot be computed for open-ended classes (Qd still can)

Key Takeaways

Mean deviation = mean of |x − x̄| — it uses ALL observations and measures them from an average, fixing both of the range’s limitations.
Absolute values are compulsory because Σ(x − x̄) = 0. So what: skip the modulus and every data set reports MD = 0 — a silent, total mistake.
Coefficient of MD = MD / x̄ turns the unit-carrying value into a comparable ratio. So what: only the coefficient lets you rank a kg-based data set against a marks-based one.
Deviations may be taken from the median or mode as the data needs, but the mean is the standard anchor.
It fails for open-ended classes — where the quartile deviation remains the only working measure. So what: income tables ending in "150 and above" must be handled by Qd, never MD.
So what: an average tells you where a crowd stands; the mean deviation tells you how tightly it stands there — which is what a Zepto, a tiffin service or an inequality report actually needs.