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Class 10 Maths Chapter 13 Statistics Exercise 13.1 NCERT Solutions

Get NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1 here. Exercise 13.1 introduces grouped data and teaches how to calculate the mean using direct and assumed mean methods. Step-by-step solutions make statistical calculations easier to understand.
authorImageNeha Tanna5 May, 2026
NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.1

NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.1 provide step-by-step answers to problems related to data handling and representation. This exercise focuses on calculating measures of central tendency, including mean, median, and mode, for grouped and ungrouped data, as covered in the CBSE Class 10 Maths syllabus

These NCERT solutions explain concepts clearly with step-by-step methods, enhancing understanding and problem-solving skills. These solutions will help you interpret and analyse data effectively.

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.1

1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of Plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14
Number of Houses 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

Solution:

To find the mean value, we will use the direct method because the numerical value of f i and x i are small. Find the midpoint of the given interval using the formula. Midpoint (x i ) = (upper limit + lower limit)/2

No. of plants

(Class interval)

No. of houses

Frequency (f i )

Mid-point (x i )

f i x i

0-2

1

1

1

2-4

2

3

6

4-6

1

5

5

6-8

5

7

35

8-10

6

9

54

10-12

2

11

22

12-14

3

13

39

 

Sum f i = 20

 

Sum f i x i = 162

The formula to find the mean is: Mean = x̄ = ∑f i x i /∑f i = 162/20 = 8.1. Therefore, the mean number of plants per house is 8.1.

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs.)

500-520

520-540

540-560

560-580

580-600

Number of workers

12

14

8

6

10

Find the mean daily wages of the workers of the factory by using an appropriate method .

Solution: Find the midpoint of the given interval using the formula. Midpoint (x i ) = (upper limit + lower limit)/2 In this case, the value of mid-point (x i ) is very large, so let us assume the mean value, a = 550. Class interval (h) = 20 So, u i = (x i – a)/h u i = (x i – 550)/20 Substitute and find the values as follows:

Daily wages

(Class interval)

Number of workers

frequency (f i )

Mid-point (x i )

u i = (x i – 550)/20

f i u i

500-520

12

510

-2

-24

520-540

14

530

-1

-14

540-560

8

550 = a

0

0

560-580

6

570

1

6

580-600

10

590

2

20

Total

Sum f i = 50

   

Sum f i u i = -12

So, the formula to find out the mean is: Mean = x̄ = a + h(∑f i u i /∑f i ) = 550 + [20 × (-12/50)] = 550 – 4.8 = 545.20 Thus, mean daily wage of the workers = Rs. 545.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Daily Pocket Allowance(in c)

11-13

13-15

15-17

17-19

19-21

21-23

23-35

Number of children

7

6

9

13

f

5

4

Solution: To find out the missing frequency, use the mean formula. Given, mean x̄ = 18

Class interval

Number of children (f i )

Mid-point (x i )

f i x i

11-13

7

12

84

13-15

6

14

84

15-17

9

16

144

17-19

13

18

234

19-21

f

20

20f

21-23

5

22

110

23-25

4

24

96

Total

f i = 44+f

 

Sum f i x i = 752+20f

The mean formula is Mean = x̄ = ∑f i x i /∑f i = (752 + 20f)/ (44 + f)

 Now substitute the values and equate to find the missing frequency (f) ⇒ 18 = (752 + 20f)/ (44 + f)

⇒ 18(44 + f) = (752 + 20f)

⇒ 792 + 18f = 752 + 20f

⇒ 792 + 18f = 752 + 20f

⇒ 792 – 752 = 20f – 18f

⇒ 40 = 2f

⇒ f = 20 So, the missing frequency, f = 20.

4. Thirty women were examined in a hospital by a doctor, and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Number of heart beats per minute

65-68

68-71

71-74

74-77

77-80

80-83

83-86

Number of women

2

4

3

8

7

4

2

Solution: From the given data, let us assume the mean as a = 75.5 x i = (Upper limit + Lower limit)/2 Class size (h) = 3. Now, find the u i and f i u i as follows:

Class Interval

Number of women (f i )

Mid-point (x i )

u i = (x i – 75.5)/h

f i u i

65-68

2

66.5

-3

-6

68-71

4

69.5

-2

-8

71-74

3

72.5

-1

-3

74-77

8

75.5 = a

0

0

77-80

7

78.5

1

7

80-83

4

81.5

2

8

83-86

2

84.5

3

6

 

Sum f i = 30

   

Sum f i u i = 4

Mean = x̄ = a + h(∑f i u i /∑f i ) = 75.5 + 3 × (4/30) = 75.5 + (4/10) = 75.5 + 0.4 = 75.9 Therefore, the mean heart beats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained a varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes

50-52

53-55

56-58

59-61

62-64

Number of boxes

15

110

135

115

25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution: The given data is not continuous, so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals is 1. Here, assumed mean (a) = 57 Class size (h) = 3 Here, the step deviation is used because the frequency values are big.

Class Interval

Number of boxes (f i )

Mid-point (x i )

u i = (x i – 57)/h

f i u i

49.5-52.5

15

51

-2

-30

52.5-55.5

110

54

-1

-110

55.5-58.5

135

57 = a

0

0

58.5-61.5

115

60

1

115

61.5-64.5

25

63

2

50

 

Sum f i = 400

   

Sum f i u i = 25

The formula to find out the Mean is: Mean = x̄ = a + h(∑f i u i /∑f i ) = 57 + 3(25/400) = 57 + 0.1875 = 57.19 Therefore, the mean number of mangoes kept in a packing box is 57.19

6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure(in c)

100-150

150-200

200-250

250-300

300-350

Number of households

4

5

12

2

2

Find the mean daily expenditure on food by a suitable method.

Solution: Find the midpoint of the given interval using the formula. Midpoint (x i ) = (upper limit + lower limit)/2 Let us assume the mean (a) = 225 Class size (h) = 50

Class Interval

Number of households (f i )

Mid-point (x i )

d i = x i – A

u i = d i /50

f i u i

100-150

4

125

-100

-2

-8

150-200

5

175

-50

-1

-5

200-250

12

225 = a

0

0

0

250-300

2

275

50

1

2

300-350

2

325

100

2

4

 

Sum f i = 25

     

Sum f i u i = -7

Mean = x̄ = a + h(∑f i u i /∑f i ) = 225 + 50(-7/25) = 225 – 14 = 211 Therefore, the mean daily expenditure on food is 211.

7. To find out the concentration of SO 2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO 2 ( in ppm)

Frequency

0.00 – 0.04

4

0.04 – 0.08

9

0.08 – 0.12

9

0.12 – 0.16

2

0.16 – 0.20

4

0.20 – 0.24

2

Find the mean concentration of SO 2 in the air.

Solution: To find out the mean, first find the midpoint of the given frequencies as follows:

Concentration of SO 2 (in ppm)

Frequency (f i )

Mid-point (x i )

f i x i

0.00-0.04

4

0.02

0.08

0.04-0.08

9

0.06

0.54

0.08-0.12

9

0.10

0.90

0.12-0.16

2

0.14

0.28

0.16-0.20

4

0.18

0.72

0.20-0.24

2

0.22

0.44

Total

Sum f i = 30

 

Sum (f i x i ) = 2.96

The formula to find out the mean is Mean = x̄ = ∑f i x i /∑f i = 2.96/30 = 0.099 ppm Therefore, the mean concentration of SO 2 in the air is 0.099 ppm.

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days

0-6

6-10

10-14

14-20

20-28

28-38

38-40

Number of students

11

10

7

4

4

3

1

Solution: Find the midpoint of the given interval using the formula. Midpoint (x i ) = (upper limit + lower limit)/2

Class interval

Frequency (f i )

Mid-point (x i )

f i x i

0-6

11

3

33

6-10

10

8

80

10-14

7

12

84

14-20

4

17

68

20-28

4

24

96

28-38

3

33

99

38-40

1

39

39

 

Sum f i = 40

 

Sum f i x i = 499

The mean formula is, Mean = x̄ = ∑f i x i /∑f i = 499/40 = 12.48 days Therefore, the mean number of days a student was absent = 12.48.

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %)

45-55

55-65

65-75

75-85

85-98

Number of cities

3

10

11

8

3

Solution: Find the midpoint of the given interval using the formula. Midpoint (x i ) = (upper limit + lower limit)/2 In this case, the value of mid-point (x i ) is very large, so let us assume the mean value, a = 70. Class interval (h) = 10 So, u i = (x i – a)/h u i = (x i – 70)/10 Substitute and find the values as follows: 

Class Interval

Frequency (f i )

(x i )

u i = (x i – 70)/10

f i u i

45-55

3

50

-2

-6

55-65

10

60

-1

-10

65-75

11

70 = a

0

0

75-85

8

80

1

8

85-95

3

90

2

6

 

Sum f i = 35

   

Sum f i u i = -2

So, Mean = x̄ = a + (∑f i u i /∑f i ) × h = 70 + (-2/35) × 10 = 69.43 Therefore, the mean literacy part = 69.43%

How to Score Better in Class 10 Maths Exam?

Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:

  • Build Strong Concepts:

Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.

  • Work on Weak Areas:

Focus on difficult topics from the  CBSE Class 10 Maths syllabus instead of skipping them to avoid losing marks.

  • Revise Formulas Daily:

Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.

  • Practise Regularly:

Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.

  • Solve Previous Year Papers:

Practising  class 10 Maths previous year questions (PYQs) helps you understand question patterns and important topics.

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.1 FAQs

What are 5 applications of statistics?

Statistics are used in business to detect market trends and sales results, in education to determine teaching method effectiveness, in government to detect changes in population demographics and effectiveness of public policy, and in sports to examine player and team successes and capabilities.

What are the two main types of statistics?

Descriptive and Inferential Statistics.

What are the 5 main statistics?

A five-number summary simply consists of the smallest data value, the first quartile, the median, the third quartile, and the largest data value. A box plot is a graphical device based on a five-number summary.

What is good knowledge of statistics?

Statistical knowledge helps to choose the proper method of collecting the data and employ those samples in the correct analysis process in order to effectively produce the results. In short, statistics is a crucial process which helps to make the decision based on the data.
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