Open App

Class 10 Exam > Class 10 Notes > Mathematics (Maths) Class 10 > RD Sharma Solutions: Statistics (Exercise 14D)

Statistics (Exercise 14D) RD Sharma Solutions | Mathematics (Maths) Class 10 PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


 
1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h
Page 2


 
1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h
 
= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50
Page 3


 
1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h
 
= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50
 
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 36 and the corresponding class is 60 80.
Thus, the median class is 60 80.
l = 60, h = 20, N = 50, f = 12 and cf = 24.
Now,
Median = l + × h
= 60 + × 20
= 60 + 1.67
= 61.67
Thus, the median is 61.67.
We know that,
Mode = 3(median) 2(mean)
= 3 × 61.67 2 × 62.4
= 185.01 124.8
= 60.21
Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21
3. Find the mean, median and mode of the following data:
Class 0 50 50 100 100 150 150 200 200 250 250 300 300 - 350
Frequency 2 3 5 6 5 3 1
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 50 2 25 50
50 100 3 75 225
100 150 5 125 625
150 200 6 175 1050
200 250 5 225 1125
250 300 3 275 825
300 350 1 325 325
Total
fi = 25 fi xi = 4225
Mean =
i i
i
i
i
f x
f
=
= 169
Thus, mean of the given data is 169.
Page 4


 
1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h
 
= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50
 
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 36 and the corresponding class is 60 80.
Thus, the median class is 60 80.
l = 60, h = 20, N = 50, f = 12 and cf = 24.
Now,
Median = l + × h
= 60 + × 20
= 60 + 1.67
= 61.67
Thus, the median is 61.67.
We know that,
Mode = 3(median) 2(mean)
= 3 × 61.67 2 × 62.4
= 185.01 124.8
= 60.21
Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21
3. Find the mean, median and mode of the following data:
Class 0 50 50 100 100 150 150 200 200 250 250 300 300 - 350
Frequency 2 3 5 6 5 3 1
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 50 2 25 50
50 100 3 75 225
100 150 5 125 625
150 200 6 175 1050
200 250 5 225 1125
250 300 3 275 825
300 350 1 325 325
Total
fi = 25 fi xi = 4225
Mean =
i i
i
i
i
f x
f
=
= 169
Thus, mean of the given data is 169.
 
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 50 2 2
50 100 3 5
100 150 5 10
150 200 6 16
200 250 5 21
250 300 3 24
300 350 1 25
Total N = = 25
Now, N = 25 = 12.5.
The cumulative frequency just greater than 12.5 is 16 and the corresponding class is 150 
200.
Thus, the median class is 150 200.
l = 150, h = 50, N = 25, f = 6 and cf = 10.
Now,
Median = l + × h
= 150 + × 50
= 150 + 20.83
= 170.83
Thus, the median is 170.83.
We know that,
Mode = 3(median) 2(mean)
= 3 × 170.83 2 × 169
= 512.49 338
= 174.49
Hence, Mean = 169, Median = 170.83 and Mode = 174.49
4. Find the mean, median and mode of the following data:
Marks 
obtained
25 - 35 35 45 45 55 55 65 65 75 75 - 85
No. of 
students
7 31 33 17 11 1
Sol: 
To find the mean let us put the data in the table given below:
Marks 
obtained
Number of students (fi) Class mark (x i) fi xi
25 35 7 30 210
35 45 31 40 1240
Page 5


 
1. Find the mean, median and mode of the following data:
Class 0 10 10 20 20 30 30 40 40 50 50 60 60 70
Frequency 4 4 7 10 12 8 5
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 10 4 5 20
10 20 4 15 60
20 30 7 25 175
30 40 10 35 350
40 50 12 45 540
50 60 8 55 440
60 70 5 65 325
Total
fi = 50 fi xi = 1910
Mean =
i i
i
i
i
f x
f
=
= 38.2
Thus, the mean of the given data is 38.2.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 10 4 4
10 20 4 8
20 30 7 15
30 40 10 25
40 50 12 37
50 60 8 45
60 70 5 50
Total N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 50.
Thus, the median class is 40 50.
l = 40, h = 10, N = 50, f = 12 and cf = 25.
Now,
Median = l + × h
 
= 40 + × 10
= 40
Thus, the median is 40.
We know that,
Mode = 3(median) 2(mean)
= 3 × 40 2 × 38.2
= 120 76.4
= 43.6
Hence, Mean = 38.2, Median = 40 and Mode = 43.6
2. Find the mean, median and mode of the following data:
Class 0 20 20 40 40 60 60 80 80 100 100 120 120 140
Frequency 6 8 10 12 6 5 3
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 20 6 10 60
20 40 8 30 240
40 60 10 50 500
60 80 12 70 840
80 100 6 90 540
100 120 5 110 550
120 140 3 130 390
Total
fi = 50 fi xi = 3120
Mean =
i i
i
i
i
f x
f
=
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 20 6 6
20 40 8 14
40 60 10 24
60 80 12 36
80 100 6 42
100 120 5 47
120 140 3 50
Total N = = 50
 
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 36 and the corresponding class is 60 80.
Thus, the median class is 60 80.
l = 60, h = 20, N = 50, f = 12 and cf = 24.
Now,
Median = l + × h
= 60 + × 20
= 60 + 1.67
= 61.67
Thus, the median is 61.67.
We know that,
Mode = 3(median) 2(mean)
= 3 × 61.67 2 × 62.4
= 185.01 124.8
= 60.21
Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21
3. Find the mean, median and mode of the following data:
Class 0 50 50 100 100 150 150 200 200 250 250 300 300 - 350
Frequency 2 3 5 6 5 3 1
Sol: 
To find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (x i) fi xi
0 50 2 25 50
50 100 3 75 225
100 150 5 125 625
150 200 6 175 1050
200 250 5 225 1125
250 300 3 275 825
300 350 1 325 325
Total
fi = 25 fi xi = 4225
Mean =
i i
i
i
i
f x
f
=
= 169
Thus, mean of the given data is 169.
 
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
0 50 2 2
50 100 3 5
100 150 5 10
150 200 6 16
200 250 5 21
250 300 3 24
300 350 1 25
Total N = = 25
Now, N = 25 = 12.5.
The cumulative frequency just greater than 12.5 is 16 and the corresponding class is 150 
200.
Thus, the median class is 150 200.
l = 150, h = 50, N = 25, f = 6 and cf = 10.
Now,
Median = l + × h
= 150 + × 50
= 150 + 20.83
= 170.83
Thus, the median is 170.83.
We know that,
Mode = 3(median) 2(mean)
= 3 × 170.83 2 × 169
= 512.49 338
= 174.49
Hence, Mean = 169, Median = 170.83 and Mode = 174.49
4. Find the mean, median and mode of the following data:
Marks 
obtained
25 - 35 35 45 45 55 55 65 65 75 75 - 85
No. of 
students
7 31 33 17 11 1
Sol: 
To find the mean let us put the data in the table given below:
Marks 
obtained
Number of students (fi) Class mark (x i) fi xi
25 35 7 30 210
35 45 31 40 1240
 
45 55 33 50 1650
55 65 17 60 1020
65 75 11 70 770
75 85 1 80 80
Total
fi = 100 fi xi = 4970
Mean =
i i
i
i
i
f x
f
=
= 49.7
Thus, mean of the given data is 49.7.
Now, to find the median let us put the data in the table given below:
Class Frequency (fi) Cumulative Frequency (cf)
25 35 7 7
35 45 31 38
45 55 33 71
55 65 17 88
65 75 11 99
75 85 1 100
Total N = = 100
Now, N = 100 = 50.
The cumulative frequency just greater than 50 is 71 and the corresponding class is 45 55.
Thus, the median class is 45 55.
l = 45, h = 10, N = 100, f = 33 and cf = 38.
Now,
Median = l + × h
= 45 + × 10
= 45 + 3.64
= 48.64
Thus, the median is 48.64.
We know that,
Mode = 3(median) 2(mean)
= 3 × 48.64 2 × 49.70
= 145.92 99.4
= 46.52
Hence, Mean = 49.70, Median = 48.64 and Mode = 46.52

	Mathematics (Maths) Class 10 124 videos\|457 docs\|77 tests

Mathematics (Maths) Class 10

124 videos|457 docs|77 tests

Join Course for Free

Top Courses for Class 10

View all

Related Exams

Class 10

About this Document

	4.86/5 Rating
	Nov 07, 2024 Last updated

Document Description: RD Sharma Solutions: Statistics (Exercise 14D) for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The notes and questions for RD Sharma Solutions: Statistics (Exercise 14D) have been prepared according to the Class 10 exam syllabus. Information about RD Sharma Solutions: Statistics (Exercise 14D) covers topics like and RD Sharma Solutions: Statistics (Exercise 14D) Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Statistics (Exercise 14D).

Introduction of RD Sharma Solutions: Statistics (Exercise 14D) in English is available as part of our Mathematics (Maths) Class 10 for Class 10 & RD Sharma Solutions: Statistics (Exercise 14D) in Hindi for Mathematics (Maths) Class 10 course. Download more important topics related with notes, lectures and mock test series for Class 10 Exam by signing up for free. Class 10: Statistics (Exercise 14D) RD Sharma Solutions | Mathematics (Maths) Class 10

Description

Full syllabus notes, lecture & questions for Statistics (Exercise 14D) RD Sharma Solutions | Mathematics (Maths) Class 10 - Class 10 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 10 | Best notes, free PDF download

Information about RD Sharma Solutions: Statistics (Exercise 14D)

In this doc you can find the meaning of RD Sharma Solutions: Statistics (Exercise 14D) defined & explained in the simplest way possible. Besides explaining types of RD Sharma Solutions: Statistics (Exercise 14D) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Statistics (Exercise 14D) tests, examples and also practice Class 10 tests

	Mathematics (Maths) Class 10 124 videos\|457 docs\|77 tests

Mathematics (Maths) Class 10

124 videos|457 docs|77 tests

Join Course for Free

Download as PDF

Explore Courses for Class 10 exam

Top Courses for Class 10

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

mock tests for examination

Statistics (Exercise 14D) RD Sharma Solutions | Mathematics (Maths) Class 10

ppt

past year papers

Important questions

MCQs

pdf

shortcuts and tricks

Free

Objective type Questions

Previous Year Questions with Solutions

Statistics (Exercise 14D) RD Sharma Solutions | Mathematics (Maths) Class 10

Sample Paper

Exam

practice quizzes

study material

video lectures

Statistics (Exercise 14D) RD Sharma Solutions | Mathematics (Maths) Class 10

Semester Notes

Extra Questions

Summary

Viva Questions

;

Additional Information about RD Sharma Solutions: Statistics (Exercise 14D) for Class 10 Preparation

RD Sharma Solutions: Statistics (Exercise 14D) Free PDF Download

The RD Sharma Solutions: Statistics (Exercise 14D) is an invaluable resource that delves deep into the core of the Class 10 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Statistics (Exercise 14D) now and kickstart your journey towards success in the Class 10 exam.

Importance of RD Sharma Solutions: Statistics (Exercise 14D)

The importance of RD Sharma Solutions: Statistics (Exercise 14D) cannot be overstated, especially for Class 10 aspirants. This document holds the key to success in the Class 10 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

RD Sharma Solutions: Statistics (Exercise 14D) Notes

RD Sharma Solutions: Statistics (Exercise 14D) Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to RD Sharma Solutions: Statistics (Exercise 14D). It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Statistics (Exercise 14D) Notes on EduRev are your ultimate resource for success.

RD Sharma Solutions: Statistics (Exercise 14D) Class 10 Questions

The "RD Sharma Solutions: Statistics (Exercise 14D) Class 10 Questions" guide is a valuable resource for all aspiring students preparing for the Class 10 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study RD Sharma Solutions: Statistics (Exercise 14D) on the App

Students of Class 10 can study RD Sharma Solutions: Statistics (Exercise 14D) alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Statistics (Exercise 14D), students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of RD Sharma Solutions: Statistics (Exercise 14D) is prepared as per the latest Class 10 syllabus.

Education Revolution