Median, Quartiles, Deciles and Perce… — Class 11 Statistics Notes (GSEB)

Positional Averages — Value Comes From Position

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

Positional averages

Median, quartiles, deciles and percentiles are called positional averages because their value is found from the value of an observation at a specific position in the ordered data — not from adding or multiplying anything.

Median (M)

The value of the middlemost observation when the data are arranged in ascending or descending order — 50% of the observations lie above it and 50% below it.

The hierarchy

Q₂ = M = D₅ = P₅₀Q₁ = P₂₅, Q₃ = P₇₅Dⱼ = P₁₀ⱼ

One family, three names: the same point on the data can be called a quartile, a decile or a percentile depending on how finely you want to slice it.

Median keeps its head

A chai stall near your college earns ₹2,400, ₹2,600, ₹2,500, ₹2,550 and ₹20,000 on the day a wedding order landed. The mean jumps to ₹6,010; the median stays at ₹2,550 — the day that was actually normal.

Mean gets carried away

In a 10-person startup paying ₹40,000 each, the founder draws ₹10,00,000. Mean salary = ₹1,36,364 — a figure nobody in the office actually earns. Median = ₹40,000 tells the truth.

Raw and Discrete Data — The Position Formulas

Median — raw & discrete

M = \text{value of the } \left(\frac{n+1}{2}\right)\text{th observation}

jth Quartile — raw & discrete

Q_j = \text{value of the } j\left(\frac{n+1}{4}\right)\text{th observation}

jth Decile — raw & discrete

D_j = \text{value of the } j\left(\frac{n+1}{10}\right)\text{th observation}

jth Percentile — raw & discrete

P_j = \text{value of the } j\left(\frac{n+1}{100}\right)\text{th observation}

Two traps in one formula

1. Order first. A position means nothing until the data are arranged in ascending order — the 25.5th observation of shuffled data is not a median.2. Use (n + 1)/2, not (n − 1)/2. The book text prints (n − 1)/2 in places, but every solved example and the formula list use (n + 1)/2 — for 50 observations the median is the value of the 25.5th observation, so you average the 25th and 26th values.

When the position falls between two observations

A position like 5.25th means: take the 5th value, then step a quarter of the way towards the 6th value.

Interpolation rule

x_{(5.25)} = x_{(5)} + 0.25\left(x_{(6)} - x_{(5)}\right)

5.5th = mean of the 5th and 6th values · 8.4th = 8th + 0.4 × (9th − 8th) · 14.7th = 14th + 0.7 × (15th − 14th). Same rule for every positional average.

Median Class Zoom — the grouped formula, seen

For continuous classes the median lives inside one class — the median class, the class that contains the (n/2)th observation. The lab builds the cumulative frequency staircase, crosses it with the n/2 line, then zooms into that one class so the formula stops being memorised and starts being seen.

Continuous frequency distribution

M = L + \frac{\frac{n}{2} - cf}{f} \times c

Median Class Zoom — watch the formula walk into one class

Number of cheques deposited in a bank per day (exclusive classes)

Step 1. Plain frequency distribution — build cf as a running total of the frequencies.

Build-up	Values
Given f	2, 14, 23, 7, 4
cf = running total	2, 16, 39, 46, 50
n = Σf	50 → n/2 = 25

The amber line below sits at height n/2 = 25.

Tap a class on the staircase to read its f, cf and boundaries. Class widths are drawn to scale.

Position Ladder — Q, D and P on one line

Quartiles cut the ordered data into 4 parts, deciles into 10, percentiles into 100. Below, the same 20 innings sit on one number line with all three ribbons stacked above them — pick a position and watch it drop onto the data.

Q₁ = 28.25 (5.25th observation = 5th + 0.25(6th − 5th))

Q₂ = MD₅ = MP₅₀ = MQ₁ = P₂₅Q₃ = P₇₅Dⱼ = P₁₀ⱼ

Reading the ribbons

Quartile cuts land every 25% of the data, decile cuts every 10%, percentile cuts every 1%. Because they are cutting the same ordered data, Q₂, D₅ and P₅₀ always fall on one another — that is the median.

Cumulative Data — Convert Before You Calculate

“Upto / less than” type

Successive subtraction turns the given cf into f — then the cf column is already there.
Expenditure	Upto 200	Upto 400	Upto 600	Upto 800	Upto 1000	Upto 1200
Given cf	2	8	17	32	57	75
f = subtraction	2	6	9	15	25	18

“More than” type

Subtract down for f, then add up from the bottom to build the less-than cf the formula wants.
Pollution	250+	270+	290+	310+	320+	330+	340+
Given	150	133	108	76	41	20	7
f = subtraction	17	25	32	35	21	13	7
less-than cf	17	42	74	109	130	143	150

The two conversions that decide the answer

cf of the class before the median class. For the pollution data the median class is 310 – 320, so cf = 74 (the total of everything below it) — not 109, which is the cf written against that class.Inclusive limits → boundaries first. 80 – 119 becomes 79.5 – 119.5 (gap type, ±0.5); classes that already touch (0 – 200, 200 – 400) share their boundary, so L is simply 800.

Common Traps

Order the data first — positions only exist after arrangement.

cf is always the cumulative frequency of the class BEFORE the median class.

Inclusive classes (80 – 119) must be converted to boundaries (79.5 – 119.5).

Raw/discrete position = (n + 1)/2 · continuous position = n/2 — do not mix them.

Interpreting the Positions

Central 50%

Everything between Q₁ and Q₃ is the middle half of the data — for 50 commuters, ₹563.46 to ₹825 of monthly travel expense.

“Top 15%” → P₈₅

If the highest 15% matter, ask for the minimum expense of that group: P₈₅ = ₹908.33.

“Lowest 10%” → D₁

The bottom decile of the same commuters spends ₹457.14 or less.

The median's two superpowers

It is the only average available for open-ended classes (“950 and above” never needs a midpoint), and it survives missing data — order what you have, take the middle. Extreme values barely move it: change the largest of 10 observations and the median stays exactly where it was.

Solved Example — Illustration 25

Solved Example

Problem

Find Q₁, D₇ and P₄₀ for the runs scored by a batsman in his 20 innings: 32, 28, 47, 63, 71, 9, 60, 10, 96, 14, 31, 148, 53, 67, 29, 10, 62, 40, 80, 54

Solution

Q₁ = 28.25, D₇ = 62.7, P₄₀ = 35.2

Worked Example Gallery

Median, Quartiles, Deciles & Percentiles — Solved Practicals

12 problems

Raw data, odd n — Illustration 17

Full question

The numbers of items produced by a factory in different weeks are 80, 85, 90, 92, 68, 80, 72, 63, 55. Find the median production.

Ordered: 55, 63, 68, 72, 80, 80, 85, 90, 92

M = value of the 5th observation = 80 items

Arrange in ascending order: 55, 63, 68, 72, 80, 80, 85, 90, 92.
n = 9 → (n + 1)/2 = 5th observation.
5th value = 80.

Raw data, even n — Illustration 18

Full question

The daily profits of a hawker (in ₹) for last 10 days are given below. Find the median profit.

261.5, 257, 258.5, 260, 265, 249, 255.5, 262.5, 264, 267

Position	5th	6th
Value (₹)	260	261.5

M = (260 + 261.5) / 2 = 260.75

Median daily profit = ₹260.75

Order: 249, 255.5, 257, 258.5, 260, 261.5, 262.5, 264, 265, 267.
n = 10 → position (10 + 1)/2 = 5.5th.
5.5th = mean of the 5th and 6th values = (260 + 261.5)/2.

Missing data — Illustration 19

Full question

There are 11 employees in an office. The monthly salaries (in ₹) of the lowest paid 7 employees among them are 4500, 2100, 3400, 3600, 2500, 4200, 1500. What is the median monthly salary of all employees?

1500, 2100, 2500, 3400, 3600, 4200, 4500, a, b, c, d

Median salary = ₹4,200

The four missing values sit after the median — they cannot move it.

Order the known salaries: 1500, 2100, 2500, 3400, 3600, 4200, 4500.
The 4 unknown salaries are higher — call them a, b, c, d in increasing order.
n = 11 → (n + 1)/2 = 6th observation = 4200.

Discrete frequency distribution — Illustration 20

Full question

The following table shows the record of absent students of a class during a month.

Find the median of number of absent days per student.

No. of absent days of student	0	1	2	3	4	5
No. of students	8	12	18	9	5	1

27th observation falls under cf 38 → x = 2
Absent days x	0	1	2	3	4	5
f	8	12	18	9	5	1
cf	8	20	38	47	52	53

Median = 2 days

n = 8 + 12 + 18 + 9 + 5 + 1 = 53 → position (53 + 1)/2 = 27th observation.
Build cf: 8, 20, 38, 47, 52, 53 — the 27th observation lies in the class x = 2 (13th to 38th are all 2).

Discrete, fractional position — Illustration 21

Full question

The time required for typing a report by different typists is given in the following data. Find the median typing time using it.

Time for typing (min.)	10	11	12	13	14
No. of typists	5	7	8	15	5

Time (min.)	10	11	12	13	14
f	5	7	8	15	5
cf	5	12	20	35	40

20.5th → (12 + 13) / 2 = 12.5

Median typing time = 12.5 min

n = 40 → position (40 + 1)/2 = 20.5th observation.
cf: 5, 12, 20, 35, 40 — the 20th value is 12 and the 21st value is 13.
20.5th = (12 + 13)/2 = 12.5.

Continuous classes (exclusive) — Illustration 22

Full question

The number of cheques deposited in a bank per day has the following frequency distribution :

No. of cheques	0 - 39	40 - 79	80 - 119	120 - 159	160 - 199
No. of days	2	14	23	7	4

Find the median number of cheques deposited.

Cheques	0 – 39	40 – 79	80 – 119	120 – 159	160 – 199
f	2	14	23	7	4
cf	2	16	39	46	50

M = 79.5 + (25 − 16) / 23 × 40 = 79.5 + 360/23 = 79.5 + 15.6522

Median = 95.15 cheques

n = 50 → n/2 = 25th observation.
cf: 2, 16, 39, 46, 50 → the 25th observation lies in 80 – 119 (median class).
Gap-type limits → boundary 79.5 – 119.5: L = 79.5, cf = 16, f = 23, c = 40.
M = 79.5 + (25 − 16)/23 × 40 = 79.5 + 15.6522 = 95.15.

“Less than” cumulative — Illustration 23

Full question

The data about monthly expenditure on petrol for 75 families is given in the following table. Find median expenditure on petrol for these families.

Expenditure (₹) on petrol	Upto 200	Upto 400	Upto 600	Upto 800	Upto 1000	Upto 1200
No. of families	2	8	17	32	57	75

Expenditure	0 – 200	200 – 400	400 – 600	600 – 800	800 – 1000	1000 – 1200
f	2	6	9	15	25	18
cf	2	8	17	32	57	75

M = 800 + (37.5 − 32) / 25 × 200 = 800 + 44

Median expenditure = ₹844

Successive subtraction gives f: 2, 6, 9, 15, 25, 18 (classes 0–200 … 1000–1200).
n = 75 → n/2 = 37.5th observation; cf = 2, 8, 17, 32, 57, 75 → median class 800 – 1000.
Classes touch, so L = 800, cf = 32, f = 25, c = 200.
M = 800 + (37.5 − 32)/25 × 200 = 800 + 44 = 844.

“More than” cumulative — Illustration 24

Full question

The level of air pollution (in ppm) in a city on different days is as follows. Find the median level of pollution.

Level of pollution (ppm)	250 and above	270 and above	290 and above	310 and above	320 and above	330 and above	340 and above
No. of days	150	133	108	76	41	20	7

Pollution	250 – 270	270 – 290	290 – 310	310 – 320	320 – 330	330 – 340	340+
f	17	25	32	35	21	13	7
cf	17	42	74	109	130	143	150

M = 310 + (75 − 74) / 35 × 10 = 310 + 0.2857

Median pollution = 310.29 ppm

f = successive subtraction: 17, 25, 32, 35, 21, 13, 7.
less-than cf (add from the bottom): 17, 42, 74, 109, 130, 143, 150.
n = 150 → n/2 = 75th observation → median class 310 – 320 (cf before = 74).
L = 310, cf = 74, f = 35, c = 10 → M = 310 + (75 − 74)/35 × 10 = 310.2857.

The cf taken is 74 — the class BEFORE 310 – 320.

Q/D/P from raw data — Illustration 25

Full question

Find Q₁, D₇, P₄₀ for the following data showing runs scored by a batsman in his 20 innings.

32, 28, 47, 63, 71, 9, 60, 10, 96, 14, 31, 148, 53, 67, 29, 10, 62, 40, 80, 54

Average	Position	Interpolation	Value
Q₁	5.25th	28 + 0.25(29 − 28)	28.25
D₇	14.7th	62 + 0.7(63 − 62)	62.7
P₄₀	8.4th	32 + 0.4(40 − 32)	35.2

Q₁ = 28.25, D₇ = 62.7, P₄₀ = 35.2

Order: 9, 10, 10, 14, 28, 29, 31, 32, 40, 47, 53, 54, 60, 62, 63, 67, 71, 80, 96, 148.
Q₁ at 1 × 21/4 = 5.25 → 28 + 0.25(29 − 28) = 28.25.
D₇ at 7 × 21/10 = 14.7 → 62 + 0.7(63 − 62) = 62.7.
P₄₀ at 40 × 21/100 = 8.4 → 32 + 0.4(40 − 32) = 35.2.

Q/D/P from continuous data — Illustration 26

Full question

The following data refer to the milk quantity in 90 bags filled by a machine in a dairy.

Milk content (ml.)	485 - 490	490 - 495	495 - 500	500 - 505	505 - 510
No. of bags	5	21	33	23	8

Find Q₃, D₂, P₅₅ and interpret them.

Milk (ml)	485 – 490	490 – 495	495 – 500	500 – 505	505 – 510
f	5	21	33	23	8
cf	5	26	59	82	90

Average	Position	Class	Value
Q₃	67.5th	500 – 505	501.85
D₂	18th	490 – 495	493.1
P₅₅	49.5th	495 – 500	498.56

Q₃ = 501.85 ml (75% of bags hold less) · D₂ = 493.1 ml (lowest 20%) · P₅₅ = 498.56 ml

cf: 5, 26, 59, 82, 90 · n = 90.
Q₃ at 3 × 90/4 = 67.5 → class 500 – 505 (cf = 59): 500 + (67.5 − 59)/23 × 5 = 501.85.
D₂ at 2 × 90/10 = 18 → class 490 – 495 (cf = 5): 490 + (18 − 5)/21 × 5 = 493.1.
P₅₅ at 55 × 90/100 = 49.5 → class 495 – 500 (cf = 26): 495 + (49.5 − 26)/33 × 5 = 498.56.

Q/D/P from discrete data — Illustration 27

Full question

We have the following information from a survey of 100 customers of a bank on their number of visits to the bank during a month.

No. of visits	0	1	2	3	4	5	6
No. of customers	12	22	40	15	6	4	1

Find the first quartile, 4^th decile and 95^th percentile.

Visits x	0	1	2	3	4	5	6
f	12	22	40	15	6	4	1
cf	12	34	74	89	95	99	100

Average	Position	Value
Q₁	25.25th	1 visit
D₄	40.4th	2 visits
P₉₅	95.95th	≈ 5 visits

Q₁ = 1, D₄ = 2, P₉₅ ≈ 5 — every value inside a class comes back as that class value.

Q₁ at 1 × 101/4 = 25.25th → cf 12, 34 → value 1.
D₄ at 4 × 101/10 = 40.4th → cf 34, 74 → value 2.
P₉₅ at 95 × 101/100 = 95.95th → cf 95, 99 → value 5.

Interpretive reads — Illustration 28

Full question

The following table shows data about monthly travelling expenses (in ₹) in a sample of 50 local bus commuters.

Monthly expenses	350 - 500	500 - 650	650 - 800	800 - 950	950 and above
No. of commuters	7	13	16	9	5

(1) Find the limits for the expense of central 50% of the commuters.

(2) Find the minimum expense of the highest spending 15% commuters.

(3) Find the maximum expense in the lowest spending 10% commuters.

Expenses (₹)	350 – 500	500 – 650	650 – 800	800 – 950	950 and above
f	7	13	16	9	5
cf	7	20	36	45	50

Read	Position	Calculation	Answer
Central 50% (Q₁ to Q₃)	12.5th & 37.5th	500 + 63.46 & 800 + 25	₹563.46 to ₹825
Top 15% (P₈₅)	42.5th	800 + 108.33	₹908.33
Bottom 10% (D₁)	5th	350 + 107.14	₹457.14

Central 50% = ₹563.46 – ₹825 · P₈₅ = ₹908.33 · D₁ = ₹457.14

cf: 7, 20, 36, 45, 50 · n = 50.
Central 50% = between Q₁ (12.5th) and Q₃ (37.5th): 500 + (12.5 − 7)/13 × 150 = 563.46 and 800 + (37.5 − 36)/9 × 150 = 825.
Highest 15% starts at P₈₅ (42.5th): 800 + (42.5 − 36)/9 × 150 = 908.33.
Lowest 10% ends at D₁ (5th): 350 + (5 − 0)/7 × 150 = 457.14.

Advantages and Limitations of the Median

Advantages

It is easy to calculate and to understand.
It can be found by inspection.
It can be located by graph.
It is the only available average when the frequency distribution has open ended classes.
It is less affected by too large or too small values.
It can be calculated even if certain data are missing.

Disadvantages

It is not rigidly defined.
It is not based on all values.
It is not suitable for further algebraic operations.
It is less stable measure of central tendency as compared to mean.

Quick Reference

Idea	Statement
Raw / discrete — median	M = value of the ((n + 1)/2)th observation
Continuous — median	M = L + ((n/2) − cf)/f × c, where cf is of the class before the median class
Raw / discrete — Q, D, P	Dⱼ = value of the j(n + 1)/10th observation · Pⱼ = value of the j(n + 1)/100th · Qⱼ = value of the j(n + 1)/4 th observation
Continuous — Q, D, P	Replace the position by j(n/4), j(n/10) and j(n/100) respectively, then interpolate inside that class
One identity	The median is also Q₂, D₅ and P₅₀
Reads	Central 50% lies between Q₁ and Q₃ · “top 15%” → P₈₅ · “lowest 10%” → D₁
True/false favourite	The median divides the data into two equal parts — the mean does not
Part count	Quartiles, deciles and percentiles split the data into 4, 10 and 100 equal parts
Outlier immunity	The median does not change when the largest observation changes
Missing data	Median survives missing data — order the observations, then take the middle
Open-ended classes	Median is the only average that can be computed for open-ended classes

Key Takeaways

Positional averages take their value from where an observation sits in ordered data — order first, always.
Raw/discrete positions: M at (n + 1)/2, Qⱼ at j(n + 1)/4, Dⱼ at j(n + 1)/10, Pⱼ at j(n + 1)/100 — then interpolate between the two bracketing values.
Continuous data: find the class containing n/2 (or jn/4, jn/10, jn/100), then walk in by (position − cf)/f of the class width.
Q₂ = M = D₅ = P₅₀ — one point, four names; Q₁ = P₂₅ and Q₃ = P₇₅.
cf always comes from the class BEFORE, and inclusive limits must become boundaries before L is read off.
So what: salary bands, expense ranges and percentile cut-offs (“top 15%”, “lowest 10%”) are all read off this ladder — P₈₅, Q₁ to Q₃ and D₁ — and open-ended classes only the median can handle.

Median, Quartiles, Deciles and Percentiles

Notes

Positional Averages — Value Comes From Position

Raw and Discrete Data — The Position Formulas

Median Class Zoom — the grouped formula, seen

Position Ladder — Q, D and P on one line

Cumulative Data — Convert Before You Calculate

Common Traps

Interpreting the Positions

Solved Example — Illustration 25

Worked Example Gallery

Median, Quartiles, Deciles & Percentiles — Solved Practicals

Advantages and Limitations of the Median

Advantages

Disadvantages

Quick Reference

Key Takeaways