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NCERT Solutions for Exercise 14.3 Class 10 Maths Chapter 14 - Statistics

NCERT Solutions for Exercise 14.3 Class 10 Maths Chapter 14 - Statistics

Edited By Ramraj Saini | Updated on Dec 02, 2023 10:37 AM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3

NCERT Solutions for Exercise 14.3 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 2023-24. Class 10 Maths ex 14.3 introduces us to the concept of finding the median. In ninth grade, we learned about the median of grouped data. The median is a measure of central tendency that represents the value of the data's middle observation. The median depends on the number of observations. The value of the median is different if the number of observations is odd and different if the number of observations is even. This exercise class 10 ex 14.3 feature a variety of questions that cover the complete topic of determining the median in grouped data.

This Story also Contains
  1. NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 14 exercise 14.3
  3. Assess NCERT Solutions for Class 10 Maths chapter 14 exercise 14.3
  4. Statistics C lass 10 Chapter 14 Statistics Exercise: 14.3
  5. More About NCERT Solutions for Class 10 Maths Exercise 14.3
  6. Benefits of NCERT Solutions for Class 10 Maths Exercise 14.3
  7. Key Features For Class 10 Maths Chapter 14 Exercise 14.3
  8. NCERT Solutions for Class 10 Subject Wise
NCERT Solutions for Exercise 14.3 Class 10 Maths Chapter 14 - Statistics
NCERT Solutions for Exercise 14.3 Class 10 Maths Chapter 14 - Statistics

Exercise 14.3 Class 10 Maths also gives us information about what is cumulative frequency column, median class and also we will know about the relation between mean, median and mode. 10th class Maths exercise 14.3 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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Statistics C lass 10 Chapter 14 Statistics Exercise: 14.3

Q1 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.

1640762287658

Answer:

Let the assumed mean be a = 130 and h = 20

Class

Number of

consumers f_i

Cumulative

Frequency

Classmark

x_i

d_i = x_i -a

u_i = \frac{d_i}{h}

f_iu_i

65-85

4

4

70

-60

-3

-12

85-105

5

9

90

-40

-2

-10

105-125

13

22

110

-20

-1

-13

125-145

20

42

130

0

0

0

145-165

14

56

150

20

1

14

165-185

8

64

170

40

2

16

185-205

4

68

190

60

3

12



\sum f_i = N

= 68




\sum f_ix_i

= 7

MEDIAN:
N= 68 \implies \frac{N}{2} = 34
\therefore Median class = 125-145; Cumulative Frequency = 42; Lower limit, l = 125;

c.f. = 22; f = 20; h = 20
Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 125 + \left (\frac{34-22}{20} \right ).20 \\ \\ = 125 + 12

= 137

Thus, the median of the data is 137

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 125-145
Lower limit (l) of modal class = 125, class size (h) = 20
Frequency ( f_1 ) of the modal class = 20; frequency ( f_0 ) of class preceding the modal class = 13, frequency ( f_2 ) of class succeeding the modal class = 14.

Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h

\\ = 125 + \left(\frac{20-13}{2(20)-13-14} \right).20 \\ \\ = 125 + \frac{7}{13}.20

= 135.76

Thus, Mode of the data is 135.76

MEAN:

Mean,
\overline x =a + \frac{\sum f_iu_i}{\sum f_i}\times h
= 130 + \frac{7}{68}\times20 = 137.05

Thus, the Mean of the data is 137.05

Q2 If the median of the distribution given below is 28.5, find the values of x and y.

1640762346146

Answer:

Class

Number of

consumers f_i

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


\sum f_i = N

= 60



N= 60 \implies \frac{N}{2} = 30

Now,
Given median = 28.5 which lies in the class 20-30

Therefore, Median class = 20-30
Frequency corresponding to median class, f = 20
Cumulative frequency of the class preceding the median class, c.f. = 5 + x
Lower limit, l = 20; Class height, h = 10

Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ \implies28.5= 20 + \left (\frac{30-5-x}{20} \right ).10 \\ \\ \implies8.5=\frac{25-x}{2} \\ \implies 25-x = 8.5(2) \\ \implies x = 25 - 17 = 8

Also,

\\ 60 = 45 + x+y \\ \implies x+y = 60-45 = 15 \\ \implies y = 15-x = 15-8 \ \ \ (\because x =8) \\ \implies y = 7

Therefore, the required values are: x=8 and y=7

Q3 A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

1640762364185

Answer:

Class

Frequency

f_i

Cumulative

Frequency

15-20

2

2

20-25

4

6

25-30

18

24

30-35

21

45

35-40

33

78

40-45

11

89

45-50

3

92

50-55

6

98

55-60

2

100


N= 100 \implies \frac{N}{2} = 50
Therefore, Median class = 35-45
Frequency corresponding to median class, f = 21
Cumulative frequency of the class preceding the median class, c.f. = 24
Lower limit, l = 35; Class height, h = 10

Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 35 + \left (\frac{50-45}{33} \right ).5 \\ \\

= 35.75

Thus, the median age is 35.75 years.

Q4 The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table :

1640762380533

Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to
117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Answer:

The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes.

Class

Frequency

f_i

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


\dpi{100} N= 40 \implies \frac{N}{2} = 20
Therefore, Median class = 144.5-153.5

Lower limit, l = 144.5; Class height, h = 9
Frequency corresponding to median class, f = 12
Cumulative frequency of the class preceding the median class, c.f. = 17

Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 144.5 + \left (\frac{20-17}{12} \right ).9 \\ \\

= 146.75

Thus, the median length of the leaves is 146.75 mm

Q5 The following table gives the distribution of the lifetime of 400 neon lamps :
1640762410074


Find the median lifetime of a lamp.

Answer:

Class

Frequency

f_i

Cumulative

Frequency

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

\dpi{100} N= 400 \implies \frac{N}{2} = 200
Therefore, Median class = 3000-3500

Lower limit, l = 3000; Class height, h = 500
Frequency corresponding to median class, f = 86
Cumulative frequency of the class preceding the median class, c.f. = 130

Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 3000 + \left (\frac{200-130}{86} \right ).500 \\ \\ = 3000+406.97

= 3406.97

Thus, the median lifetime of a lamp is 3406.97 hours

= 146.75

Thus, the median length of the leaves is 146.75 mm

Q6 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:

1640762454336

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.

Answer:

Class

Number of

surnames f_i

Cumulative

Frequency

Classmark

x_i

f_ix_i

1-4

6

6

2.5

15

4-7

30

36

5.5

165

7-10

40

76

8.5

340

10-13

16

92

11.5

184

13-16

4

96

14.5

51

16-19

4

100

17.5

70



\sum f_i = N

= 100


\sum f_ix_i

= 825

MEDIAN:
N= 100 \implies \frac{N}{2} = 50
\therefore Median class = 7-10; Lower limit, l = 7;

Cumulative frequency of preceding class, c.f. = 36; f = 40; h = 3
Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 7+ \left (\frac{50-36}{40} \right ).3 \\ \\

= 8.05

Thus, the median of the data is 8.05

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 40 and hence the modal class = 7-10
Lower limit (l) of modal class = 7, class size (h) = 3
Frequency ( f_1 ) of the modal class = 40; frequency ( f_0 ) of class preceding the modal class = 30, frequency ( f_2 ) of class succeeding the modal class = 16

Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h

\\ = 7 + \left(\frac{40-30}{2(40)-30-16} \right).3 \\ \\ = 125 + \frac{10}{34}.3

= 7.88

Thus, Mode of the data is 7.88

MEAN:

Mean,
\overline x =\frac{\sum f_ix_i}{\sum f_i}
= \frac{825}{100} = 8.25

Thus, the Mean of the data is 8.25

Q7 The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

1640762471389

Answer:

Class

Number of

students f_i

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

MEDIAN:
N= 30 \implies \frac{N}{2} = 15
\therefore Median class = 55-60; Lower limit, l = 55;

Cumulative frequency of preceding class, c.f. = 13; f = 6; h = 5
Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 55+ \left (\frac{15-13}{6} \right ).5 \\ \\ = 55+\frac{2}{6}.5

= 56.67

Thus, the median weight of the student is 56.67 kg



More About NCERT Solutions for Class 10 Maths Exercise 14.3

Class 10 Maths chapter 14 exercise 14 includes some of the basic questions in which we have to find the median of the grouped data. Along with this NCERT solutions for Class 10 Maths exercise 14.3 includes questions in which there is no class and we have to form the class of data ourselves. We also have to find all the central tendencies for some of the questions.

Exercise 14.3 Class 10 Maths covers all types of questions that can be formed in the median of grouped data and it has the question in which we have to find the frequency of the specific modal class. Moreover, there are few examples before NCERT solutions for Class 10 Maths exercise 14.3 that have questions that are based on the relation of mean, median and mode.

Also Read| Statistics Class 10 Notes

Benefits of NCERT Solutions for Class 10 Maths Exercise 14.3

  • Chapter 14 exercise 14.3 in Class 10 Maths presents a wide range of issues that might be based on a central tendency median, particularly on grouped data.
  • NCERT Statistics is a big element of the NCERT curriculum for Class 10 Mathematics chapter 14 exercise 14.3. It will be useful in JEE Main (joint admission exam) as statistics is a major part of the syllabus.
  • Class 10 Mathematics chapter 14 exercise 14.3 will serve as a foundation for Class 11 statistics chapter 15 (which is a continuation of this chapter).
  • The median can be used as a measure of location when one attaches reduced importance to extreme values

Key Features For Class 10 Maths Chapter 14 Exercise 14.3

  • Step-by-Step Solutions: Clear and detailed explanations for each problem in Ex 14.3 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 14.3.
  • Variety of Problems: In class 10 ex 14.3, 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 14.3 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 Questions (FAQs)

1. How does the median depend on the number of observations?

Median gives the value of the observation which is at the center so it is dependent on either the observation is odd number or even number. Practice ex 14.3 class 10 to command these concepts.

2. What is the formula of median of odd observation ?

The concept related to discusses in class 10 ex 14.3. Formula of median of odd observation is  n/2th observation. Students can practice ex 14.3 class 10 to get deeper understanding of concepts.

3. What is the formula of median of even observation?

The concepts related to median is discussed in class 10 maths ex 14.3. Formula of median of even observation is  (n/2+ 1 )th observation. Also practice class 10 ex 14.3 which is discussed in this article to get deeper understanding of the concepts.

4. What is the relation between mean, medium and mode as mentioned in Class 10 Maths chapter 14 exercise 14.3?

3 Median = Mode + 2 Mean  is the relation between mean, medium and mode as mentioned in Class 10 maths chapter 14 exercise 14.3. 

5. What do you mean by Cumulative Frequency Table ?

These concepts are discussed in 10th class maths exercise 14.3 answers. practice them to command the concepts. Cumulative Frequency Table is the cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors.

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sadhu ashram ramgart road aligarh

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sai raja 01 September,2024
show the previous year exams ?

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best of luck!

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Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

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