Class 11 Statistics Notes · GSEB

Mode

Mode — find the most frequent value by inspection, by formula and by histogram diagonals, plus the empirical fallback. GSEB Class 11 Statistics notes with solved sums, the formula anatomy and a construction lab.

Last updated: 5 Oct 2026

Notes

What Is a Mode?

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

Mode (Mo)

The value which gets repeated the maximum number of times, or the value occurring with the maximum frequency in the given data. It is denoted by Mo.

Mode is the average businesses actually quote: the dish ordered by the most customers on Zomato, the kurta size that walks out of the store most often, the shoe size a footwear brand keeps restocking. No calculations — just the most popular choice.

Raw data — count the repeats

1, 0, 2, 2, 3, 4, 2, 7, 2, 2, 5, 4, 2, 1, 2

Mo = 2

The value 2 appears 7 times — more than anything else (Illustration 29).

Discrete — highest frequency

Defects x = 0–5 with f = 45, 22, 18, 10, 6, 4

Mo = 0

Maximum frequency 45 sits on x = 0 — at the very start of the data (Illustration 30).

Bimodal — two peaks

Trips x = 1, 2, 4, 5, 6, 7 with f = 3, 7, 4, 7, 2, 1

Mo = 2 and 5

Two values share the maximum frequency 7 — the distribution is bimodal (Illustration 31).

No mode at all

Patients per hour: 3, 5, 4, 2, 7, 8

Mode cannot be found

Mode may not exist at all — every value appears once, so no value repeats (Illustration 32).

A mode at the start is not a representative

By definition the defects data above has Mo = 0 — but 0 sits at the edge of the distribution, so it does not represent a “typical” TV set. For such data choose the mean or the median, or find the mode with the empirical formula once you know the mean and median.

Grouped Data — The Modal Class Formula

For continuous classes, the class with the maximum frequency is the modal class, and the mode is read inside it:

Grouped (continuous) mode

M_o = L + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times c

The formula needs two neighbours — and equal class lengths.
Symbol	Meaning
L	Lower boundary point of the modal class
fm	Frequency of the modal class
f₁	Frequency of the class prior to the modal class
f₂	Frequency of the class succeeding the modal class
c	Class length of the modal class

Two conditions, strictly

The formula works only when (1) all classes have equal class lengths and (2) the maximum frequency belongs to one single class. Break either rule and the flowchart below tells you where to go.

Solved Example

Problem

The following table gives output of workers in a factory. Find the modal output. Output (items): 150–160, 160–170, 170–180, 180–190, 190–200, 200–210, 210–220, 220–230 Workers: 4, 5, 19, 33, 48, 22, 12, 6

Solution

Modal output = 193.66 items

Histogram Mode Finder — build the cross-diagonals

The graphical method works for equal and unequal class lengths, but only for unimodal distributions. Watch the bars grow, the diagonals cross at P, and the perpendicular land at Q.

Histogram Mode Finder — build the cross-diagonals yourself

Profits (₹ thousand) of 50 small-scale industrial units — equal class lengths, graphical method.

Data as given

Equal class lengths (20 each) → the raw frequencies can go straight onto the histogram.

Profit (₹ thousand)	20 – 40	40 – 60	60 – 80	80 – 100	100 – 120
No. of companies	2	13	22	9	4

Tap any bar to read its f, height and boundaries. Steps 3 and 4 replay the construction from the book, one line at a time.

Formula Anatomy — where the numbers come from

Same numbers as Illustration 33 (worker output 190 – 200): f₁ = 33, fm = 48, f₂ = 22. The two gaps at the top of the modal bar are the numerator and the denominator of the formula.

f₁

prior class

26fm = 48

modal class

f₂

succeeding class

Numerator (amber)

fm − f₁ = 48 − 33 = 15

Denominator (amber + emerald)

15 + (48 − 22) = 15 + 26 = 41

Result

190 + (15 / 41) × 10 = 193.66

Read the gaps, not the heights

The formula never uses f₁ or f₂ on their own — it uses how far they sit below fm. If both neighbours are equally far below (a perfectly symmetric peak), the two gaps are equal, the fraction becomes ½, and the mode sits exactly at the centre of the modal class.

When the Formula Fails → Empirical Formula

Every observation appears only once — no mode exists

More than one class shares the maximum frequency (bimodal)

Class lengths are unequal

Mixed distribution (discrete values + classes together)

One tail is too extended — the peak is not representative

any of these firesabandon the L + … formula

Use the empirical formula instead

Karl Pearson's empirical formula

M_o = 3M - 2\bar{x}

It comes from Karl Pearson's relation (Mean − Mode) = 3 (Mean − Median), rearranged for the mode. The value it returns can even be negative — that is a valid answer, not an error.

What Mo = 3M − 2x̄ is saying

In moderately skewed data the mean is pulled furthest from the peak, the median sits in between, and the mode lags behind: Mean − Mode = 3 (Mean − Median). Rearranged, the mode is what is left after you take the median three times and the mean twice.

Mean 38.36, Median 38.5 → Mo = 3(38.5) − 2(38.36) = 38.78 hours

Illustration 35 — two classes tie at frequency 7, so the empirical formula takes over.

Graphical Construction, Step by Step

Tall bar + neighbours

1/4

Draw the histogram. The tallest rectangle belongs to the modal class — label its top AB (A on the left, B on the right).

The two neighbouring rectangles get CD (left neighbour, D against A) and EF (right neighbour, E against B).

Worked Example Gallery

Mode — Solved Practicals

11 problems

Raw data, by inspection — Illustration 29

Full question

The numbers of books purchased by each of the 15 persons from a book store are as follows.

1, 0, 2, 2, 3, 4, 2, 7, 2, 2, 5, 4, 2, 1, 2

Find the modal value for the number of books purchased.

Value	0	1	2	3	4	5	7
Times repeated	1	2	7	1	2	1	1

Mo = 2 books

Count repetitions: 2 appears 7 times, 4 twice, 1 twice, and the rest once.
7 is the highest repetition → Mo = 2.

Discrete, max frequency — Illustration 30

Full question

TV sets assembled by a TV manufacturing company in a month are tested. The following table shows the numbers of defects per TV set. Find the mode for the number of defects.

No. of defects	0	1	2	3	4	5
No. of TV sets	45	22	18	10	6	4

Defects x	0	1	2	3	4	5
f	45	22	18	10	6	4

Mo = 0 (by definition)

Note: it cannot be used as the measure of central tendency here — use mean/median, or the empirical formula.

Highest frequency 45 belongs to x = 0 → Mo = 0.
But 0 sits at the beginning of the data, so it is not a representative average.

Bimodal — Illustration 31

Full question

The number of trips made by 24 taxi drivers in a day are shown in the following data. Find the modal number of trips.

No. of trips	1	2	4	5	6	7
No. of drivers	3	7	4	7	2	1

Trips x	1	2	4	5	6	7
f	3	7	4	7	2	1

Mo = 2 and 5 (bimodal distribution)

Maximum frequency = 7, shared by x = 2 and x = 5.
The distribution is bimodal → Mo = 2 and Mo = 5.

No mode possible — Illustration 32

Full question

The number of patients arriving at a clinic each hour during working hours of a day are recorded as follows.

3, 5, 4, 2, 7, 8

Find the mode for number of patients.

3, 5, 4, 2, 7, 8 — all frequencies are 1.

Mode cannot be found from the given data

Each value appears exactly once.
No value repeats → the most common observation does not exist.

A data set with all frequencies equal simply has no mode.

Continuous, by formula — Illustration 33

Full question

The following table gives output of workers in a factory. Find the modal output.

Output (no. of items)	150 - 160	160 - 170	170 - 180	180 - 190	190 - 200	200 - 210	210 - 220	220 - 230
No. of workers	4	5	19	33	48	22	12	6

Output	150–160	160–170	170–180	180–190	190–200	200–210	210–220	220–230
f	4	5	19	33	48	22	12	6

Mo = 190 + (48 − 33) / (96 − 33 − 22) × 10 = 190 + 150/41

Modal output = 193.66 items

Modal class = 190 – 200 (max f = 48) → L = 190, fm = 48, f₁ = 33, f₂ = 22, c = 10.
Denominator = 2(48) − 33 − 22 = 41.
Mo = 190 + (15 / 41) × 10 = 190 + 3.6585 = 193.6585.

Inclusive classes — Illustration 34

Full question

The number of cold drink bottles sold by a shopkeeper on different days is shown in the following table. Find the mode for the number of cold drink bottles sold.

No. of bottles	0 - 3	4 - 7	8 - 11	12 - 15	16 - 19	20 - 23
No. of days	3	11	16	20	18	12

Bottles	0 – 3	4 – 7	8 – 11	12 – 15	16 – 19	20 – 23
f	3	11	16	20	18	12
Boundary	−0.5 – 3.5	3.5 – 7.5	7.5 – 11.5	11.5 – 15.5	15.5 – 19.5	19.5 – 23.5

Mo = 11.5 + (20 − 16) / (40 − 16 − 18) × 4 = 11.5 + 16/6

Mode = 14.17 bottles

Modal class 12 – 15 (f = 20) is inclusive → boundaries 11.5 – 15.5, so L = 11.5, c = 4.
f₁ = 16, fm = 20, f₂ = 18 → denominator = 40 − 16 − 18 = 6.
Mo = 11.5 + (4 / 6) × 4 = 11.5 + 2.6666 = 14.1666.

Convert inclusive limits to boundaries before reading L and c.

Empirical (two classes tie) — Illustration 35

Full question

The time between placing an order and its delivery was noted for a certain wholesaler as follows. Find the mode for this time.

Time (hours)	20 - 25	25 - 30	30 - 35	35 - 40	40 - 45	45 - 50	50 - 55
No. of orders	2	5	7	5	6	7	3

Hours	20–25	25–30	30–35	35–40	40–45	45–50	50–55
f	2	5	7	5	6	7	3
cf	2	7	14	19	25	32	35

Mo = 3M − 2x̄ = 3(38.5) − 2(38.36) = 38.78

Mode = 38.78 hours

Max frequency 7 appears for two classes → the L + … formula is unusable.
Mean (short-cut, A = 37.5, c = 5): 37.5 + (6 / 35) × 5 = 38.36.
Median class of the 17.5th observation is 35 – 40 → M = 35 + (17.5 − 14)/5 × 5 = 38.5.
Mo = 3(38.5) − 2(38.36) = 115.5 − 76.72 = 38.78.

Empirical (mixed + unequal) — Illustration 36

Full question

The following table shows the number of visits to the dentist of the persons having toothache. Find the mode for number of visits.

No. of visits	1	2	3	4 - 7	7 - 10	10 - 15
No. of patients	7	11	18	9	4	1

Visits	1	2	3	4 – 7	7 – 10	10 – 15
f	7	11	18	9	4	1
Mid x	1	2	3	5.5	8.5	12.5
cf	7	18	36	45	49	50

Mo = 3(3) − 2(3.58) = 1.84

Mode = 1.84 visits

Mixed distribution with unequal class lengths → empirical formula.
Mean = Σfx / n = 179 / 50 = 3.58.
Median = value of the 25th observation = 3 (cf reaches 36 at x = 3).
Mo = 3(3) − 2(3.58) = 9 − 7.16 = 1.84.

Graphical, equal classes

Full question

We shall understand this method with the help of the following data which are about the profits (in thousand ₹) of 50 small scale industrial units. Find the mode of profits of these companies using the graphical method.

Profit (thousand ₹)	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120
No. of companies	2	13	22	9	4

Profit (₹ thousand)	20 – 40	40 – 60	60 – 80	80 – 100	100 – 120
Companies	2	13	22	9	4

Construction: A(60, 22), B(80, 22), D(60, 13), E(80, 9) → AE × BD cross at x = 68.18

OQ = ₹68 thousand (read from the histogram)

Tallest rectangle = 60 – 80, top AB; neighbours 40 – 60 (CD) and 80 – 100 (EF).
Draw AE and BD; they cross at P.
Perpendicular from P meets the x-axis at Q.

The formula gives 68.18; the graph is read as 68 — ₹68 thousand.

Graphical, inclusive classes — Illustration 37

Full question

The following data show the number of empty seats per bus for the buses leaving from a depot. Find the modal number of empty seats using the graphical method.

No. of empty seats x	1 - 4	5 - 8	9 - 12	13 - 16	17 - 20	21 - 24
No. of buses	4	7	12	19	9	3

Seats	1 – 4	5 – 8	9 – 12	13 – 16	17 – 20	21 – 24
f	4	7	12	19	9	3
Boundary	0.5 – 4.5	4.5 – 8.5	8.5 – 12.5	12.5 – 16.5	16.5 – 20.5	20.5 – 24.5

OQ = 14.2 empty seats

Inclusive limits → boundaries 0.5 – 4.5, 4.5 – 8.5, …, 20.5 – 24.5 before drawing.
Tallest rectangle = 12.5 – 16.5 (f = 19); diagonals cross over the modal class.
The perpendicular lands at Q → OQ = 14.2.

Graphical, unequal classes — Illustration 38

Full question

The distribution of the share prices of a company on different days is given in the following table. Find the mode for the share prices using the graphical method.

Share price (₹)	200 - 210	210 - 220	220 - 240	240 - 260	260 - 300
No. of days	4	13	36	16	8

Share price	200 – 210	210 – 220	220 – 240	240 – 260	260 – 300
Length	10	10	20	20	40
f	4	13	36	16	8
Proportionate f = f/length × 10	4	13	18	8	2

OQ = ₹227

Construction gives 226.67 — the graph reads 227.

Class lengths are unequal → proportionate frequencies with minimum length 10: 4, 13, 18, 8, 2.
Draw the histogram on those heights; tallest bar = 220 – 240.
Cross-diagonals → P → perpendicular → Q.

Advantages and Limitations

Advantages

It is easy to understand and calculate.
It can be found merely by inspection.
It is not affected by too large or too small values.
Its value can be found using graph.

Disadvantages

It is not rigidly defined.
There can be more than one mode for the given variable whereas sometimes the mode cannot be found.
It is not based on all observations.
It has less stability in sampling as compared to mean.
It is not suitable for further mathematical calculations.

Quick Reference

Case	Statement
Raw data	Mo = value repeated the most number of times
How you find it	By inspection for raw data and for discrete frequency distributions — no calculation needed
Discrete frequency	Mo = value with maximum frequency
Continuous (equal classes, one peak)	Mo = L + (fm − f₁)/(2fm − f₁ − f₂) × c
When the formula fails	Mode = 3 × Median − 2 × Mean
Unequal classes	Use the graphical method, or compute proportionate frequencies = (f / class length) × minimum class length first
Graphical method	Diagonals AE and BD cross at P → drop a perpendicular to Q → OQ = mode

Key Takeaways

Mode is the value that repeats the most — found by inspection for raw and discrete data.
Grouped data: the class with maximum frequency is the modal class; Mo = L + (fm − f₁)/(2fm − f₁ − f₂) × c needs equal classes and one single peak.
The gaps (fm − f₁) and (fm − f₂) are the whole formula — the anatomy diagram shows them as the coloured segments on the tallest bar.
Five situations break the formula (no repeats, bimodal, unequal, mixed, long tail) — each routes you to Mo = 3M − 2x̄, Karl Pearson’s empirical formula.
Graphical method: AB, CD, EF → AE × BD → P → perpendicular → Q → OQ = mode; it survives unequal classes.
So what: restaurants size their menus, garment brands pick production sizes and shops reorder stock on mode — the most popular choice, not the average one.