Statistics: Class 10 Maths · Chapter 13

Mean, mode and median when the data come in classes

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

60 min
31 practice questions
14 MCQs

What you will learn in Statistics

Find the mean of grouped data by the direct, assumed-mean and step-deviation methods, and pick the quickest one
Explain why a mean worked out from class marks is only an estimate of the true mean
Identify the modal class and calculate the mode of grouped data with the mode formula
Build less-than and more-than cumulative frequency tables and use them to find the median class
Calculate the median of grouped data, including from a less-than table and with missing frequencies
Decide whether the mean, the median or the mode best answers a question, and use the empirical relation between them

Original companion text aligned to NCERT Mathematics Class 10, Chapter 13, pp. 171–201. Not affiliated with or endorsed by NCERT.

13.1 Introduction

A board result sheet lists the marks of thousands of students. A weather office logs the rainfall at hundreds of stations every day. A cricket board stores the total of every match ever played. Nobody reads such data one value at a time. We sort the values into classes first, and then ask one question: which single number best stands for the whole lot?

In Class 9 you met three answers, the measures of central tendency (केंद्रीय प्रवृत्ति के माप): the mean, the median and the mode. You found them for ungrouped data, where every value is known. This chapter finds all three for grouped data, where we know only the classes and how many values fall in each. Along the way you will use cumulative frequency, a running total that tells you where the middle of the data lies.

Warm-up: averages of ungrouped data

1 of 3

Find the mean of 6, 9, 9, 11 and 15.

Show answer

Answer 10

The sum is 50 and there are 5 values, so the mean is

50 \div 5 = 10.

Dividing by 4 (the number of different values) would give 12.5, which is wrong.

13.2 Mean of Grouped Data

Suppose five workers each earn ₹ 450 a day. Together they earn $5 \times 450 = 2250$ rupees: a value that repeats is added in one go by multiplying it by its frequency. The same shortcut works for any table: multiply each value $x_{i}$ by its frequency $f_{i},$ add the products to get the grand total, and divide by the total frequency. We write $Σ$ (the Greek capital sigma) for “add up all of these”, which keeps the formula short.

Mean of data with frequencies

\overset{ˉ}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}

(13.1)

where

$\overset{ˉ}{x}$: the mean, read “x bar”
$f_{i}$: the frequency of the value $x_{i}$
$\sum$: the sum over $i = 1, 2, \dots, n$

Definition

Mean

The sum of all the observations divided by the number of observations. For data with frequencies it is $Σ f_{i} x_{i} \div Σ f_{i} .$

ExampleThree 4s and one 8 have mean $(3 \times 4 + 8) \div 4 = 5.$

Example 13.1NCERT Example 1 · p. 172

Daily wages on a building site

A contractor in Lucknow pays the 40 workers on a building site different daily wages, depending on their skill. The table lists each wage and the number of workers who get it. Find the mean daily wage.

Table 13.1Daily wages of 40 workers

Daily wage (₹)	Workers
360	1
380	2
400	2
420	4
450	5
480	7
500	6
520	3
550	2
580	4
600	3
640	1

Try it yourself first. Then reveal the solution one step at a time.

Real registers hold hundreds of wages, so they are first grouped into classes. Let us group the same 40 wages into classes of width ₹ 50: 350–400, 400–450, and so on up to 600–650. One rule settles the edges: a wage equal to an upper limit belongs to the next class, so the five workers on ₹ 450 go into 450–500.

2/3Lines every ₹ 50 split the wages into six classes.

Fig. 13.1Grouping keeps the counts but forgets the exact values.NCERT Table 13.2 · p. 173

A number line runs from 350 to 650 rupees. Step 1: forty dots, one per worker, are stacked above the wage each worker earns, from 360 up to 640. Step 2: dashed lines at 350, 400, 450 and so on split the line into six classes of width 50, and each class is labelled with its count of workers: 3, 6, 12, 9, 6 and 4. Step 3: the dots of each class slide sideways into one column above the middle of the class, at 375, 425, 475, 525, 575 and 625, showing that grouped data treats every value as if it sat at the class mark.

Table 13.2The 40 wages in classes of width ₹ 50

Daily wage (₹)	Workers
350–400	3
400–450	6
450–500	12
500–550	9
550–600	6
600–650	4
Total	40

Once the wages are grouped, the individual values are gone: the table only says that 12 workers earn something from ₹ 450 up to ₹ 500. Formula (13.1) still needs a value $x_{i}$ for every class. The fairest single guess is the middle of the class, its class mark (वर्ग चिह्न). Using it means pretending that the workers in each class are spread evenly about the middle: the mid-point assumption.

Definition

Class mark

The mid-point of a class: $\frac{1}{2} (lower limit + upper limit) .$

ExampleThe class mark of 450–500 is $\frac{1}{2} (450 + 500) = 475.$

With the class marks as the $x_{i},$ the mean is worked out exactly as before. Finding the mean straight from $Σ f_{i} x_{i}$ and $Σ f_{i}$ like this is called the direct method.

Table 13.3Direct method for the grouped wages

Class	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
350–400	3	375	1125
400–450	6	425	2550
450–500	12	475	5700
500–550	9	525	4725
550–600	6	575	3450
600–650	4	625	2500
Total	40	—	20,050

So $\overset{ˉ}{x} = 20,050 \div 40 = 501.25.$ By the direct method the mean wage is ₹ 501.25.

Assumed mean method

The direct method soon means big multiplications: $12 \times 475$ is already 5700. Here is a trick. Choose a convenient class mark near the middle, call it the assumed mean $a$ (कल्पित माध्य), and measure every class mark from it. The distances $d_{i} = x_{i} - a,$ called deviations, are small numbers, some of them negative, while the frequencies stay exactly as they were.

DerivationWhy the deviations lead back to the mean

Average the deviations, weighting each by its frequency:

\overset{ˉ}{d} = \frac{\sum f_{i} (x_{i} - a)}{\sum f_{i}}

Split the top into two sums. The constant $a$ comes out of the second sum, and then $Σ f_{i}$ cancels:

\begin{aligned} \overset{ˉ}{d} & = \frac{Σ f_{i} x_{i} - a Σ f_{i}}{Σ f_{i}} \\ = \frac{\sum f_{i} x_{i}}{\sum f_{i}} - a \\ = \overset{ˉ}{x} - a \end{aligned}

So the average deviation is the mean shifted down by $a .$ Adding $a$ back gives the assumed mean method:

Assumed mean method

\overset{ˉ}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}

(13.2)

For the wages, take $a = 525.$ The deviations are −150, −100, −50, 0, 50 and 100. Multiplying by the frequencies 3, 6, 12, 9, 6 and 4 and adding gives $Σ f_{i} d_{i} = - 950.$ So $\overset{ˉ}{x} = 525 - 950 \div 40 = 525 - 23.75 = 501.25,$ the same ₹ 501.25 as before, with much smaller products.

Activity 13.1

Does the choice of a matter?

You’ll need

the grouped wages table
a calculator (optional)

Predict first

Ravi takes

a = 375

and Sana takes

a = 625

for the grouped wages. What will their means be?

Ravi's will be smallerSana's will be smallerboth will be ₹ 501.25

Take $a = 375$ and find each deviation $d_{i} = x_{i} - 375.$
Multiply each deviation by its frequency and add, to get $Σ f_{i} d_{i} .$
Work out $a + Σ f_{i} d_{i} \div 40.$
Repeat with each of the other class marks as $a .$

Table 13.4

$a$	$Σ f_{i} d_{i}$	Mean
375
425
475
525
575
625

Your entries are saved automatically.

Step-deviation method

Look at the deviations again: −150, −100, −50, 0, 50, 100. Every one is a multiple of 50, the class size $h$ (वर्ग माप). Dividing by $h$ shrinks them to tiny whole numbers:

u_{i} = \frac{x_{i} - a}{h}

DerivationFrom the average of the u's to the mean

Average the $u_{i}$ with the frequencies, and take the factor $\frac{1}{h}$ outside:

\begin{aligned} \overset{ˉ}{u} & = \frac{\sum f_{i} u_{i}}{\sum f_{i}} \\ = \frac{1}{h} \cdot \frac{\sum f_{i} d_{i}}{\sum f_{i}} \\ = \frac{1}{h} (\overset{ˉ}{x} - a) \end{aligned}

The second line uses the result of the assumed mean derivation. Multiply by $h$ and add $a$ to get $\overset{ˉ}{x} = a + h \overset{ˉ}{u} :$

Step-deviation method

\overset{ˉ}{x} = a + h \frac{\sum f_{i} u_{i}}{\sum f_{i}}

(13.3)

Table 13.5Step-deviation method for the wages, with

a = 525

and

h = 50

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
375	3	−3	−9
425	6	−2	−12
475	12	−1	−12
525	9	0	0
575	6	1	6
625	4	2	8
Total	40	—	−19

So $\overset{ˉ}{u} = - 19 \div 40 = - 0.475$ and $\overset{ˉ}{x} = 525 + 50 \times (- 0.475) = 501.25.$ The products are now single digits, and the answer has not moved. Fig. 13.2 shows the two shortcuts as a change of scale on the number line.

3/4Dividing by $h = 50$ gives tiny whole numbers $u_{i} .$

Fig. 13.2Subtract

a,

divide by

h,

average, then undo both steps.NCERT Table 13.5 · p. 176

Six columns of dots show the frequencies 3, 6, 12, 9, 6 and 4 of the wage classes. Step 1 labels the columns with their class marks 375 to 625. Step 2 subtracts the assumed mean 525, so the labels become the deviations −150, −100, −50, 0, 50 and 100, with 0 under the fourth column. Step 3 divides by 50, giving −3, −2, −1, 0, 1 and 2. Step 4 marks the balance point: the average of the u-values is −0.475, which converts back to a mean of 501.25 rupees, drawn as a marker just left of 525.

Example 13.2NCERT Example 2 · p. 178

T20 totals, three ways

The table shows the first-innings totals in 40 matches of a state T20 league. Find the mean total by all three methods.

Table 13.6

Total (runs)	Matches
100–120	3
120–140	5
140–160	9
160–180	12
180–200	7
200–220	4

Try it yourself first. Then reveal the solution one step at a time.

Example 13.3NCERT Example 3 · p. 180

Rainfall with unequal classes

Rain gauges at 50 stations in Maharashtra recorded the following rainfall in the first week of July. The classes are not all the same width. Find the mean rainfall.

Table 13.7

Rainfall (mm)	Stations
0–40	4
40–80	7
80–120	12
120–200	15
200–320	8
320–400	4

Try it yourself first. Then reveal the solution one step at a time.

Activity 13.2

Collect, group and average

You’ll need

a notebook
a measuring tape
a bathroom or luggage scale

Form three teams.
Team A asks every classmate how many minutes the journey to school takes.
Team B weighs the school bag of every classmate, in kg to one decimal place.
Team C measures the arm span of every classmate in cm, fingertip to fingertip with arms stretched out.
Each team chooses 5 to 7 classes of equal width, tallies its data and makes a grouped frequency table.
Each team finds the mean by the method that suits its numbers, and says why it chose that method.

Table 13.8

Team	Method	Mean
A
B
C

Your entries are saved automatically.

Mean of grouped data

1 of 4

What is the class mark of the class 24.5–29.5?

Show answer

Answer 27

(24.5 + 29.5) \div 2 = 27.

The class size, 5, is a different quantity.

Practice · 9 questions

Practice 13.1: questions with answers

1
NumericalEasyNCERT Exercise 13.1, Q1
A panchayat recorded the hours of power cut in one week in 25 villages.
Power cut (hours) Villages
0–2 3
2–4 6
4–6 8
6–8 4
8–10 3
10–12 1
1. Find the mean number of hours of power cut per village.
  h
  Show solution
  Class $f_{i}$ $x_{i}$ $f_{i} x_{i}$
  0–2 3 1 3
  2–4 6 3 18
  4–6 8 5 40
  6–8 4 7 28
  8–10 3 9 27
  10–12 1 11 11
  Total 25 — 127
  $\overset{ˉ}{x} = 127 \div 25 = 5.08$ hours.
2. Which method did you use, and why?
  Show solution
  The direct method. The class marks and frequencies are small, so the products $f_{i} x_{i}$ are easy and a shortcut saves nothing.
NumericalMediumNCERT Exercise 13.1, Q2
The monthly rents paid by 40 students for paying-guest rooms near a college are grouped below. Find the mean monthly rent by a suitable method.
Rent (₹) Students
4000–5000 5
5000–6000 9
6000–7000 12
7000–8000 8
8000–9000 6
₹
Show solution
The class marks 4500, 5500, …, 8500 are large and evenly spaced, so use step deviation with $a = 6500,$ $h = 1000.$
$x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
4500 5 −2 −10
5500 9 −1 −9
6500 12 0 0
7500 8 1 8
8500 6 2 12
Total 40 — 1
$\overset{ˉ}{x} = 6500 + 1000 \times \frac{1}{40} = 6525.$ The mean rent is ₹ 6525.
NumericalMediumNCERT Exercise 13.1, Q3
A group of senior citizens noted the minutes they spend walking in a park each morning. The mean time is 37 minutes. Find the missing frequency $k .$
Time (min) People
10–20 5
20–30 8
30–40 $k$
40–50 12
50–60 7
Show solution
Class marks: 15, 25, 35, 45, 55. So $Σ f_{i} = 32 + k$ and $Σ f_{i} x_{i} = 75 + 200 + 35 k + 540 + 385 = 1200 + 35 k .$
$1200 + 35 k = 37 (32 + k)$
$1200 + 35 k = 1184 + 37 k$
So $2 k = 16$ and $k = 8.$
NumericalMediumNCERT Exercise 13.1, Q4
On sports day, 36 students ran the 100 m. Their times are grouped below. Find the mean time, choosing a suitable method.
Time (s) Students
12–13 2
13–14 4
14–15 7
15–16 9
16–17 8
17–18 4
18–19 2
s
Show solution
The class marks 12.5, 13.5, …, 18.5 are decimals 1 apart, so step deviation with $a = 15.5,$ $h = 1$ turns them into −3 to 3.
$x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
12.5 2 −3 −6
13.5 4 −2 −8
14.5 7 −1 −7
15.5 9 0 0
16.5 8 1 8
17.5 4 2 8
18.5 2 3 6
Total 36 — 1
$\overset{ˉ}{x} = 15.5 + 1 \times \frac{1}{36} \approx 15.53$ s.
5
NumericalMediumNCERT Exercise 13.1, Q5
A quality inspector counted the pages in 160 notebooks.
Pages Notebooks
96–98 12
99–101 38
102–104 60
105–107 34
108–110 16
1. Find the mean number of pages per notebook.
  Show solution
  Class marks: 97, 100, 103, 106, 109, which are 3 apart. Take $a = 103,$ $h = 3.$
  $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  97 12 −2 −24
  100 38 −1 −38
  103 60 0 0
  106 34 1 34
  109 16 2 32
  Total 160 — 4
  $\overset{ˉ}{x} = 103 + 3 \times \frac{4}{160} = 103.075 \approx 103.08$ pages.
2. Which method did you choose, and why?
  Show solution
  Step deviation: the frequencies are large and the class marks are equally spaced, so dividing the deviations by 3 keeps every product small.
NumericalMediumNCERT Exercise 13.1, Q6
The litres of milk sold in a day at 30 dairy booths in a town are grouped below. Find the mean quantity sold per booth.
Milk (L) Booths
40–60 3
60–80 5
80–100 9
100–120 7
120–140 4
140–160 2
L
Show solution
$x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
50 3 −2 −6
70 5 −1 −5
90 9 0 0
110 7 1 7
130 4 2 8
150 2 3 6
Total 30 — 10
$\overset{ˉ}{x} = 90 + 20 \times \frac{10}{30} \approx 96.67$ litres.
7
NumericalMediumNCERT Exercise 13.1, Q7
The fluoride content of the water in 40 village wells is grouped below, in mg per litre.
Fluoride (mg/L) Wells
0.0–0.3 5
0.3–0.6 9
0.6–0.9 12
0.9–1.2 7
1.2–1.5 4
1.5–1.8 3
1. Find the mean fluoride content.
  mg/L
  Show solution
  $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  0.15 5 −2 −10
  0.45 9 −1 −9
  0.75 12 0 0
  1.05 7 1 7
  1.35 4 2 8
  1.65 3 3 9
  Total 40 — 5
  $\overset{ˉ}{x} = 0.75 + 0.3 \times \frac{5}{40} = 0.7875 \approx 0.79$ mg/L.
2. The safe limit for drinking water is 1.5 mg/L. The mean is below it. Does that mean every well is safe?
  Show solution
  No. The mean only describes the wells as a group. Three wells are in the class 1.5–1.8, at or above the limit, so they need checking even though the mean is low.
NumericalHardNCERT Exercise 13.1, Q8
The days of leave taken in a year by 50 employees of a bank branch are grouped below. The classes are not equal. Find the mean number of days of leave.
Days of leave Employees
0–4 6
4–8 12
8–12 14
12–20 10
20–30 5
30–36 3
days
Show solution
The class marks 2, 6, 10, 16, 25, 33 are small and their deviations share no useful factor, so the direct method is simplest.
Class $f_{i}$ $x_{i}$ $f_{i} x_{i}$
0–4 6 2 12
4–8 12 6 72
8–12 14 10 140
12–20 10 16 160
20–30 5 25 125
30–36 3 33 99
Total 50 — 608
$\overset{ˉ}{x} = 608 \div 50 = 12.16$ days.
NumericalEasyNCERT Exercise 13.1, Q9
The Class 10 pass percentages of 40 schools in a district are grouped below. Find the mean pass percentage.
Pass percentage Schools
50–60 3
60–70 8
70–80 12
80–90 11
90–100 6
%
Show solution
$x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
55 3 −2 −6
65 8 −1 −8
75 12 0 0
85 11 1 11
95 6 2 12
Total 40 — 9
$\overset{ˉ}{x} = 75 + 10 \times \frac{9}{40} = 77.25,$ so the mean pass percentage is 77.25%.

Power cut (hours)	Villages
0–2	3
2–4	6
4–6	8
6–8	4
8–10	3
10–12	1

Class	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
0–2	3	1	3
2–4	6	3	18
4–6	8	5	40
6–8	4	7	28
8–10	3	9	27
10–12	1	11	11
Total	25	—	127

Rent (₹)	Students
4000–5000	5
5000–6000	9
6000–7000	12
7000–8000	8
8000–9000	6

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
4500	5	−2	−10
5500	9	−1	−9
6500	12	0	0
7500	8	1	8
8500	6	2	12
Total	40	—	1

Time (min)	People
10–20	5
20–30	8
30–40	$k$
40–50	12
50–60	7

Time (s)	Students
12–13	2
13–14	4
14–15	7
15–16	9
16–17	8
17–18	4
18–19	2

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
12.5	2	−3	−6
13.5	4	−2	−8
14.5	7	−1	−7
15.5	9	0	0
16.5	8	1	8
17.5	4	2	8
18.5	2	3	6
Total	36	—	1

Pages	Notebooks
96–98	12
99–101	38
102–104	60
105–107	34
108–110	16

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
97	12	−2	−24
100	38	−1	−38
103	60	0	0
106	34	1	34
109	16	2	32
Total	160	—	4

Milk (L)	Booths
40–60	3
60–80	5
80–100	9
100–120	7
120–140	4
140–160	2

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
50	3	−2	−6
70	5	−1	−5
90	9	0	0
110	7	1	7
130	4	2	8
150	2	3	6
Total	30	—	10

Fluoride (mg/L)	Wells
0.0–0.3	5
0.3–0.6	9
0.6–0.9	12
0.9–1.2	7
1.2–1.5	4
1.5–1.8	3

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
0.15	5	−2	−10
0.45	9	−1	−9
0.75	12	0	0
1.05	7	1	7
1.35	4	2	8
1.65	3	3	9
Total	40	—	5

Days of leave	Employees
0–4	6
4–8	12
8–12	14
12–20	10
20–30	5
30–36	3

Class	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
0–4	6	2	12
4–8	12	6	72
8–12	14	10	140
12–20	10	16	160
20–30	5	25	125
30–36	3	33	99
Total	50	—	608

Pass percentage	Schools
50–60	3
60–70	8
70–80	12
80–90	11
90–100	6

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
55	3	−2	−6
65	8	−1	−8
75	12	0	0
85	11	1	11
95	6	2	12
Total	40	—	9

13.3 Mode of Grouped Data

A shoe shop has no use for the fact that its customers' average size is 7.4; there is no such shoe. It wants the size that sells most often. That is the mode (बहुलक), the value with the highest frequency. Sometimes two or more values tie for the top frequency, and the data are called bimodal or multimodal. Every data set in this chapter has a single mode.

Example 13.4NCERT Example 4 · p. 183

Shoe sizes

A shop sold 15 pairs of school shoes on a Saturday, in these sizes: 8, 6, 9, 7, 8, 10, 7, 8, 5, 9, 8, 6, 7, 8, 9. Find the modal size.

Try it yourself first. Then reveal the solution one step at a time.

Once data are grouped, no single value can be seen repeating. What the table does show is the busiest class, the one with the highest frequency: the modal class (बहुलक वर्ग). The mode is somewhere in it, and the neighbouring classes decide where. If the class just before the modal class is nearly as full and the class after it is thin, the values are probably crowded towards the lower end, so the mode sits nearer the lower limit.

Mode of grouped data

\begin{aligned} Mode = l + \\ \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times h \end{aligned}

(13.4)

where

$l$: lower limit of the modal class
$h$: class size (all classes equal)
$f_{1}$: frequency of the modal class
$f_{0}$: frequency of the class before it
$f_{2}$: frequency of the class after it

Fig. 13.3 shows where the formula comes from. Join each top corner of the modal bar to the top of the neighbouring bar on the far side, so that the two lines cross. The crossing point lies above the mode. The steeper the climb from $f_{0}$ to $f_{1}$ compared with the drop to $f_{2},$ the further right the mode.

3/4Join each top corner to the far neighbour's top.

Fig. 13.3The two crossing lines lean towards the taller neighbour.

A histogram shows finishing times from 20 to 50 minutes in classes of 5 minutes, with heights 4, 9, 16, 11, 6 and 4. Step 1 shows only the bars. Step 2 highlights the tallest bar, 30 to 35, as the modal class and labels the heights f0 = 9 before it, f1 = 16 and f2 = 11 after it. Step 3 draws one line from the top of the 25 to 30 bar at 30 up to the top right corner of the modal bar, and another from the top left corner of the modal bar down to the top of the 35 to 40 bar at 35; they cross inside the modal bar. Step 4 drops a dashed line from the crossing to the axis at about 32.9 minutes, the mode.

Example 13.5NCERT Example 5 · p. 184

Fun-run finishing times

Fifty runners finished a 5 km fun run in Bengaluru. Their times are grouped below. Find the modal finishing time.

Table 13.9

Time (min)	Runners
20–25	4
25–30	9
30–35	16
35–40	11
40–45	6
45–50	4

Try it yourself first. Then reveal the solution one step at a time.

Example 13.6NCERT Example 6 · p. 185

Wages: mode against mean

Find the mode of the grouped wages in Table 13.2. Compare it with the mean, ₹ 501.25, and say what each tells the contractor.

Try it yourself first. Then reveal the solution one step at a time.

Do not expect the mode to sit below the mean every time: data that trail off to the left put it above the mean, and symmetric data make the two equal. Choosing between them depends on the question being asked. A contractor budgeting the total wage bill needs the mean. A labour officer asking what a typical worker on the site is paid would quote the mode. One more limit: the formula assumes all classes have the same width. A mode can be found for unequal classes too, but that is beyond this chapter.

Mode of grouped data

1 of 3

Marks of a class: 0–10: 5 students, 10–20: 12, 20–30: 9, 30–40: 3. Which is the modal class?

10–2030–4020–300–10

Show answer

Answer (A) 10–20

10–20 has the highest frequency, 12. 30–40 has the highest marks but the fewest students.

Practice · 6 questions

Practice 13.2: questions with answers

1
NumericalMediumNCERT Exercise 13.2, Q1
The ages of 80 people examined at a free eye camp are grouped below.
Age (years) People
10–20 5
20–30 8
30–40 10
40–50 14
50–60 19
60–70 16
70–80 8
1. Find the mode.
  years
  Show solution
  Modal class 50–60: $l = 50,$ $f_{1} = 19,$ $f_{0} = 14,$ $f_{2} = 16,$ $h = 10.$
  $\begin{aligned} Mode \\ = 50 + \frac{5}{38 - 30} \times 10 \\ = 56.25 \end{aligned}$
2. Find the mean.
  years
  Show solution
  $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  15 5 −3 −15
  25 8 −2 −16
  35 10 −1 −10
  45 14 0 0
  55 19 1 19
  65 16 2 32
  75 8 3 24
  Total 80 — 34
  $\overset{ˉ}{x} = 45 + 10 \times \frac{34}{80} = 49.25$ years.
3. Compare and interpret the two values.
  Show solution
  More people came from around age 56 than from any other age, while the average age was about 49. The younger visitors pull the mean below the mode. For planning reading-glasses stock, the mode is the more useful figure.
NumericalEasyNCERT Exercise 13.2, Q2
A shop tested the backup time of 180 inverter batteries. Find the modal backup time.
Backup (min) Batteries
60–90 12
90–120 28
120–150 46
150–180 52
180–210 30
210–240 12
min
Show solution
$l = 150,$ $h = 30,$ $f_{1} = 52,$ $f_{0} = 46,$ $f_{2} = 30.$
$\begin{aligned} Mode \\ = 150 + \frac{6}{104 - 76} \times 30 \end{aligned}$
$= 150 + \frac{180}{28} \approx 156.43$ minutes.
3
NumericalMediumNCERT Exercise 13.2, Q3
The monthly electricity bills of 150 homes in a colony are grouped below.
Bill (₹) Homes
500–1000 14
1000–1500 30
1500–2000 38
2000–2500 27
2500–3000 20
3000–3500 13
3500–4000 8
1. Find the modal monthly bill.
  ₹
  Show solution
  Modal class 1500–2000: $l = 1500,$ $f_{1} = 38,$ $f_{0} = 30,$ $f_{2} = 27,$ $h = 500.$
  $\begin{aligned} Mode \\ = 1500 + \frac{8}{76 - 57} \times 500 \end{aligned}$
  $= 1500 + \frac{4000}{19} \approx 1710.53,$ so about ₹ 1710.53.
2. Find the mean monthly bill.
  ₹
  Show solution
  $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
  750 14 −3 −42
  1250 30 −2 −60
  1750 38 −1 −38
  2250 27 0 0
  2750 20 1 20
  3250 13 2 26
  3750 8 3 24
  Total 150 — −70
  $\overset{ˉ}{x} = 2250 + 500 \times \frac{- 70}{150} \approx 2016.67,$ so about ₹ 2016.67.
4
NumericalMediumNCERT Exercise 13.2, Q4
A survey of 40 schools recorded the number of students sharing each computer. Two classes are empty.
Per computer Schools
5–10 4
10–15 9
15–20 13
20–25 8
25–30 3
30–35 0
35–40 0
40–45 3
1. Find the mode.
  Show solution
  Modal class 15–20: $l = 15,$ $f_{1} = 13,$ $f_{0} = 9,$ $f_{2} = 8,$ $h = 5.$
  $\begin{aligned} Mode \\ = 15 + \frac{4}{26 - 17} \times 5 \\ \approx 17.22 \end{aligned}$
2. Find the mean.
  Show solution
  Class marks 7.5, 12.5, …, 42.5. The products $f_{i} x_{i}$ are 30, 112.5, 227.5, 180, 82.5, 0, 0 and 127.5, which add up to 760.
  $\overset{ˉ}{x} = 760 \div 40 = 19.$
3. Interpret the two measures.
  Show solution
  In the most common situation about 17 students share a computer, but the average is 19. The three very crowded schools (40–45) pull the mean up; they are the ones that most need new computers.
NumericalEasyNCERT Exercise 13.2, Q5
The distance covered in a day by 60 autorickshaws in Pune is grouped below. Find the mode.
Distance (km) Autos
40–60 5
60–80 11
80–100 18
100–120 14
120–140 8
140–160 4
km
Show solution
$l = 80,$ $h = 20,$ $f_{1} = 18,$ $f_{0} = 11,$ $f_{2} = 14.$
$\begin{aligned} Mode \\ = 80 + \frac{7}{36 - 25} \times 20 \end{aligned}$
$= 80 + \frac{140}{11} \approx 92.73$ km.
NumericalEasyNCERT Exercise 13.2, Q6
At a busy metro station, a student counted the passengers boarding one coach of each of 120 trains. Find the mode.
Passengers Trains
0–10 6
10–20 11
20–30 18
30–40 27
40–50 24
50–60 17
60–70 10
70–80 7
Show solution
$l = 30,$ $h = 10,$ $f_{1} = 27,$ $f_{0} = 18,$ $f_{2} = 24.$
$\begin{aligned} Mode \\ = 30 + \frac{9}{54 - 42} \times 10 \\ = 37.5 \end{aligned}$
The most common number of passengers boarding a coach is about 37 or 38.

Age (years)	People
10–20	5
20–30	8
30–40	10
40–50	14
50–60	19
60–70	16
70–80	8

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
15	5	−3	−15
25	8	−2	−16
35	10	−1	−10
45	14	0	0
55	19	1	19
65	16	2	32
75	8	3	24
Total	80	—	34

Backup (min)	Batteries
60–90	12
90–120	28
120–150	46
150–180	52
180–210	30
210–240	12

Bill (₹)	Homes
500–1000	14
1000–1500	30
1500–2000	38
2000–2500	27
2500–3000	20
3000–3500	13
3500–4000	8

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
750	14	−3	−42
1250	30	−2	−60
1750	38	−1	−38
2250	27	0	0
2750	20	1	20
3250	13	2	26
3750	8	3	24
Total	150	—	−70

Per computer	Schools
5–10	4
10–15	9
15–20	13
20–25	8
25–30	3
30–35	0
35–40	0
40–45	3

Distance (km)	Autos
40–60	5
60–80	11
80–100	18
100–120	14
120–140	8
140–160	4

Passengers	Trains
0–10	6
10–20	11
20–30	18
30–40	27
40–50	24
50–60	17
60–70	10
70–80	7

13.4 Median of Grouped Data

The median (माध्यक) is the middle value once the data are in order: half the observations lie below it and half above. For $n$ ungrouped values in ascending order, the median is the $(\frac{n + 1}{2})$ th value when $n$ is odd. When $n$ is even it is the average of the $\frac{n}{2}$ th and the $(\frac{n}{2} + 1)$ th values.

With a frequency table you need not write out every value. Keep a running total of the frequencies, the cumulative frequency (संचयी बारंबारता), and watch where it passes the middle. Here are the hours of sleep reported by 60 Class 10 students, rounded to the nearest hour.

Table 13.10Hours of sleep of 60 students, with cumulative frequency (cf)

Hours	Students	cf
5	4	4
6	9	13
7	17	30
8	16	46
9	10	56
10	4	60

As $n = 60$ is even, we need the 30th and 31st values. The first 13 students sleep 6 hours or less. The next 17, the 14th to the 30th, sleep 7 hours, so the 30th value is 7. The 31st is in the next group, 8 hours. The median is $\frac{7 + 8}{2} = 7.5$ hours: roughly half the students sleep less than 7.5 hours and half sleep more. The first and last columns on their own form a cumulative frequency table.

Cumulative frequency of grouped data

Now take grouped data: the marks out of 80 scored by 70 students in a pre-board Science paper.

Table 13.11Marks of 70 students, with cumulative frequency (cf)

Marks	Students	cf
0–10	3	3
10–20	5	8
20–30	8	16
30–40	11	27
40–50	15	42
50–60	13	55
60–70	9	64
70–80	6	70

The cf column keeps a running total. Its entry for 20–30 is $3 + 5 + 8 = 16 :$ sixteen students scored below 30, so 16 is the cumulative frequency of that class. Each entry adds the class's own frequency to the entry above, so the last entry is always $n .$ Each total counts the students below an upper limit (less than 10, less than 20, …), so the table is called a less than cumulative frequency distribution.

3/3The line at 35 first falls short of the 40–50 bar.

Fig. 13.4The first running total above

\frac{n}{2}

marks the median class.NCERT Table 13.13 · p. 191

Eight bars stand over the classes 0–10 to 70–80. Step 1: the bars show the class frequencies 3, 5, 8, 11, 15, 13, 9 and 6. Step 2: each bar grows to the running total of all the frequencies up to that class, 3, 8, 16, 27, 42, 55, 64 and 70, so the last bar reaches n = 70. Step 3: a horizontal dashed line at half of 70, which is 35, passes above the fourth bar (27) but below the fifth (42), so the fifth class, 40–50, is shaded as the median class.

Counting the other way works too: start from all 70 and peel off one class at a time. Remove the 3 students in 0–10 and 67 are left who scored at least 10; remove the next 5 and 62 scored at least 20, and so on. These totals count upwards from the lower limits, giving a more than or equal to cumulative frequency distribution.

Table 13.12The same marks as a “more than or equal to” distribution

Marks	Students
0 or more	70
10 or more	67
20 or more	62
30 or more	54
40 or more	43
50 or more	28
60 or more	15
70 or more	6

The median formula

Half of 70 is 35. The less-than totals run 3, 8, 16, 27, 42, …: the 35th student is not among the first 27 but is among the first 42, so the 35th mark lies in 40–50. This is the median class (माध्यक वर्ग), the first class whose cumulative frequency is greater than $\frac{n}{2} .$

To place the median inside the class, assume its 15 students are spread evenly from 40 to 50 (Fig. 13.5). We need $35 - 27 = 8$ of them, which takes us $\frac{8}{15}$ of the way across the class: $40 + \frac{8}{15} \times 10 \approx 45.33.$ So about half the students scored below 45.33 and half above.

3/38 of 15 is $\frac{8}{15}$ of the width: median ≈ 45.33.

Fig. 13.5The median is a fraction

\frac{n / 2 - c f}{f}

of the way across the median class.NCERT Median formula · p. 193

A strip runs from 40 to 50 marks. Step 1: fifteen evenly spaced dots stand for the 15 students in the median class, and a note on the left says that 27 students scored below 40. Step 2: the first 8 dots are highlighted, because 35 minus 27 is 8 more students needed to reach the 35th. Step 3: an arrow runs from 40 to the eighth dot, labelled 8/15 of the class width of 10, and the point is marked as the median, about 45.33.

Median of grouped data

\begin{aligned} Median = l + \\ \frac{\frac{n}{2} - c f}{f} \times h \end{aligned}

(13.5)

where

$l$: lower limit of the median class
$n$: number of observations
$c f$: cumulative frequency of the class before the median class
$f$: frequency of the median class
$h$: class size (all classes equal)

Like the mode formula, this one assumes equal class sizes. The median of data with unequal classes can also be found, but that is beyond this chapter.

Example 13.7NCERT Example 7 · p. 194

School-bag weights

On a bag-check day, students weighed 60 school bags. No bag was lighter than 2 kg. Find the median weight.

Table 13.13

Weight (kg)	Bags
less than 3	5
less than 4	14
less than 5	28
less than 6	44
less than 7	54
less than 8	60

Try it yourself first. Then reveal the solution one step at a time.

Example 13.8NCERT Example 8 · p. 195

Two smudged frequencies

The waiting times of 60 patients at a hospital OPD are grouped below. Two frequencies were smudged. The median waiting time is 35 minutes. Find $x$ and $y .$

Table 13.14

Waiting time (min)	Patients
0–10	3
10–20	$x$
20–30	12
30–40	16
40–50	$y$
50–60	7
60–70	4

Try it yourself first. Then reveal the solution one step at a time.

2/2Subtract 0.5 below and add 0.5 above.

Fig. 13.6Move each limit half a gap outwards so that the classes meet.NCERT Exercise 13.3 Q4 hint · p. 199

Step 1 shows three histogram bars labelled 41–45, 46–50 and 51–55 with narrow empty gaps between 45 and 46 and between 50 and 51. Step 2 widens every bar by half a unit on each side, so the bars now touch, and the new boundaries 40.5, 45.5, 50.5 and 55.5 are written underneath. A note says the class width is now 5.

Choosing a measure

You can now find three averages for any grouped data. Which one should you report? Start with the mean's weak spot. In Fig. 13.7, nine daily wages balance at ₹ 500. Replace one of them with a manager's ₹ 1300 and the mean jumps to ₹ 580, a wage that eight of the nine people do not earn. Even so, the mean is the workhorse: every observation feeds into it, so it is the fair way to compare whole groups, such as two sections, two years or two districts.

2/3One wage becomes ₹ 1300: the mean jumps to ₹ 580.

Fig. 13.7One extreme value drags the mean; the median and mode stay put.NCERT Which measure · p. 197

A number line runs from 400 to 1350 rupees. Step 1 shows nine dots for the wages 420, 450, 480, 500, 500, 500, 520, 550 and 580; a mean marker above and a median-and-mode marker below both point at 500. Step 2 moves the 580 dot far to the right, to 1300; the mean marker slides to 580 while the median and mode markers stay at 500. Step 3 adds a note that eight of the nine wages are now below the mean, so the median and mode describe the typical worker better.

The median depends only on the middle position, so the ₹ 1300 does not move it. When a few values are extreme, as with incomes, house prices or the waiting time in a queue, the median describes the typical case better. The mode answers “which is most common?”: the best-selling shoe size, the favourite flavour at an ice-cream counter, the busiest hour at a ration shop. It is the only one of the three that also works for categories that are not numbers.

Compare

Which average to report

Mean

uses every value
best for comparing groups
dragged by extreme values

Median

the middle position
not moved by extremes
best for incomes and prices

Mode

the most common value
best for “most popular” questions
works for categories too

For data that are not too lopsided, the three measures are tied together by an empirical relation, one found from experience rather than proved:

Empirical relation

\begin{aligned} 3 Median & = Mode \\ + 2 Mean \end{aligned}

(13.6)

Check it on the grouped wages. Their median is $450 + \frac{20 - 9}{12} \times 50 \approx 495.83,$ so $3 \times 495.83 \approx 1487.5.$ The mode and mean give $483.33 + 2 \times 501.25 \approx 1485.8.$ Close, but not equal. In exam questions you use the relation as an equation, to find one measure from the other two.

InteractiveGrouped-data explorer

A histogram of a grouped frequency table with four data sets to choose from: daily wages, T20 totals, fun-run times and marks. Tap or drag a bar, or use the minus and plus buttons, to change the frequency of a class. Vertical markers show the mean as a solid line, the median as a dashed line and the mode as a dotted line. A table lists each class with its class mark, frequency and cumulative frequency, and a panel works out the mean by the direct method, the median and mode with their formulas, and compares 3 times the median with the mode plus twice the mean.

Fig. 13.8Change the frequencies and watch the mean, median and mode move.

Median and choosing a measure

1 of 4

Classes 0–10, 10–20, 20–30, … have cumulative frequencies 6, 15, 24, 37, 46, 50. The median class is

20–3030–4010–2040–50

Show answer

Answer (B) 30–40

\frac{n}{2} = 25.

The first cumulative frequency above 25 is 37, for 30–40. The total 24 is close to 25 but below it, so 20–30 is the tempting wrong answer.

Practice · 7 questions

Practice 13.3: questions with answers

1
Long answerMediumNCERT Exercise 13.3, Q1
The daily water use of 70 households in a housing society is grouped below.
Water used (L) Households
100–150 4
150–200 9
200–250 15
250–300 19
300–350 12
350–400 7
400–450 4
1. Find the median.
  L
  Show solution
  Cumulative frequencies: 4, 13, 28, 47, 59, 66, 70. $\frac{n}{2} = 35,$ so the median class is 250–300 with $c f = 28,$ $f = 19.$
  $\begin{aligned} Median \\ = 250 + \frac{35 - 28}{19} \times 50 \\ \approx 268.42 \end{aligned}$
2. Find the mean.
  L
  Show solution
  With $a = 275,$ $h = 50,$ the $u_{i}$ are −3 to 3 and the products $f_{i} u_{i}$ are −12, −18, −15, 0, 12, 14, 12, so $Σ f_{i} u_{i} = - 7.$
  $\overset{ˉ}{x} = 275 + 50 \times \frac{- 7}{70} = 270$ litres.
3. Find the mode.
  L
  Show solution
  Modal class 250–300: $f_{1} = 19,$ $f_{0} = 15,$ $f_{2} = 12.$
  $\begin{aligned} Mode \\ = 250 + \frac{4}{38 - 27} \times 50 \\ \approx 268.18 \end{aligned}$
4. Compare the three measures.
  Show solution
  All three are close to 270 litres, so the data are nearly symmetric with no extreme households. Any of them describes a typical household's daily use.
2
NumericalHardNCERT Exercise 13.3, Q2
The marks out of 70 of 80 students are grouped below. The median is 36. Find $x$ and $y .$
Marks Students
0–10 6
10–20 $x$
20–30 14
30–40 20
40–50 $y$
50–60 10
60–70 5
1. Find $x .$
  Show solution
  Total: $55 + x + y = 80,$ so $x + y = 25.$ The median 36 lies in 30–40, so $l = 30,$ $f = 20,$ $h = 10$ and $c f = 6 + x + 14 = 20 + x .$
  $36 = 30 + \frac{40 - (20 + x)}{20} \times 10$
  So $6 = \frac{20 - x}{2},$ giving $x = 8.$
2. Find $y .$
  Show solution
  $y = 25 - 8 = 17.$
NumericalMediumNCERT Exercise 13.3, Q3
An RTO office recorded the ages of 150 people who applied for a learner's licence. Nobody under 18 can apply. Find the median age.
Age (years) Applicants
below 20 22
below 25 61
below 30 92
below 35 112
below 40 126
below 45 136
below 50 143
below 55 147
below 60 150
years
Show solution
The classes are 18–20, 20–25, 25–30, …, 55–60. $\frac{n}{2} = 75.$ The first running total above 75 is 92, so the median class is 25–30, with $c f = 61,$ $f = 92 - 61 = 31,$ $h = 5.$
$\begin{aligned} Median \\ = 25 + \frac{75 - 61}{31} \times 5 \\ \approx 27.26 \end{aligned}$
The median age is about 27.26 years.
NumericalMediumNCERT Exercise 13.3, Q4
A poultry farm weighed 50 eggs to the nearest gram. Find the median weight.
Weight (g) Eggs
41–45 4
46–50 9
51–55 15
56–60 12
61–65 7
66–70 3
g
Show solution
Class $f$ $c f$
40.5–45.5 4 4
45.5–50.5 9 13
50.5–55.5 15 28
55.5–60.5 12 40
60.5–65.5 7 47
65.5–70.5 3 50
$\frac{n}{2} = 25,$ so the median class is 50.5–55.5: $l = 50.5,$ $c f = 13,$ $f = 15,$ $h = 5.$
$\begin{aligned} Median \\ = 50.5 + \frac{25 - 13}{15} \times 5 \\ = 54.5 \end{aligned}$
NumericalMediumNCERT Exercise 13.3, Q5
A dealer tested how far each of 250 electric scooters ran on a full charge. Find the median range.
Range (km) Scooters
60–70 12
70–80 28
80–90 45
90–100 62
100–110 51
110–120 34
120–130 18
km
Show solution
Cumulative frequencies: 12, 40, 85, 147, …. The first above 125 is 147, so the median class is 90–100 with $c f = 85,$ $f = 62.$
$\begin{aligned} Median \\ = 90 + \frac{125 - 85}{62} \times 10 \\ \approx 96.45 \end{aligned}$
Half the scooters ran less than about 96.45 km.
6
Long answerMediumNCERT Exercise 13.3, Q6
A school surveyed 100 students on how long they take to reach school.
Time (min) Students
0–10 8
10–20 22
20–30 34
30–40 20
40–50 10
50–60 6
1. Find the median time.
  min
  Show solution
  Cumulative frequencies: 8, 30, 64, 84, 94, 100. The median class is 20–30 with $c f = 30,$ $f = 34.$
  $\begin{aligned} Median \\ = 20 + \frac{50 - 30}{34} \times 10 \\ \approx 25.88 \end{aligned}$
2. Find the mean time.
  min
  Show solution
  Class marks 5, 15, …, 55. The products $f_{i} x_{i}$ are 40, 330, 850, 700, 450, 330, which add up to 2700.
  $\overset{ˉ}{x} = 2700 \div 100 = 27$ minutes.
3. Find the modal time.
  min
  Show solution
  Modal class 20–30: $f_{1} = 34,$ $f_{0} = 22,$ $f_{2} = 20.$
  $\begin{aligned} Mode \\ = 20 + \frac{12}{68 - 42} \times 10 \\ \approx 24.62 \end{aligned}$
NumericalEasyNCERT Exercise 13.3, Q7
The daily maximum temperatures in Jaipur over the 31 days of May are grouped below. Find the median maximum temperature.
Temperature (°C) Days
36–38 2
38–40 5
40–42 9
42–44 8
44–46 5
46–48 2
°C
Show solution
Cumulative frequencies: 2, 7, 16, 24, 29, 31. The first above 15.5 is 16, so the median class is 40–42 with $c f = 7,$ $f = 9,$ $h = 2.$
$\begin{aligned} Median \\ = 40 + \frac{15.5 - 7}{9} \times 2 \\ \approx 41.89 \end{aligned}$

Water used (L)	Households
100–150	4
150–200	9
200–250	15
250–300	19
300–350	12
350–400	7
400–450	4

Marks	Students
0–10	6
10–20	$x$
20–30	14
30–40	20
40–50	$y$
50–60	10
60–70	5

Age (years)	Applicants
below 20	22
below 25	61
below 30	92
below 35	112
below 40	126
below 45	136
below 50	143
below 55	147
below 60	150

Weight (g)	Eggs
41–45	4
46–50	9
51–55	15
56–60	12
61–65	7
66–70	3

Class	$f$	$c f$
40.5–45.5	4	4
45.5–50.5	9	13
50.5–55.5	15	28
55.5–60.5	12	40
60.5–65.5	7	47
65.5–70.5	3	50

Range (km)	Scooters
60–70	12
70–80	28
80–90	45
90–100	62
100–110	51
110–120	34
120–130	18

Time (min)	Students
0–10	8
10–20	22
20–30	34
30–40	20
40–50	10
50–60	6

Temperature (°C)	Days
36–38	2
38–40	5
40–42	9
42–44	8
44–46	5
46–48	2

13.5 Summary

Recap

Statistics: key points and summary

Direct method: $\overset{ˉ}{x} = Σ f_{i} x_{i} \div Σ f_{i},$ with each class pictured at its class mark (Section 13.2).
Assumed mean method: $\overset{ˉ}{x} = a + Σ f_{i} d_{i} \div Σ f_{i},$ where $d_{i} = x_{i} - a$ (assumed mean).
Step-deviation method: $\overset{ˉ}{x} = a + h \overset{ˉ}{u},$ where $u_{i} = (x_{i} - a) \div h$ (step deviation).
All three methods give the same mean; it is an estimate because of the mid-point assumption.
The modal class has the highest frequency, and the mode lies inside it: formula (13.4).
The cumulative frequency of a class is its own frequency plus those of all the classes before it (cumulative frequency).
Less-than tables use upper limits; more-than-or-equal tables use lower limits.
The median class is the first class whose cumulative frequency exceeds $\frac{n}{2};$ the median comes from formula (13.5).
Make gapped classes continuous before using the mode or median formula (close the gaps).
Mean for comparing groups, median when there are extremes, mode for the most common value (choosing a measure).
Empirical relation (13.6): $3 Median = Mode + 2 Mean .$

2/3The median splits the area into two equal halves.

Fig. 13.9Mean: balance point. Median: halves the area. Mode: under the crossing.

The histogram of the 40 grouped wages, from 350 to 650 rupees, with bar heights 3, 6, 12, 9, 6 and 4. Step 1 marks about 501 rupees with a pointer and a dashed line: the mean is the point where the histogram would balance. Step 2 draws a dashed line at about 496 rupees and shades the area to its left; 20 workers lie on each side, so the median splits the area in half. Step 3 draws the two crossing lines inside the tallest bar and drops a line from the crossing to about 483 rupees, the mode.

Flash cards · 7

Quick recall

1 / 7

Board-exam practice

These questions follow the CBSE board-paper formats: multiple choice, assertion–reason, short answer, long answer and a case study. Show the working table in every calculation question; examiners award marks for it.

Practice · 9 questions

Practice questions with answers (NCERT “Board-exam practice”)

Quick answerEasy
The class mark of the class 15.5–20.5 is
51817.518.5
Show answer
Answer (B) 18
$(15.5 + 20.5) \div 2 = 18.$ 5 is the class size.
Quick answerEasy
In the step-deviation method, $u_{i}$ equals
$(x_{i} + a) \div h$ $h (x_{i} - a)$ $(x_{i} - a) \div h$ $(a - x_{i}) \div h$
Show answer
Answer (C) $(x_{i} - a) \div h$
$u_{i} = (x_{i} - a) \div h .$ The last option has the sign reversed.
Quick answerEasy
The mode of a distribution is 12 and its mean is 15. By the empirical relation, its median is
13.5161413
Show answer
Answer (C) 14
$3 Median = 12 + 30 = 42,$ so the median is 14.
Quick answerEasy
In a less-than cumulative frequency table, the last cumulative frequency is equal to
the frequency of the last classthe upper limit of the last class $\frac{n}{2}$ the total frequency $n$
Show answer
Answer (D) the total frequency $n$
Every observation is counted by the end, so the last running total is $n .$
Quick answerMedium
Assertion (A): For the classes 1–5, 6–10, 11–15, …, the lower limit used in the median formula for the median class 6–10 is 5.5. Reason (R): The median formula needs classes with no gaps between them.
Both A and R are true, and R is the correct explanation of A.Both A and R are true, but R is not the correct explanation of A.A is true but R is false.A is false but R is true.
Show answer
Answer (A) Both A and R are true, and R is the correct explanation of A.
Closing the gaps turns 6–10 into 5.5–10.5, so $l = 5.5.$ Both are true and R is exactly why.
Quick answerMedium
Assertion (A): The modal class is the class with the largest cumulative frequency. Reason (R): The mode is the value that occurs most often.
Both A and R are true, and R is the correct explanation of A.Both A and R are true, but R is not the correct explanation of A.A is true but R is false.A is false but R is true.
Show answer
Answer (D) A is false but R is true.
The last class always has the largest cumulative frequency, $n .$ The modal class has the largest frequency. A is false and R is true.
Short answerMedium
Find the mean weight of 40 Class 10 students by the step-deviation method.
Weight (kg) Students
35–40 3
40–45 7
45–50 12
50–55 9
55–60 6
60–65 3
kg
Show solution
$x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$
37.5 3 −2 −6
42.5 7 −1 −7
47.5 12 0 0
52.5 9 1 9
57.5 6 2 12
62.5 3 3 9
Total 40 — 17
$\overset{ˉ}{x} = 47.5 + 5 \times \frac{17}{40} = 49.625 \approx 49.63$ kg.
8
Long answerHard
The weekly distance cycled by the 120 members of a cycling club is grouped below. The mean distance is 50 km. Find $f_{1}$ and $f_{2} .$
Distance (km) Members
0–20 17
20–40 $f_{1}$
40–60 32
60–80 $f_{2}$
80–100 19
1. Find $f_{1} .$
  Show solution
  Total: $68 + f_{1} + f_{2} = 120,$ so $f_{1} + f_{2} = 52.$
  Class marks 10, 30, 50, 70, 90, so $Σ f_{i} x_{i} = 170 + 30 f_{1} + 1600 + 70 f_{2} + 1710.$ This must equal $50 \times 120 = 6000.$
  $30 f_{1} + 70 f_{2} = 2520$
  Divide by 10: $3 f_{1} + 7 f_{2} = 252.$ Put $f_{2} = 52 - f_{1}$ : $3 f_{1} + 364 - 7 f_{1} = 252,$ so $4 f_{1} = 112$ and $f_{1} = 28.$
2. Find $f_{2} .$
  Show solution
  $f_{2} = 52 - 28 = 24.$ Check: $Σ f_{i} x_{i} = 170 + 840 + 1600 + 1680 + 1710 = 6000.$
9
Long answerMedium
Case study. A cricket academy in Chennai recorded the runs scored by each of its 50 players in a season, to pick a squad of batters.
Runs Players
0–100 6
100–200 9
200–300 15
300–400 11
400–500 6
500–600 3
1. Which is the median class?
  100–200200–300300–400400–500
  Show answer
  Answer (B) 200–300
  $\frac{n}{2} = 25;$ the first cumulative frequency above 25 is 30, for 200–300.
2. Find the median runs.
  runs
  Show solution
  $\begin{aligned} Median \\ = 200 + \frac{25 - 15}{15} \times 100 \\ \approx 266.67 \end{aligned}$
3. Find the mean runs by the step-deviation method.
  runs
  Show solution
  With $a = 250,$ $h = 100,$ the $u_{i}$ are −2 to 3 and the products $f_{i} u_{i}$ are −12, −9, 0, 11, 12, 9, so $Σ f_{i} u_{i} = 11.$
  $\overset{ˉ}{x} = 250 + 100 \times \frac{11}{50} = 272$ runs.
4. Find the modal runs.
  runs
  Show solution
  $\begin{aligned} Mode \\ = 200 + \frac{15 - 9}{30 - 9 - 11} \times 100 \\ = 260 \end{aligned}$
  Check: $3 \times 266.67 = 800$ and $260 + 2 \times 272 = 804,$ close, as the empirical relation predicts.

Weight (kg)	Students
35–40	3
40–45	7
45–50	12
50–55	9
55–60	6
60–65	3

$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$
37.5	3	−2	−6
42.5	7	−1	−7
47.5	12	0	0
52.5	9	1	9
57.5	6	2	12
62.5	3	3	9
Total	40	—	17

Distance (km)	Members
0–20	17
20–40	$f_{1}$
40–60	32
60–80	$f_{2}$
80–100	19

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In your NCERT textbook	Practice question here
NCERT Exercise 13.1, Q1	Practice 13.1 · Q1
NCERT Exercise 13.1, Q2	Practice 13.1 · Q2
NCERT Exercise 13.1, Q3	Practice 13.1 · Q3
NCERT Exercise 13.1, Q4	Practice 13.1 · Q4
NCERT Exercise 13.1, Q5	Practice 13.1 · Q5
NCERT Exercise 13.1, Q6	Practice 13.1 · Q6
NCERT Exercise 13.1, Q7	Practice 13.1 · Q7
NCERT Exercise 13.1, Q8	Practice 13.1 · Q8
NCERT Exercise 13.1, Q9	Practice 13.1 · Q9
NCERT Exercise 13.2, Q1	Practice 13.2 · Q1
NCERT Exercise 13.2, Q2	Practice 13.2 · Q2
NCERT Exercise 13.2, Q3	Practice 13.2 · Q3
NCERT Exercise 13.2, Q4	Practice 13.2 · Q4
NCERT Exercise 13.2, Q5	Practice 13.2 · Q5
NCERT Exercise 13.2, Q6	Practice 13.2 · Q6
NCERT Exercise 13.3, Q1	Practice 13.3 · Q1
NCERT Exercise 13.3, Q2	Practice 13.3 · Q2
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NCERT Exercise 13.3, Q5	Practice 13.3 · Q5
NCERT Exercise 13.3, Q6	Practice 13.3 · Q6
NCERT Exercise 13.3, Q7	Practice 13.3 · Q7

Quick check: MCQs on Statistics

Warm-up: averages of ungrouped data3 questions
Mean of grouped data4 questions
Mode of grouped data3 questions
Median and choosing a measure4 questions

Glossary · 12

Key terms (English–Hindi)

assumed meanकल्पित माध्यkalpit maadhya: A convenient value $a,$ usually a middle class mark, from which the class marks are measured to make the arithmetic smaller.First seen in the chapter →

class markवर्ग चिह्नvarg chihn: The mid-point of a class, (lower limit + upper limit) ÷ 2. It stands for every observation in the class.First seen in the chapter →
class sizeवर्ग मापvarg maap: The width of a class, upper limit minus lower limit, written $h .$ First seen in the chapter →
cumulative frequencyसंचयी बारंबारताsanchayi baarambaarta: The running total of the frequencies: the frequency of a class added to the frequencies of all the classes before it.First seen in the chapter →

deviationविचलनvichalan: How far a class mark is from the assumed mean: $d_{i} = x_{i} - a .$ It is negative below $a .$ First seen in the chapter →

frequencyबारंबारताbaarambaarta: The number of times a value occurs, or the number of observations that fall in a class.First seen in the chapter →

meanमाध्यmaadhya: The sum of all the observations divided by their number; for data with frequencies, $Σ f_{i} x_{i} \div Σ f_{i} .$
measure of central tendencyकेंद्रीय प्रवृत्ति का मापkendriya pravritti ka maap: A single number that represents a whole set of data: the mean, the median or the mode.First seen in the chapter →
medianमाध्यकmaadhyak: The middle value of data arranged in order, so that half the observations lie below it and half above.First seen in the chapter →
median classमाध्यक वर्गmaadhyak varg: The first class whose cumulative frequency is greater than $\frac{n}{2};$ the median lies inside it.First seen in the chapter →
modal classबहुलक वर्गbahulak varg: The class with the highest frequency in a grouped frequency distribution.First seen in the chapter →
modeबहुलकbahulak: The value that occurs most often. For grouped data it is estimated inside the modal class with the mode formula.First seen in the chapter →

Runs	Players
0–100	6
100–200	9
200–300	15
300–400	11
400–500	6
500–600	3

13.1 Introduction

13.2 Mean of Grouped Data

Assumed mean method

Step-deviation method

13.3 Mode of Grouped Data

13.4 Median of Grouped Data

Cumulative frequency of grouped data

The median formula

Choosing a measure

13.5 Summary

Board-exam practice

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