Chapter 13 · Statistics

13.2 Mode of Grouped Data

The mean uses every observation, but sometimes you want the value that is most typical: the shoe size a shop should stock most of, or the time most customers wait. For grouped data, find the modal class and then use the mode formula to place the mode inside it.

A shoe shop owner does not care about the average shoe size of her customers. The average might be $7.4$, and no one wears a shoe of size $7.4$. What she needs to know is the size that is bought most often, so that she stocks plenty of it. A bus company asks the same question about waiting times, and a doctor about the age group that falls ill most. The most frequent value is the mode.

For a plain list the mode is easy to spot. For grouped data we can only say which class is the most crowded, but we can do better: the exact position of the mode inside that class depends on how crowded the two neighbouring classes are. A neat formula captures this, and you will see why it works from a picture of the histogram.

Mode and modal class

📘 Definition 13.3 — Mode
The mode of a set of observations is the value that occurs most often, that is, the value with the highest frequency.
📘 Definition 13.4 — Modal class
In a grouped frequency distribution, the modal class is the class with the highest frequency. The mode of the grouped data lies in this class.
✏️ Example 1 — Mode of a list
The shoe sizes of nine students are $6,\ 7,\ 7,\ 8,\ 7,\ 9,\ 6,\ 7,\ 8$. Find the mode.
Tally Size $6$ appears $2$ times, size $7$ appears $4$ times, size $8$ appears $2$ times, size $9$ appears once.
Mode The highest frequency is $4$, so the mode is shoe size 7.

With grouped data we need a way to pick a single number from the modal class. Think of three neighbouring bars of a histogram. Let the modal class have lower limit $l$ and class size $h$. Write $f_1$ for the frequency of the modal class, $f_0$ for the frequency of the class just before it, and $f_2$ for the frequency of the class just after it. Two ideas decide where the mode sits:

📐 Theorem 13.4 — Mode of grouped data
Let the classes be continuous (touching) with equal class size $h$. If the modal class has lower limit $l$, frequency $f_1$, and the classes before and after it have frequencies $f_0$ and $f_2$ (where $f_1$ is larger than both), then $$\text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)h.$$

Why it works. Draw the histogram. Join the top-right corner of the modal bar to the top-right corner of the bar before it, and join the top-left corner of the modal bar to the top-left corner of the bar after it. The two lines cross above the modal class (you can see this in Interactive 1), and the mode is the $x$-coordinate of the crossing. Measure $t=\dfrac{x-l}{h}$ along the modal class. The first line rises from height $f_0$ to $f_1$, so its height at $t$ is $f_0+(f_1-f_0)t$. The second falls from $f_1$ to $f_2$, so its height is $f_1+(f_2-f_1)t$. At the crossing the heights are equal: $$f_0+(f_1-f_0)t=f_1+(f_2-f_1)t\ \Longrightarrow\ t\,(2f_1-f_0-f_2)=f_1-f_0\ \Longrightarrow\ t=\frac{f_1-f_0}{2f_1-f_0-f_2}.$$ So the mode is $l+th$, which is the formula. In words, the modal class is split in the ratio $(f_1-f_0):(f_1-f_2)$, and each neighbour pulls in proportion to how crowded it is. The denominator $2f_1-f_0-f_2=(f_1-f_0)+(f_1-f_2)$ is positive because $f_1$ is the largest frequency.

🧭 Problem-solving strategy — mode of grouped data
  1. Make sure the classes touch. If they are written as $10$–$19$, $20$–$29$, subtract $0.5$ from each lower limit and add $0.5$ to each upper limit ($9.5$–$19.5$, $19.5$–$29.5$).
  2. Find the highest frequency $f_1$: that row is the modal class.
  3. Read off $l$ (its lower limit), $h$ (class size), $f_0$ (the class listed just before it) and $f_2$ (class after it).
  4. Substitute in $l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\,h$. Check that the answer lies between $l$ and $l+h$.
⚠️ Common mistakes

1. Swapping $f_0$ and $f_2$. $f_0$ is the class before the modal class (smaller values), $f_2$ the class after. A swap gives a wrong answer that still looks reasonable.

2. Reading $l$ from the wrong class. $l$ is the lower limit of the modal class, not of the class before.

3. Using the formula for non-touching classes. Convert $10$–$19$ type classes first.

4. Forgetting $h$ at the end, or writing $2f_1-f_0+f_2$ in the denominator. The answer must fall inside the modal class; if not, recheck.

✏️ Example 2 — Mode from the formula
The travel times (in minutes) of $56$ commuters are: $0$–$10$: $4$; $10$–$20$: $12$; $20$–$30$: $20$; $30$–$40$: $14$; $40$–$50$: $6$. Find the modal class and the mode.
Modal class The highest frequency is $20$, so the modal class is $20$–$30$.
Values $l=20$, $h=10$, $f_1=20$, $f_0=12$ (class $10$–$20$), $f_2=14$ (class $30$–$40$).
Substitute $\text{Mode}=20+\dfrac{20-12}{2(20)-12-14}\times10=20+\dfrac{8}{14}\times10=20+5.71=25.71$.
Answer The mode is about 25.71 minutes.
Check $25.71$ is inside $20$–$30$ ✓, and a little above the middle ($25$ is the midpoint): the class after ($14$) is more crowded than the class before ($12$), so it pulls the mode up, past $25$ ✓.
✏️ Example 3 — Museum visitors
A museum counted its daily visitors over $60$ days: $100$–$150$: $5$; $150$–$200$: $13$; $200$–$250$: $22$; $250$–$300$: $12$; $300$–$350$: $8$. Find the mode.
Values Highest frequency $22$, so the modal class is $200$–$250$: $l=200$, $h=50$, $f_1=22$, $f_0=13$, $f_2=12$.
Substitute $\text{Mode}=200+\dfrac{22-13}{44-13-12}\times50=200+\dfrac{9}{19}\times50=200+23.68=223.68$.
Answer About 224 visitors a day is the most typical figure.
Interactive 1 — Modal-Class Finder

Change the frequencies with the sliders (or press Play). The tallest bar is the modal class ($f_1$, amber); the bar before it is $f_0$ (red) and the bar after it is $f_2$ (blue). The green lines are the construction from the proof: where they cross, the mode is read straight down onto the axis. Check that this equals the formula. (If two bars tie for the top, change a slider so that one class is clearly highest.)

modal class—
$f_0,\ f_1,\ f_2$—
formula—
mode—

Special cases and missing frequencies

Two situations come up often. First, the neighbours are equal: if $f_0=f_2$ the formula gives $t=\dfrac{f_1-f_0}{2(f_1-f_0)}=\dfrac12$, so the mode is exactly the middle of the modal class, its class mark. Second, the modal class is the first or the last class, so one neighbour is missing. A missing neighbour is treated as an empty class, with frequency $0$: use $f_0=0$ for the first class (or $f_2=0$ for the last).

✏️ Example 4 — Classes that do not touch
The typing speeds (words per minute) of $45$ trainees are: $30$–$39$: $4$; $40$–$49$: $9$; $50$–$59$: $18$; $60$–$69$: $11$; $70$–$79$: $3$. Find the mode.
Make continuous The gap between $39$ and $40$ is $1$, so subtract $0.5$ from every lower limit and add $0.5$ to every upper limit. The classes become $29.5$–$39.5$, $39.5$–$49.5$, $49.5$–$59.5$, $59.5$–$69.5$, $69.5$–$79.5$. The class size is still $h=10$.
Values Modal class $49.5$–$59.5$: $l=49.5$, $f_1=18$, $f_0=9$, $f_2=11$.
Substitute $\text{Mode}=49.5+\dfrac{18-9}{36-9-11}\times10=49.5+\dfrac{9}{16}\times10=49.5+5.625=55.125$.
Answer The mode is about 55.13 words per minute.
✏️ Example 5 — The modal class is the first class
Daily rainfall (mm) over $35$ days: $0$–$5$: $15$; $5$–$10$: $10$; $10$–$15$: $6$; $15$–$20$: $4$. Find the mode.
Modal class Highest frequency $15$: class $0$–$5$, so $l=0$, $h=5$, $f_1=15$.
Neighbours There is no class before it, so $f_0=0$. The class after has $f_2=10$.
Substitute $\text{Mode}=0+\dfrac{15-0}{30-0-10}\times5=\dfrac{15}{20}\times5=3.75$ mm.
✏️ Example 6 — Finding a missing frequency from the mode
The mode of the distribution below is $34$. Find the missing frequency $p$.
Class$10$–$20$$20$–$30$$30$–$40$$40$–$50$$50$–$60$
Frequency51016$p$3
Modal class The mode $34$ lies in $30$–$40$, so this is the modal class: $l=30$, $h=10$, $f_1=16$, $f_0=10$, $f_2=p$.
Equation $34=30+\dfrac{16-10}{32-10-p}\times10$, so $4=\dfrac{60}{22-p}$ and $22-p=15$.
Solve $p=7$.
Check With $p=7$, $16$ is still the largest frequency ✓, and $30+\dfrac{6}{15}\times10=34$ ✓.
Interactive 2 — The Tug of War Inside the Modal Class

The modal class is $20$–$30$ with $f_1=20$. Slide the frequency $f_0$ of the class before and $f_2$ of the class after. The class is cut into two pieces in the ratio $(f_1-f_0):(f_1-f_2)$: the red piece shows how far the mode is from the lower limit, the blue piece how far from the upper limit. A crowded neighbour (frequency close to $20$) pulls the mode towards itself. When $f_0=f_2$ the mode sits exactly in the middle, at $25$.

1214
$f_1-f_0$—
$f_1-f_2$—
fraction of the class—
mode—

Exercises

I
Mode and modal class.
  • 1 Find the mode of $3,5,5,7,8,5,9,3$.
  • 2 Find the mode of $10,12,12,14,15,15,15,16$.
  • 3 Find the modal class: $0$–$5$: $3$; $5$–$10$: $9$; $10$–$15$: $14$; $15$–$20$: $8$.
  • 4 Evaluate $l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\,h$ for $l=40$, $h=5$, $f_0=6$, $f_1=14$, $f_2=10$ (to $2$ d.p.).
  • 5 Why can the denominator $2f_1-f_0-f_2$ never be zero when $f_1$ is the largest frequency?
1 $5$ occurs three times, more than any other value: mode $=5$.
2 $15$ occurs three times, $12$ twice: mode $=15$.
3 The highest frequency is $14$: modal class $10$–$15$.
4 $40+\dfrac{14-6}{28-6-10}\times5=40+\dfrac{8}{12}\times5=40+3.33=43.33$.
5 $2f_1-f_0-f_2=(f_1-f_0)+(f_1-f_2)$, and both brackets are positive because $f_1$ is larger than $f_0$ and than $f_2$. A sum of two positive numbers is not zero.
II
Mode of grouped data. Find the mode of each distribution.
  • 1 $10$–$20$: $2$; $20$–$30$: $6$; $30$–$40$: $11$; $40$–$50$: $8$; $50$–$60$: $3$.
  • 2 Heights (cm): $140$–$145$: $4$; $145$–$150$: $9$; $150$–$155$: $15$; $155$–$160$: $10$; $160$–$165$: $2$.
  • 3 Daily wages (₹): $400$–$500$: $6$; $500$–$600$: $10$; $600$–$700$: $14$; $700$–$800$: $12$; $800$–$900$: $8$.
  • 4 Pocket money (₹): $0$–$50$: $8$; $50$–$100$: $14$; $100$–$150$: $20$; $150$–$200$: $12$; $200$–$250$: $6$.
  • 5 Pages read: $1$–$10$: $6$; $11$–$20$: $9$; $21$–$30$: $10$; $31$–$40$: $5$.
  • 6 Rainfall (mm): $0$–$4$: $18$; $4$–$8$: $10$; $8$–$12$: $7$; $12$–$16$: $5$.
1 Modal class $30$–$40$: $l=30$, $h=10$, $f_1=11$, $f_0=6$, $f_2=8$. Mode $=30+\dfrac{5}{22-14}\times10=30+6.25=36.25$.
2 Modal class $150$–$155$: $l=150$, $h=5$, $f_1=15$, $f_0=9$, $f_2=10$. Mode $=150+\dfrac{6}{30-19}\times5=150+2.73=152.73$ cm.
3 Modal class $600$–$700$: $l=600$, $h=100$, $f_1=14$, $f_0=10$, $f_2=12$. Mode $=600+\dfrac{4}{28-22}\times100=600+66.67=$ ₹ 666.67.
4 Modal class $100$–$150$: $l=100$, $h=50$, $f_1=20$, $f_0=14$, $f_2=12$. Mode $=100+\dfrac{6}{40-26}\times50=100+21.43=$ ₹ 121.43.
5 Make the classes continuous: $0.5$–$10.5$, $10.5$–$20.5$, $20.5$–$30.5$, $30.5$–$40.5$. Modal class $20.5$–$30.5$: $l=20.5$, $h=10$, $f_1=10$, $f_0=9$, $f_2=5$. Mode $=20.5+\dfrac{1}{20-14}\times10=20.5+1.67=22.17$ pages.
6 Modal class $0$–$4$ (first class): $l=0$, $h=4$, $f_1=18$, $f_0=0$, $f_2=10$. Mode $=\dfrac{18}{36-10}\times4=\dfrac{72}{26}=2.77$ mm.
III
Missing frequencies, special cases and applications.
  • 1 The mode of: $20$–$30$: $5$; $30$–$40$: $9$; $40$–$50$: $15$; $50$–$60$: $q$; $60$–$70$: $4$ is $46$. Find $q$.
  • 2 A modal class $10$–$20$ has $f_0=14$, $f_1=20$, $f_2=6$. Find the mode and say whether it is above or below the class mark $15$, and why.
  • 3 A modal class $40$–$50$ has $f_0=f_2=8$ and $f_1=20$. Show that the mode is $45$.
  • 4 Ages of $80$ patients in a clinic: $5$–$15$: $6$; $15$–$25$: $11$; $25$–$35$: $21$; $35$–$45$: $23$; $45$–$55$: $14$; $55$–$65$: $5$. Find the modal age (to $2$ d.p.).
  • 5 Speeds of $55$ vehicles on a highway (km/h): $40$–$50$: $6$; $50$–$60$: $14$; $60$–$70$: $23$; $70$–$80$: $9$; $80$–$90$: $3$. Find the modal speed.
1 Modal class $40$–$50$: $l=40$, $h=10$, $f_1=15$, $f_0=9$, $f_2=q$. $46=40+\dfrac{6}{30-9-q}\times10$, so $\dfrac{60}{21-q}=6$, $21-q=10$, $q=11$. (Check: $15$ is still the largest ✓.)
2 Mode $=10+\dfrac{20-14}{40-20}\times10=10+3=13$. It is below $15$: the class before ($14$) is much more crowded than the class after ($6$), so it pulls the mode down.
3 Mode $=40+\dfrac{20-8}{40-16}\times10=40+\dfrac{12}{24}\times10=45$.
4 Modal class $35$–$45$: $l=35$, $h=10$, $f_1=23$, $f_0=21$, $f_2=14$. Mode $=35+\dfrac{2}{46-35}\times10=35+1.82=36.82$ years.
5 Modal class $60$–$70$: $l=60$, $h=10$, $f_1=23$, $f_0=14$, $f_2=9$. Mode $=60+\dfrac{9}{46-23}\times10=60+3.91=63.91$ km/h.
IV
Think harder.
  • 1 Show that the mode given by the formula always lies in the modal class, that is, $l\le\text{Mode}\le l+h$.
  • 2 A modal class $40$–$50$ has $f_0=10$ and $f_2=12$ and the mode is $45.5$. Find $f_1$.
  • 3 If the first class is the modal class, $f_0=0$ and $f_2=0$ (a single bar), where is the mode? Explain.
1 Mode $-\,l=\dfrac{A}{A+B}h$ with $A=f_1-f_0\gt0$ and $B=f_1-f_2\gt0$. The fraction $\dfrac{A}{A+B}$ is strictly between $0$ and $1$, so $l\lt\text{Mode}\lt l+h$.
2 $45.5=40+\dfrac{f_1-10}{2f_1-22}\times10$, so $5.5(2f_1-22)=10(f_1-10)$, i.e. $11f_1-121=10f_1-100$, so $f_1=21$. (Check: $\dfrac{11}{20}\times10=5.5$ ✓; $21$ is larger than $10$ and $12$ ✓.)
3 Mode $=l+\dfrac{f_1}{2f_1}h=l+\dfrac h2$, the middle of the class: with no neighbour pulling either way, the mode is the class mark.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE Main

How this is examined: Boards ask for the mode of a table in 3 marks (identify the modal class, list $l,h,f_0,f_1,f_2$, substitute), a missing-frequency item in 2–3 marks, and 1-mark MCQs on the formula. Many marks are lost by swapping $f_0$ and $f_2$ or by forgetting to make classes continuous. JEE Main problems tend to give frequencies as a ratio.

E1. Board · 1 mark The modal class of a distribution is $20$–$30$. The class before it has frequency $8$, the modal class has frequency $15$, and the class after it has frequency $11$. The mode is  (A) $26.36$  (B) $25$  (C) $23.64$  (D) $30$

Formula $20+\dfrac{15-8}{30-8-11}\times10=20+\dfrac{7}{11}\times10=26.36$.
Answer (A). Option (C) comes from swapping $f_0$ and $f_2$ ($20+\frac{4}{11}\times10$); option (B) is just the class mark.

E2. Board · 1 mark Assertion (A): If the modal class is $30$–$40$ and the frequencies of the classes before and after it are equal, the mode is $35$. Reason (R): When $f_0=f_2$, the mode formula gives $l+\dfrac h2$.
(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.

Check R With $f_0=f_2$: $\dfrac{f_1-f_0}{2f_1-2f_0}=\dfrac12$, so mode $=l+\dfrac h2$: true.
Check A $l=30$, $h=10$: mode $=35$: true, and it follows from R.
Answer (A).

E3. Board · 3 marks Find the mode: $15$–$20$: $4$; $20$–$25$: $11$; $25$–$30$: $19$; $30$–$35$: $14$; $35$–$40$: $2$.

Modal class (1 mark) Highest frequency $19$: modal class $25$–$30$, with $l=25$, $h=5$, $f_1=19$, $f_0=11$, $f_2=14$.
Formula (1 mark) $\text{Mode}=l+\dfrac{f_1-f_0}{2f_1-f_0-f_2}h=25+\dfrac{19-11}{38-11-14}\times5$.
Answer (1 mark) $=25+\dfrac{8}{13}\times5=25+3.08=28.08$.

E4. Board · 2 marks The mode of the following distribution is $56$. Find $p$: $30$–$40$: $5$; $40$–$50$: $12$; $50$–$60$: $18$; $60$–$70$: $p$; $70$–$80$: $6$.

Set up (1 mark) Mode $56$ lies in $50$–$60$: $l=50$, $h=10$, $f_1=18$, $f_0=12$, $f_2=p$. $56=50+\dfrac{6}{36-12-p}\times10$.
Solve (1 mark) $\dfrac{60}{24-p}=6$, so $24-p=10$ and $p=14$. (Check: $18$ is still the largest frequency ✓.)

E5. Board · Case study A bank manager recorded the waiting times (in minutes) of $60$ customers: $0$–$5$: $5$; $5$–$10$: $14$; $10$–$15$: $21$; $15$–$20$: $12$; $20$–$25$: $8$. (i) Write the modal class and the values of $l$, $h$, $f_0$, $f_1$, $f_2$. (2) (ii) Find the mode. (1) (iii) How many customers waited $15$ minutes or more, and what fraction of all customers is that? (1)

(i) (2 marks) Modal class $10$–$15$; $l=10$, $h=5$, $f_0=14$, $f_1=21$, $f_2=12$.
(ii) (1 mark) $\text{Mode}=10+\dfrac{21-14}{42-14-12}\times5=10+\dfrac{7}{16}\times5=12.19$ minutes.
(iii) (1 mark) $12+8=20$ customers, and $\dfrac{20}{60}=\dfrac13$ of all customers.

E6. JEE Main For a grouped distribution with class size $10$, the modal class is $20$–$30$ and the frequencies of the preceding, modal and succeeding classes are in the ratio $2:5:3$. The mode is  (A) $24$  (B) $25$  (C) $26$  (D) $27.5$

Set up Let $f_0=2k$, $f_1=5k$, $f_2=3k$ with $k\gt0$. The factor $k$ cancels in the formula.
Substitute $20+\dfrac{5k-2k}{10k-2k-3k}\times10=20+\dfrac{3}{5}\times10=26$.
Answer (C). Swapping $f_0$ and $f_2$ gives $20+\frac{2}{5}\times10=24$, option (A); (B) is the class mark.
🔬 Research–Teaching NexusMini project · optional, about 40 minutes

Research question. If you measure the same thing for 40 people, how does the mode you find from the raw values compare with the mode from the grouped formula, and how does it change with the class size?

Why it matters. Shops decide what stock to order, cities decide when to run extra buses, and schools decide how many sections to open by looking at the most common value, not the average. When the raw data are unavailable, the grouped mode is the best estimate.

What to do.

  1. Collect one whole-number measurement from 40 people, such as minutes spent travelling to school.
  2. Make a tally of the raw values and note the most frequent value (if several values tie, group first).
  3. Group the values with class size $5$, find the modal class and compute the mode with the formula.
  4. Repeat with class size $10$.
  5. Draw both histograms (a spreadsheet chart will do) and mark the mode on each.

Tools: A notebook or a spreadsheet, a calculator

Hand in: The raw data, both frequency tables, the two modes, the histograms, and a short comment on which class size told you more.

Going further. Compare the mode with the mean of the same grouped data (Section 13.1). Which value would the bus company prefer, and which would a student planning her morning prefer?

Teacher/parent note: raw data often have no single most frequent value, which is exactly why the grouped formula is useful. Wide classes hide detail, narrow classes make the histogram ragged; there is no single best class size.

🧪 Self-check quiz

0/5 correct

Q1. The mode of $4,6,6,7,6,9,4$ is

Q2. In the table $0$–$10$: $3$; $10$–$20$: $11$; $20$–$30$: $8$; $30$–$40$: $5$, the modal class is

Q3. A modal class $40$–$50$ has $f_0=f_2=7$ and $f_1=15$. The mode is

Q4. For $l=10$, $h=10$, $f_0=4$, $f_1=10$, $f_2=6$ the mode is

Q5. Which statement about the mode of grouped data is correct?

← 13.1 Mean of Grouped Data13.3 Median of Grouped Data →