• NEET
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • Class 6-10
      • Class 6th
      • Class 7th
      • Class 8th
      • Class 9th
      • Class 10th
    • View All Options
      • Online Courses
      • Offline Courses
      • Distance Learning
      • Hindi Medium Courses
      • International Olympiad
    • NEET
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE (Main+Advanced)
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE Main
      • Class 11th
      • Class 12th
      • Class 12th Plus
  • NEW
    • JEE 2025
    • NEET
      • 2024
      • 2023
      • 2022
    • Class 6-10
    • JEE Main
      • Previous Year Papers
      • Sample Papers
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • JEE Advanced
      • Previous Year Papers
      • Sample Papers
      • Mock Test
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • NEET
      • Previous Year Papers
      • Sample Papers
      • Mock Test
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • NCERT Solutions
      • Class 6
      • Class 7
      • Class 8
      • Class 9
      • Class 10
      • Class 11
      • Class 12
    • CBSE
      • Notes
      • Sample Papers
      • Question Papers
    • Olympiad
      • NSO
      • IMO
      • NMTC
    • TALLENTEX
    • AOSAT
    • ALLEN e-Store
    • ALLEN for Schools
    • About ALLEN
    • Blogs
    • News
    • Careers
    • Request a call back
    • Book home demo
NCERT Solutions
Class 10
Maths
Chapter 13 Statistics
Exercise 13.3

NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3

NCERT Solutions Class 10 Maths Chapter 13 Exercise 13.3 focuses on solving the questions related to the median of a dataset, a crucial concept of statistics. To calculate the median, the exercise includes forming cumulative frequency tables to interpret the grouped data. This exercise is a great help for solving the Median for simple problems as well as forming a solid base for complex real-world data analysis. In this exercise, we will also learn about the relationship between different measures of central tendencies. So, let’s start exploring this interesting topic of statistics. 

1.0Download NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3: Free PDF

NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3

2.0Introduction to Median

Before we get started on the formulas and procedures for obtaining the median of a data set, let's discuss the general concept of the median. The median is the middle number in a data set when it is ordered from least to greatest or from greatest to least. For ungrouped data, determining the median is simply done by identifying the middle figure. However, for grouped data, it's a little complicated since data is tabulated in class intervals.

3.0Class 10 Maths Chapter 13 Statistics Exercise 13.3 : Key Concepts 

Now that we have understood the basic idea of the median let’s explore some of the important formulas and key concepts used to solve various questions in the exercise:  

Median of Ungrouped Data

The steps to determine the median of ungrouped data are quite straightforward. However, it can differ based on the number of terms, either odd or even, in a dataset. To find this, simply arrange the data in ascending or descending order. After arranging the data, simply use the following formulas to determine the median in each case: 

Median if the number of observations (n) is odd

Median for odd number of terms=(2n+1​)thobservation

Median if the number of observations (n) is even

Median for even number of terms=2(2n​)thobservation+(2n​+1)thobservation​

Median of Grouped Data

To determine the median for grouped data, we use the principle of cumulative frequency and apply it to determine the median class, wherein lies the middle-most value. The formula to determine the median for grouped data can be established after determining the median class. To find the median of grouped data, we simply need to construct a cumulative frequency table for each class interval. After the formation of the table, simply use the following formula to find the median: 

Median=l+f(2n​−cf)​×h

Here, 

  • l = lower limit of the median class
  • n = total number of observations
  • h = class size (assuming equal class sizes), calculated as: 

h=Upper Limit -Lower limit

  • cf = cumulative frequency of the class before the median class
  • f = frequency of the median class

Median, Mean, and Mode: Relation

In statistics, mean and mode are the primary measures of central tendency other than the median. Each of these measures gives a different insight into a dataset. For example, the mean is the average of all values of a dataset, the mode is the most frequent value of data, and the median is the middle value of data when arranged in ascending or descending order. These three values are related to one another by the following equation: 

3×Median=Mode+2×Mean

To further comprehend these principles and enhance your expertise, ensure practising constantly with the NCERT Solutions Class 10 Maths Chapter 13 Exercise.

4.0Benefits of Solving NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3

  • Helps in learning how to calculate the mode of grouped data.
  • Practicing mode-related questions improves logical thinking and mathematical accuracy.
  • Strengthens the basics of statistics for higher studies and competitive exams.
  • This exercise helps in analyzing frequency distribution tables efficiently.

5.0NCERT Class 10 Maths Chapter 13 Statistics Exercise 13.3 : Detailed Solution

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly  consumption  (in units)

Number of  consumers

65−85

4

85−105

5

105−125

13

125−145

20

145−165

14

165−185

8

185−205

4

Sol.

(i)

Monthly  Consumpti on  (in units)

Number  of  consumers

Cumulative  frequency

65−85

4

4

85−105

5

9

105−125

13

22

125−145

20

42

145−165

14

56

165−185

8

64

185−205

4

68

Total

N=68


n=68 gives 2N​=34
So, we have the median class (125-145)
ℓ=125, N=68,f=20,cf=22, h=20
Median =ℓ+{f2N​−cf​}×h
=125+{2034−22​}×20=137 units.
(ii) Modal class is (125 - 145) having maximum frequency f1​=20,f0​=13,f2​=14, ℓ=125 and h=20
Mode =ℓ+{2f1​−f0​−f2​f1​−f0​​}×h=125+{40−13−1420−13​}×20=125+
137×20​
=125+13140​=125+10.76=135.76 units
(iii)

Monthly  consumption  (in units)

Number  of  consumers 

fi​fi​

Class  marks 

xi​xi​

ui​=xi​ui​=xi​

fi​xi​fi​xi​

65−85

4

75

-3

-12

85−105

5

95

-2

-10

105−125

13

115

-1

-13

125−145

20

135=a

0

0

145−165

14

155

1

14

165−185

8

175

2

16

185−205

4

195

3

12

Total

N=68



77

N=68,a=135, h=20 and Σfi​ui​=7
By step-deviation method.
Mean =a+h× N1​×Σfi​ui​
=135+20×681​×7
=135+1735​=135+2.05
=137.05 units

2. If the median of the distribution given below is 28.5 , find the values of x and y.

Class interval

Frequency

0−10

5

10−20

x

20−30

20

30−40

15

40−50

y

50−60

5

Total

60

Sol.

Class  interval

Frequency

Cumulative  frequency

0−10

5

5

10−20

x

5+x

20−30

20

25+x

30−40

15

40+x

40−50

y

40+x+y

50−60

5

45+x+y

Total

60


Median =28.5 lies in the class-interval (20-30).
Then the median class is (20-30).
So, we have ℓ=20,f=20,cf=5+x,h=10,
N=60
Median =ℓ+{f2N​−cf​}×h=28.5
28.5=20+{2030−(5+x)​}×10
⇒8.5=225−x​⇒17=25−x⇒x=8
Find the given table, we have
i.e., x+y+45=60 or x+y=15
⇒y=15−x=15−8=7, i.e., y=7

3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years)

No. of policy holders

Below 20

2

Below 25

6

Below 30

24

Below 35

45

Below 40

78

Below 45

89

Below 50

92

Below 55

98

Below 60

100

Sol.

Age  (in  years)

Number  of  policy  holders 

fi​fi​

Cumulative  frequency

Below  20

2=2

2

20−25

(6−2)=4

6

Median



class

25−30

(24−6)=18

30−35

(45−24)=21

24

40−40

(78−45)=21

78

40−45

(89−78)=11

89

45−50

(92−89)=3

92

50−55

(98−92)=6

98

55−60

(100−98)=2

100

Total

N=100


Here, ℓ=35, N=100,f=33,cf=45, h=5
Median =ℓ+{f2N​−cf​}×h
=35+{3350−45​}×5
=35+3325​=35+0.76
=35.76 years.

4. The length of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table.

Length (in mm)

No. of leaves

118−126

3

127−135

5

136−144

9

145−153

12

154−162

5

163−171

4

172−180

2

Find the median length of the leaves.

Sol.

The given series is in inclusive form. We may prepare the table in exclusive form and prepare the cumulative frequency table as given below :

Length  (in mm)

No. of  leaves 

(fi​)

Cumulative  frequency

117.5−126.5

3

3

126.5−135.5

5

8

135.5−144.5

9

17

144.5−153.5

12

29

153.5−162.5

5

34

162.5−171.5

4

38

171.5−180.5

2

40


N=40


Here, N=40
⇒2N​=20
The cumulative frequency just greater than 20 is 29 and the corresponding class is 144.5-153.5.
So, the median class is 144.5-153.5.
∴ℓ=144.5, N=40,cf=17,f=12 and h=9
Therefore, median =ℓ+{f2N​−cf​}×h
=144.5+12(20−17)​×9
=144.5+123×9​
=144.5+2.25=146.75
Hence, median length of leaves is 146.75 mm .

5. The following table gives the distribution of the life time of 400 neon lamps :

Life time (in hours)

No. of lamps

1500−2000

14

2000−2500

56

2500−3000

60

3000−3500

86

3500−4000

74

4000−4500

62

4500−5000

48

Find the median life time of a lamp.

Sol.

Life time  (in hours)

No. of lamps

cf

1500−2000

14

14

2000−2500

56

70

2500−3000

60

130

3000−3500

86

216

3500−4000

74

290

4000−4500

62

352

4500−5000

48

400

2N​=2400​=200
Median class =3000−3500
Median =ℓ+{f2N​−cf​}×h
=3000+{86200−130​}×500=3406.98
hours
Hence, median life time of a lamp 3406.98 hours.

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

No. of letters

No. of surnames

1−4

6

4−7

30

7−10

40

10−13

16

13−16

4

16−19

4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Sol.

No.  of  letters

No. of  Surnames 

fi​

Cumulative  frequency


1−4

6

6=6


4−7

30

6+30=36


7−10

40

36+40=76

50=2N​

10−13

16

76+16=92


13−16

4

92+4=96


16−19

4

96+4=100


Total

N=100



(i) Here,
ℓ=7, N=100,f=40,cf=36, h=3
Median =ℓ+{f2N​−cf​}×h
=7+{4050−36​}×3
=7+2021​=8.05
(ii) Modal class is (7-10).
ℓ=7,f1​=40,f0​=30,f2​=16, h=3 Mode =ℓ+{2f1​−f0​−f2​f1​−f0​​}×h=7+{80−30−1640−30​}×3=7+3430​=7.88
(iii) Here, a=8.5, h=3,n=100 and Σfi​ui​=−6.

Number  of  letters

(fi​)

mark (xi)

ui​=3xi​−8​3

fi​ui​

1−4

6

2.5

-2

-12

4−7

30

5.5

-1

-30

7−10

40

8.5=a

0

0

10−13

16

11.5

1

16

13−16

4

14.5

2

8

16−19

4

17.5

3

12

Total

N=100



−6

Mean =a+h× N1​×Σfi​ui​
=8.5+3×1001​×(−6).
=8.5−10018​=8.5−0.18=8.32

7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg)

No. of students

40−45

2

45−50

3

50−55

8

55−60

6

60−65

6

65−70

3

70−75


Sol.

Weight  (in kg)

No. of  students

Cumulative  frequency

40−45

2

2

45−50

3

5

50−55

8

13

55−60

6

19

60−65

6

25

65−70

3

28

70−75

2

30

2N​=230​=15
Median class =55−60
Median =ℓ+{f2N​−cf​}×h=55+{615−13​}×5=56.67 kg

NCERT Class 10 Maths Ch 13 Statistics Other Exercises:

Exercise 13.1

Exercise 13.2

Exercise 13.3

Exercise 13.4


NCERT Solutions Class 10 Maths All Chapters:-

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Chapter 3 - Linear Equations in Two Variables

Chapter 4 - Quadratic Equations

Chapter 5 - Arithmetic Progressions

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometery

Chapter 8 - Introdction to Trigonometry

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Areas Related to Circles

Chapter 12 - Surface Areas and Volumes

Chapter 13 - Statistics

Chapter 14 - Probability

Frequently Asked Questions

Exercise 13.3 focuses on finding the mode of grouped data using a specific formula and frequency distribution tables.

The modal class is the class interval with the highest frequency in a grouped frequency distribution table.

Mode represents the most frequent value, mean gives the average, and median represents the middle value of the dataset.

The median is the middle value of a dataset when arranged in ascending order, dividing it into two equal halves.

Common mistakes include incorrect cumulative frequency calculation, choosing the wrong median class, and calculation errors in applying the formula.

Join ALLEN!

(Session 2025 - 26)


Choose class
Choose your goal
Preferred Mode
Choose State
  • About
    • About us
    • Blog
    • News
    • MyExam EduBlogs
    • Privacy policy
    • Public notice
    • Careers
    • Dhoni Inspires NEET Aspirants
    • Dhoni Inspires JEE Aspirants
  • Help & Support
    • Refund policy
    • Transfer policy
    • Terms & Conditions
    • Contact us
  • Popular goals
    • NEET Coaching
    • JEE Coaching
    • 6th to 10th
  • Courses
    • Online Courses
    • Distance Learning
    • Online Test Series
    • International Olympiads Online Course
    • NEET Test Series
    • JEE Test Series
    • JEE Main Test Series
    • CUET Test Series
  • Centers
    • Kota
    • Bangalore
    • Indore
    • Delhi
    • More centres
  • Exam information
    • JEE Main
    • JEE Advanced
    • NEET UG
    • CBSE
    • NCERT Solutions
    • NEET Mock Test
    • CUET
    • Olympiad
    • NEET 2025 Answer Key

ALLEN Career Institute Pvt. Ltd. © All Rights Reserved.

ISO