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NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14 - Statistics

NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14 - Statistics

Edited By Ramraj Saini | Updated on Mar 26, 2024 10:45 AM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1

NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14 Statistics are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2024-25. Class 10 Maths ex 14.1 covers the classification of ungrouped and grouped frequency distributions. In the prior class, students should have learned how to visualise data in a variety of graphical styles, such as bar graphs, histograms, and frequency polygons. They must understand concepts like numerical representations of ungrouped data and measures of central tendencies like mean, median, and mode. In NCERT solutions Class 10 Maths chapter 14.1, students will learn to transition from ungrouped to grouped data and study all three measures: mean, median, and mode.

This Story also Contains
  1. NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.1
  2. Assess NCERT Solutions for Class 10 Maths chapter 14 exercise 14.1
  3. More About NCERT Solutions for Class 10 Maths Exercise 14.1 –
  4. Benefits of NCERT Solutions for Class 10 Maths Exercise 14.1
  5. Key Features For Class 10 Maths Chapter 14 Exercise 14.1
  6. NCERT Solutions for Class 10 Subject Wise
NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14 - Statistics
NCERT Solutions for Exercise 14.1 Class 10 Maths Chapter 14 - Statistics

Students will be introduced to new statistical concepts including cumulative frequency, cumulative frequency distribution, cumulative frequency curves, or 'ogives,' among others. We will also come across many statistical terms and their methods like class mark, direct method, assumed method for finding mean, etc. 10th class Maths exercise 14.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Assess NCERT Solutions for Class 10 Maths chapter 14 exercise 14.1

Statistics C lass 10 Chapter 14 Statistics Exercise: 14.1

Q1 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.

1

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

Answer:

Number of plants


Number of houses

f_i

Classmark

x_i

f_ix_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_ix_i

=162

Mean,

\overline x = \frac{\sum f_ix_i}{\sum f_i}
= \frac{162}{20} = 8.1
We used the direct method in this as the values of x_i\ and\ f_i are small.

Q2 Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.
2

Answer:

Let the assumed mean be a = 550

Daily

Wages

Number of

workers f_i

Classmark

x_i

d_i = x_i -a

f_id_i

500-520

12

510

-40

-480

520-540

14

530

-20

-280

540-560

8

550

0

0

560-580

6

570

20

120

580-600

10

590

40

400


\sum f_i

=50



\sum f_ix_i

=-240

Mean,

\overline x =a + \frac{\sum f_id_i}{\sum f_i}
= 550 + \frac{-240}{50} = 550-4.8 = 545.20
Therefore, the mean daily wages of the workers of the factory is Rs. 545.20

Q3 following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

3

Answer:

Daily pocket

allowance

Number of

children f_i

Classmark

x_i

f_ix_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


\sum f_i

=44+f


\sum f_ix_i

=752+20f

Mean,

\overline x = \frac{\sum f_ix_i}{\sum f_i}
\implies 18 = \frac{752+20f}{44+f}

\\ \implies 18(44+f) =( 752+20f) \\ \implies 2f = 40 \\ \implies f = 20
Therefore the missing f = 20

Q4 Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

1636093324773

Answer:

Let the assumed mean be a = 75.5


No. of heartbeats

per minute

Number of

women f_i

Classmark

x_i

d_i = x_i -a

f_id_i

65-68

2

66.5

-9

-18

68-71

4

69.5

-6

-24

71-74

3

72.5

-3

-9

74-77

8

75.5

0

0

77-80

7

78.5

3

21

80-83

4

81.5

6

24

83-86

2

84.5

9

18


\sum f_i

=30



\sum f_ix_i

=12

Mean,

\overline x =a + \frac{\sum f_id_i}{\sum f_i}
= 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9
Therefore, the mean heartbeats per minute of these women are 75.9

Q5 In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution of mangoes according to the number of boxes.

5

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

Answer:

Let the assumed mean be a = 57

Number of

mangoes

Number of

boxes f_i

Classmark

x_i

d_i = x_i -a

f_id_i

50-52

15

51

-6

-90

53-55

110

54

-3

-330

56-58

135

57

0

0

59-61

115

60

3

345

62-64

25

63

6

150


\sum f_i

=400



\sum f_ix_i

=75

Mean,

\overline x =a + \frac{\sum f_id_i}{\sum f_i}
= 57+ \frac{75}{400} = 57+0.1875 = 57.1875 \approx 57.19
Therefore, the mean number of mangoes kept in a packing box is approx 57.19

Q6 The table below shows the daily expenditure on the food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

Daily expenditure in rupees

100-150150-200200-250250-300300-350
Number of households451222


Answer:

Let the assumed mean be a = 225 and h = 50

Daily

Expenditure

Number of

households f_i

Classmark

x_i

d_i = x_i -a

u_i = \frac{d_i}{h}

f_iu_i

100-150

4

125

-100

-2

-8

150-200

5

175

-50

-1

-5

200-250

12

225

0

0

0

250-300

2

275

50

1

2

300-350

2

325

100

2

4


\sum f_i

=25




\sum f_ix_i

= -7

Mean,

\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h
= 225 + \frac{-7}{25}\times50 = 225 -14 = 211
Therefore, the mean daily expenditure on food is Rs. 211

Q7 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:
7

Find the mean concentration of SO_2 in the air.

Answer:

Class

Interval

Frequency

f_i

Classmark

x_i

f_ix_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


\sum f_i

=30


\sum f_ix_i

=2.96

Mean,

\overline x = \frac{\sum f_ix_i}{\sum f_i}
= \frac{2.96}{30} = 0.099
Therefore, the mean concentration of SO_2 in the air is 0.099 ppm

Q8 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.

1636093576677

Answer:

Number of

days

Number of

Students f_i

Classmark

x_i

f_ix_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_ix_i

=499

Mean,

\overline x = \frac{\sum f_ix_i}{\sum f_i}
= \frac{499}{40} = 12.475 = \frac{499}{40} = 12.475\approx 12.48
Therefore, the mean number of days a student was absent is 12.48 days.

Q9 The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

9

Answer:

Let the assumed mean be a = 75 and h = 10

Literacy

rates

Number of

cities f_i

Classmark

x_i

d_i = x_i -a

u_i = \frac{d_i}{h}

f_iu_i

45-55

3

50

-20

-2

-6

55-65

10

60

-10

-1

-10

65-75

11

70

0

0

0

75-85

8

80

10

1

8

85-95

3

90

20

2

6


\sum f_i

= 35




\sum f_ix_i

= -2

Mean,

\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h
= 70 + \frac{-2}{35}\times10 = 70 -0.57 = 69.43
Therefore, the mean mean literacy rate is 69.43%



More About NCERT Solutions for Class 10 Maths Exercise 14.1 –

In this exercise we will now go through the definition of some basics terms like mean, median mode etc.

The mean (or average) of observations is the sum of all the observations' values divided by the total number of observations.

1639043472375

1639043472017

Where

1639043471526- sum of the values of all the observations

1639043471744 - sum of the number of observations

1639043472601

In statistics, the median is the middle value of a set of data when ordered in a certain order. Data or observations might be arranged in ascending or descending order.

The mode is the observation with the greatest frequency in grouped data without class intervals.

Also Read| Statistics Class 10 Notes

Benefits of NCERT Solutions for Class 10 Maths Exercise 14.1

  • Exercise 14.1 Class 10 Maths, is based on Statistics and its uses.
  • From Class 10 Maths chapter 14 exercise 14.1 we learn new formulas to calculate the mean and class mark.
  • Understanding the concepts from Class 10 Maths chapter 14 exercise 14.1 will allow us to understand the concepts related to statistics of higher class.
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Key Features For Class 10 Maths Chapter 14 Exercise 14.1

  • Step-by-Step Solutions: Clear and detailed explanations for each problem in Ex 14.1 class 10, ensuring students understand the methodology.
  • Conceptual Understanding: Explanation of statistical concepts involved in the exercises, aiding students in grasping the underlying principles are discussed in class 10 maths ex 14.1.
  • Variety of Problems: In class 10 ex 14.1, diverse types of problems, offering a broad spectrum of challenges to reinforce understanding.
  • Visual Aids, Diagrams, and Examples: Use of visual representations, diagrams, and real-life examples to simplify complex concepts and aid visual learners.
  • Application-Based Problems: In10th class maths exercise 14.1 answers, application-oriented problems that relate statistics to real-world scenarios, fostering practical understanding and problem-solving skills.

Also, See

NCERT Solutions for Class 10 Subject Wise

Frequently Asked Question (FAQs)

1. What is the main concept of Class 10 Maths chapter 14 exercise 14.1?

To find what is the basics of statistic. Definition of mean and its formula.  To learn about different methods of finding mean like Assumed mean method, step deviation method etc. Go through the ex 14.1 class 10 to command these concepts.

2. What do you understand by the mean of a set of data?

The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. Practice class 10 maths ex 14.1 to command these concepts.

3. What is class mark?

The class midpoint (or class mark) is a specific point in the centre of the bins (categories) in a frequency distribution table. Go through exercise 14.1 class 10 maths to get deeper understanding of concepts.

4. Find the mean of the data?

This class 10 ex 14.1 discusses the concept of mean in detail. to find mean of data use formula discussed in this exercise. mean = sum of total data / number of data

5. The centre of a bar in a histogram is known as?

The centre of a bar in a histogram is known as class mark.

6. What will happen to the mean of the data if every data set has increased by the value of 5?

The mean of the data will also increase by 5.

7. A data set has few intervals, one of such intervals is between 20 and 40. Find the class mark of this interval?

The mean of the data will also increase by 5.

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