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NCERT Solutions
Class 9
Maths
Chapter 12 Statistics
Exercise 12.1

NCERT Solutions Class 9 Maths Chapter 12 Statistics Exercise 12.1

Our NCERT Solutions Class 9 Maths Chapter 12 Statistics Exercise 12.1 introduces students to the basics of data collection and organization. This exercise focuses on creating frequency tables to represent data clearly and systematically. By working through these problems, students learn how to arrange raw data in a meaningful way, making it easier to analyze and draw conclusions. Understanding these concepts is crucial as statistics plays a vital role in everyday life and higher studies.

The solutions provided for Exercise 12.1 are detailed and easy to follow, helping students avoid confusion and build confidence in handling statistical problems. Each question is solved step by step as per the latest NCERT guidelines, ensuring students grasp both the methods and logic behind each answer. These solutions are an excellent tool for thorough practice and effective exam preparation.

1.0Download NCERT Solutions of Class 9 Maths Chapter 12 statistics Exercise 12.1 Free PDF

Gain access to free and well-structured and accurate NCERT Solutions for Class 9 Maths Chapter 12 Exercise 12.1. These solutions will help clarify concepts of data collection and data representation. You will find answers for all questions in the textbook along with step-by-step answers.

NCERT Solutions Class 9 Maths Chapter 12 Ex 12.1

Key Concepts of Exercise 12.1

  • An Introduction to Statistics
  • Understanding Data Collection
  • Presentation of Data using tables
  • Understanding Frequency and Frequency Distribution
  • Making a Tally Chart and Frequency Table

2.0NCERT Solutions Class 9 Maths Chapter 12: All Exercises

Here is a quick overview and access to solutions for all exercises from Chapter 12, Surface Area and Volume.

NCERT Solutions for Class 9 Maths Chapter 12: Exercise 12.1

3.0NCERT Class 9 Maths Chapter 12 Exercise 12.1 : Detailed Solutions

1. A survey was conducted by a group of students as a part of their environment awareness programme, 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

Number of Houses

0−2

1

2−4

2

4−6

1

6−8

5

8−10

6

10−12

2

12−14

3

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

Sol.

To find the mean, we can create the following table:

Number of Plants (Class)

Number of Houses (Frequency, fᵢ)

Class Marks (xᵢ)

fᵢxᵢ

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

Total

N = Σfᵢ = 20


Σfᵢxᵢ = 162

We have, N = Σfᵢ = 20 and Σfᵢxᵢ = 162.

Then the mean of the data is x̄ = (1/N) × Σfᵢxᵢ = (1/20) × 162 = 8.1.

Hence, the required mean of the data is 8.1 plants.

We find the mean of the data by the direct method because the figures (frequencies and class marks) are small, making calculations straightforward.

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

Daily Wages (in Rs.)

Number of Workers

500−520

12

520−540

14

540−560

8

560−580

6

580−600

10

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

Sol.

To find the mean, we can prepare the following table:

Daily Wages (In Rs.)

No. of Workers (fᵢ)

Class Marks (xᵢ)

fᵢxᵢ

500−520

12

510

6120

520−540

14

530

7420

540−560

8

550

4400

560−580

6

570

3420

580−600

10

590

5900

Total

N = Σfᵢ = 50


Σfᵢxᵢ = 27260

We have Σfᵢ = 50 and Σfᵢxᵢ = 27260.

Mean = Σfᵢxᵢ / Σfᵢ = 27260 / 50 = 545.2 Rs.

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 Rs.)

Number of Children

11−13

7

13−15

6

15−17

9

17−19

13

19−21

f

21−23

5

23−25

4

Sol. We may prepare the table as given below, using the Assumed Mean Method since we have a missing frequency and a given mean. Let the assumed mean (a) be 18.

Daily Pocket Allowance (in Rs.)

Number of Children (fᵢ)

Class Marks (xᵢ)

dᵢ = xᵢ − 18

fᵢdᵢ

11−13

7

12

−6

−42

13−15

6

14

−4

−24

15−17

9

16

−2

−18

17−19

13

18 = a

0

0

19−21

f

20

2

2f

21−23

5

22

4

20

23−25

4

24

6

24

Total

Σfᵢ = 44 + f



Σfᵢdᵢ = 2f − 40

It is given that mean = 18.

From the table, we have a = 18, N = 44 + f, and Σfᵢdᵢ = 2f − 40.

Now, Mean = a + (1/N) × Σfᵢdᵢ

Substituting the values as given above, we have:

18 = 18 + (1/(44+f)) × (2f−40)

0 = (2f−40) / (44+f)

Since 44+f cannot be zero (as frequency is always positive), the numerator must be zero:

2f − 40 = 0

2f = 40

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

Number of Women

65−68

2

68−71

4

71−74

3

74−77

8

77−80

7

80−83

4

83−86

2

Sol.

To find the mean, we can prepare the following table using the Direct Method:

No. of Heart Beats per Min

No. of Women (fᵢ)

Class Marks (xᵢ)

fᵢxᵢ

65−68

2

66.5

133.0

68−71

4

69.5

278.0

71−74

3

72.5

217.5

74−77

8

75.5

604.0

77−80

7

78.5

549.5

80−83

4

81.5

326.0

83−86

2

84.5

169.0

Total

N = Σfᵢ = 30


Σfᵢxᵢ = 2277.0

Mean = Σfᵢxᵢ / Σfᵢ = 2277 / 30 = 75.9.

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

No. of Mangoes

Number of Boxes

50−52

15

53−55

110

56−58

135

59−61

115

62−64

25

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

Sol.

First, notice that the class intervals are inclusive. To make them continuous (exclusive), we adjust the boundaries by 0.5. For example, 50-52 becomes 49.5-52.5.

Then we choose the Step Deviation Method as the class marks are relatively large, simplifying calculations. Let assumed mean (a) = 57 and class size (h) = 3.

Number of Mangoes (Adjusted Class)

Number of Boxes (fᵢ)

Class Mark (xᵢ)

uᵢ = (xᵢ − 57)/3

fᵢuᵢ

49.5−52.5

15

51

−2

−30

52.5−55.5

110

54

−1

−110

55.5−58.5

135

57

0

0

58.5−61.5

115

60

1

115

61.5−64.5

25

63

2

50

Total

N = Σfᵢ = 400



Σfᵢuᵢ = 25

We have a = 57, h = 3, N = 400, and Σfᵢuᵢ = 25.

By the Step Deviation Method:

Mean = a + h × (1/N) × Σfᵢuᵢ

= 57 + 3 × (1/400) × 25

= 57 + (75/400)

= 57 + 0.1875

= 57.1875 (or approximately 57.19)

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

Daily Expenditure (in Rs.)

No. of Households

100−150

4

150−200

5

200−250

12

250−300

2

300−350

2

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

Sol.

To find the mean, we can use the Direct Method:

Daily Exp. (in Rs.)

No. of Households (fᵢ)

Class Marks (xᵢ)

fᵢxᵢ

100−150

4

125

500

150−200

5

175

875

200−250

12

225

2700

250−300

2

275

550

300−350

2

325

650

Total

Σfᵢ = 25


Σfᵢxᵢ = 5275

Mean = Σfᵢxᵢ / Σfᵢ = 5275 / 25 = 211 Rs.

8. To find out the concentration of SO₂ 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 SO₂ (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 SO₂ in the air.

Sol.

To find the mean, we can use the Direct Method:

Concentration of SO₂ (in ppm)

Class Marks (xᵢ)

Frequency (fᵢ)

fᵢxᵢ

0.00−0.04

0.02

4

0.08

0.04−0.08

0.06

9

0.54

0.08−0.12

0.10

9

0.90

0.12−0.16

0.14

2

0.28

0.16−0.20

0.18

4

0.72

0.20−0.24

0.22

2

0.44

Total


Σfᵢ = 30

Σfᵢxᵢ = 2.96

Mean = Σfᵢxᵢ / Σfᵢ = 2.96 / 30 = 0.0986 ppm (approximately).


9. 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.

No. of Days

No. of Students

0−6

11

6−10

10

10−14

7

14−20

4

20−28

4

28−38

3

38−40

1

Sol.

To find the mean, we can use the Direct Method:

No. of Days

No. of Students (fᵢ)

Class Marks (xᵢ)

fᵢxᵢ

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

Total

Σfᵢ = 40


Σfᵢxᵢ = 499

Mean = Σfᵢxᵢ / Σfᵢ = 499 / 40 = 12.475 days.


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

Literacy Rate (in %)

No. of Cities

45−55

3

55−65

10

65−75

11

75−85

8

85−95

3

Sol.

To find the mean, we can use the Direct Method:

Literacy Rate (in %)

No. of Cities (fᵢ)

Class Marks (xᵢ)

fᵢxᵢ

45−55

3

50

150

55−65

10

60

600

65−75

11

70

770

75−85

8

80

640

85−95

3

90

270

Total

Σfᵢ = 35


Σfᵢxᵢ = 2430

Mean = Σfᵢxᵢ / Σfᵢ = 2430 / 35 = 69.43% (approximately).

4.0Key Features and Benefits Class 9 Maths Chapter 12 statistics Exercise 12.1

  • NCERT-Solved Solutions: Solutions based on NCERT guidelines, and solved by subject experts.
  • Clear Understanding of Data Handling: It provides students the correct methodology to organize data and counts frequency.
  • Stop Step-by-Step Solved Examples: Every question is broken down into steps and explained in detail.
  • Enhances Analytical Reasoning: It assists in honing logical thinking skills and abilities to make reasoning and conclusions.
  • Useful for Exams & Projects: A strong foundation in the area of data handling is important when completing examinations or applying it to an actual project, such as a survey or statistics based project.

NCERT Class 9 Maths Ch. 12 Statistics Other Exercises:-

Exercise 12.1


NCERT Solutions for Class 9 Maths Other Chapters:-

Chapter 1: Number Systems

Chapter 2: Polynomials

Chapter 3: Coordinate Geometry

Chapter 4: Linear Equations in Two Variables

Chapter 5: Introduction to Euclid’s Geometry

Chapter 6: Lines and Angles

Chapter 7: Triangles

Chapter 8: Quadrilaterals

Chapter 9: Circles

Chapter 10: Heron’s Formula

Chapter 11: Surface Areas and Volumes

Chapter 12: Statistics

Frequently Asked Questions

Grouping data into class intervals simplifies large datasets, making patterns and trends easier to analyze and interpret.

Yes. Methods include: Direct Method Assumed Mean Method Step-Deviation Method The choice depends on data size and convenience of calculation.

Absolutely! Students learn to handle real-life data, like measuring wages, heartbeats, expenditure, air pollution levels, and survey data.

You can download free NCERT Solutions PDFs for Class 9 Maths Chapter 12 Exercise 12.1 from trusted educational websites or resources.

Statistics helps develop analytical thinking and problem-solving skills, essential for handling real-world information and making decisions.

It’s a table showing how often each value or range of values occurs in a dataset. It makes data easier to interpret and visualize.

Exercise 12.1 includes: Organizing raw data Creating frequency tables Calculating mean Solving real-life data problems like wages, allowances, and survey results

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