Open App

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

Statistics (Exercise 14B) 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. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age 
(in years)
0 15 15 30 30 45 45 60 60 - 75
No. of patients 5 20 40 50 25
Sol:
We prepare the cumulative frequency table, as shown below:
Age (in years) Number of patients (fi) Cumulative Frequency (cf)
0 15 5 5
15 30 20 25
30 45 40 65
45 60 50 115
60 75 25 140
Total N = = 140
Now, N = 140 = 70.
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45 
60.
Thus, the median class is 45 60.
l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Median = l + × h
= 45 + × 15
= 45 + × 15
= 45 + 1.5
= 46.5
Page 2


1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age 
(in years)
0 15 15 30 30 45 45 60 60 - 75
No. of patients 5 20 40 50 25
Sol:
We prepare the cumulative frequency table, as shown below:
Age (in years) Number of patients (fi) Cumulative Frequency (cf)
0 15 5 5
15 30 20 25
30 45 40 65
45 60 50 115
60 75 25 140
Total N = = 140
Now, N = 140 = 70.
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45 
60.
Thus, the median class is 45 60.
l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Median = l + × h
= 45 + × 15
= 45 + × 15
= 45 + 1.5
= 46.5
 
Hence, the median age is 46.5 years.
2. Compute mean from the following data:
Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49
Number of 
Students
3 4 7 11 0 16 9
Sol: 
Class Frequency (f) Cumulative Frequency (cf)
0 7 3 3
7 14 4 7
14 21 7 14
21 - 28 11 25
28 35 0 25
35 42 16 41
42 49 9 50
N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42.
Thus, the median class is 35 42.
l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25.
Now,
Median = l + × h
= 35 + 7 ×
= 35 + 0
= 35
Hence, the median age is 46.5 years.
3. The following table shows the daily wages of workers in a factory:
0 100 100 200 200 300 300 400 400 500
Number of 
workers
40 32 48 22 8
Find the median daily wage income of the workers.
Sol: 
Class Frequency (f) Cumulative Frequency (cf)
0 100 40 40
100 200 32 72
200 300 48 120
300 400 22 142
400 500 8 150
N = = 150
Page 3


1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age 
(in years)
0 15 15 30 30 45 45 60 60 - 75
No. of patients 5 20 40 50 25
Sol:
We prepare the cumulative frequency table, as shown below:
Age (in years) Number of patients (fi) Cumulative Frequency (cf)
0 15 5 5
15 30 20 25
30 45 40 65
45 60 50 115
60 75 25 140
Total N = = 140
Now, N = 140 = 70.
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45 
60.
Thus, the median class is 45 60.
l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Median = l + × h
= 45 + × 15
= 45 + × 15
= 45 + 1.5
= 46.5
 
Hence, the median age is 46.5 years.
2. Compute mean from the following data:
Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49
Number of 
Students
3 4 7 11 0 16 9
Sol: 
Class Frequency (f) Cumulative Frequency (cf)
0 7 3 3
7 14 4 7
14 21 7 14
21 - 28 11 25
28 35 0 25
35 42 16 41
42 49 9 50
N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42.
Thus, the median class is 35 42.
l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25.
Now,
Median = l + × h
= 35 + 7 ×
= 35 + 0
= 35
Hence, the median age is 46.5 years.
3. The following table shows the daily wages of workers in a factory:
0 100 100 200 200 300 300 400 400 500
Number of 
workers
40 32 48 22 8
Find the median daily wage income of the workers.
Sol: 
Class Frequency (f) Cumulative Frequency (cf)
0 100 40 40
100 200 32 72
200 300 48 120
300 400 22 142
400 500 8 150
N = = 150
 
Now, N = 150 
= 75.
The cumulative frequency just greater than 75 is 120 and the corresponding class is 200 
300.
Thus, the median class is 200 300.
l = 200, h = 100, f = 48, cf = c.f. of preceding class = 72 and = 75.
Now,
Median, M = l +
= 200 + 
= 200 + 6.25
= 206.25
Hence, the median daily wage income of the workers is Rs 206.25.
4. Calculate the median from the following frequency distribution table:
Class 5 10 10 15 15 20 20 25 25 30 30 35 35 40 40 45
Frequency 5 6 15 10 5 4 2 2
Sol: 
Class Frequency (f) Cumulative Frequency 
(cf)
5 10 5 5
10 15 6 11
15 20 15 26
20 25 10 36
25 30 5 41
30 35 4 45
35 40 2 47
40 45 2 49
N = = 49
Now, N = 49 
= 24.5.
The cumulative frequency just greater than 24.5 is 26 and the corresponding class is 15 -
20.
Thus, the median class is 15 20.
l = 15, h = 5, f = 15, cf = c.f. of preceding class = 11 and = 24.5.
Now,
Median, M = l +
= 15 + 
Page 4


1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age 
(in years)
0 15 15 30 30 45 45 60 60 - 75
No. of patients 5 20 40 50 25
Sol:
We prepare the cumulative frequency table, as shown below:
Age (in years) Number of patients (fi) Cumulative Frequency (cf)
0 15 5 5
15 30 20 25
30 45 40 65
45 60 50 115
60 75 25 140
Total N = = 140
Now, N = 140 = 70.
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45 
60.
Thus, the median class is 45 60.
l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Median = l + × h
= 45 + × 15
= 45 + × 15
= 45 + 1.5
= 46.5
 
Hence, the median age is 46.5 years.
2. Compute mean from the following data:
Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49
Number of 
Students
3 4 7 11 0 16 9
Sol: 
Class Frequency (f) Cumulative Frequency (cf)
0 7 3 3
7 14 4 7
14 21 7 14
21 - 28 11 25
28 35 0 25
35 42 16 41
42 49 9 50
N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42.
Thus, the median class is 35 42.
l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25.
Now,
Median = l + × h
= 35 + 7 ×
= 35 + 0
= 35
Hence, the median age is 46.5 years.
3. The following table shows the daily wages of workers in a factory:
0 100 100 200 200 300 300 400 400 500
Number of 
workers
40 32 48 22 8
Find the median daily wage income of the workers.
Sol: 
Class Frequency (f) Cumulative Frequency (cf)
0 100 40 40
100 200 32 72
200 300 48 120
300 400 22 142
400 500 8 150
N = = 150
 
Now, N = 150 
= 75.
The cumulative frequency just greater than 75 is 120 and the corresponding class is 200 
300.
Thus, the median class is 200 300.
l = 200, h = 100, f = 48, cf = c.f. of preceding class = 72 and = 75.
Now,
Median, M = l +
= 200 + 
= 200 + 6.25
= 206.25
Hence, the median daily wage income of the workers is Rs 206.25.
4. Calculate the median from the following frequency distribution table:
Class 5 10 10 15 15 20 20 25 25 30 30 35 35 40 40 45
Frequency 5 6 15 10 5 4 2 2
Sol: 
Class Frequency (f) Cumulative Frequency 
(cf)
5 10 5 5
10 15 6 11
15 20 15 26
20 25 10 36
25 30 5 41
30 35 4 45
35 40 2 47
40 45 2 49
N = = 49
Now, N = 49 
= 24.5.
The cumulative frequency just greater than 24.5 is 26 and the corresponding class is 15 -
20.
Thus, the median class is 15 20.
l = 15, h = 5, f = 15, cf = c.f. of preceding class = 11 and = 24.5.
Now,
Median, M = l +
= 15 + 
 
= 15 + 4.5
= 19.5
Hence, the median = 19.5.
5. Given below is the number of units of electricity consumed in a week in a certain locality:
Calculate the median.
Sol: 
Class Frequency (f) Cumulative Frequency (cf)
65- 85 4 4
85 105 5 9
105 125 13 22
125 145 20 42
145 165 14 56
165 185 7 63
185 205 4 67
N = = 67
Now, N = 67
= 33.5.
The cumulative frequency just greater than 33.5 is 42 and the corresponding class is 125 -
145.
Thus, the median class is 125 145.
l = 125, h = 20, f = 20, cf = c.f. of preceding class = 22 and = 33.5.
Now,
Median, M = l +
= 125 + 
= 125 + 11.5
= 136.5
Hence, the median = 136.5.
6. Calculate the median from the following data:
Class 65 85 85 105 105 125 125 145 145 165 165 185 185 200
Frequency 4 5 13 20 14 7 4
Height(in 
cm)
135 -
140
140 -
145
145 -
150
150 -
155
155 -
160
160 -
165
165 -
170
170 -
175
Frequency 6 10 18 22 20 15 6 3
Page 5


1. In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
Age 
(in years)
0 15 15 30 30 45 45 60 60 - 75
No. of patients 5 20 40 50 25
Sol:
We prepare the cumulative frequency table, as shown below:
Age (in years) Number of patients (fi) Cumulative Frequency (cf)
0 15 5 5
15 30 20 25
30 45 40 65
45 60 50 115
60 75 25 140
Total N = = 140
Now, N = 140 = 70.
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45 
60.
Thus, the median class is 45 60.
l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Median = l + × h
= 45 + × 15
= 45 + × 15
= 45 + 1.5
= 46.5
 
Hence, the median age is 46.5 years.
2. Compute mean from the following data:
Marks 0 7 7 14 14 21 21 28 28 35 35 42 42 49
Number of 
Students
3 4 7 11 0 16 9
Sol: 
Class Frequency (f) Cumulative Frequency (cf)
0 7 3 3
7 14 4 7
14 21 7 14
21 - 28 11 25
28 35 0 25
35 42 16 41
42 49 9 50
N = = 50
Now, N = 50 = 25.
The cumulative frequency just greater than 25 is 41 and the corresponding class is 35 42.
Thus, the median class is 35 42.
l = 35, h = 7, f = 16, cf = c.f. of preceding class = 25 and = 25.
Now,
Median = l + × h
= 35 + 7 ×
= 35 + 0
= 35
Hence, the median age is 46.5 years.
3. The following table shows the daily wages of workers in a factory:
0 100 100 200 200 300 300 400 400 500
Number of 
workers
40 32 48 22 8
Find the median daily wage income of the workers.
Sol: 
Class Frequency (f) Cumulative Frequency (cf)
0 100 40 40
100 200 32 72
200 300 48 120
300 400 22 142
400 500 8 150
N = = 150
 
Now, N = 150 
= 75.
The cumulative frequency just greater than 75 is 120 and the corresponding class is 200 
300.
Thus, the median class is 200 300.
l = 200, h = 100, f = 48, cf = c.f. of preceding class = 72 and = 75.
Now,
Median, M = l +
= 200 + 
= 200 + 6.25
= 206.25
Hence, the median daily wage income of the workers is Rs 206.25.
4. Calculate the median from the following frequency distribution table:
Class 5 10 10 15 15 20 20 25 25 30 30 35 35 40 40 45
Frequency 5 6 15 10 5 4 2 2
Sol: 
Class Frequency (f) Cumulative Frequency 
(cf)
5 10 5 5
10 15 6 11
15 20 15 26
20 25 10 36
25 30 5 41
30 35 4 45
35 40 2 47
40 45 2 49
N = = 49
Now, N = 49 
= 24.5.
The cumulative frequency just greater than 24.5 is 26 and the corresponding class is 15 -
20.
Thus, the median class is 15 20.
l = 15, h = 5, f = 15, cf = c.f. of preceding class = 11 and = 24.5.
Now,
Median, M = l +
= 15 + 
 
= 15 + 4.5
= 19.5
Hence, the median = 19.5.
5. Given below is the number of units of electricity consumed in a week in a certain locality:
Calculate the median.
Sol: 
Class Frequency (f) Cumulative Frequency (cf)
65- 85 4 4
85 105 5 9
105 125 13 22
125 145 20 42
145 165 14 56
165 185 7 63
185 205 4 67
N = = 67
Now, N = 67
= 33.5.
The cumulative frequency just greater than 33.5 is 42 and the corresponding class is 125 -
145.
Thus, the median class is 125 145.
l = 125, h = 20, f = 20, cf = c.f. of preceding class = 22 and = 33.5.
Now,
Median, M = l +
= 125 + 
= 125 + 11.5
= 136.5
Hence, the median = 136.5.
6. Calculate the median from the following data:
Class 65 85 85 105 105 125 125 145 145 165 165 185 185 200
Frequency 4 5 13 20 14 7 4
Height(in 
cm)
135 -
140
140 -
145
145 -
150
150 -
155
155 -
160
160 -
165
165 -
170
170 -
175
Frequency 6 10 18 22 20 15 6 3
 
Sol: 
Class Frequency (f) Cumulative Frequency (cf)
135 140 6 6
140 145 10 16
145 150 18 34
150 155 22 56
155 160 20 76
160 165 15 91
165 170 6 97
170 175 3 100
N = = 100
Now, N = 100
= 50.
The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 -
155.
Thus, the median class is 150 155.
l = 150, h = 5, f = 22, cf = c.f. of preceding class = 34 and = 50.
Now,
Median, M = l +
= 150 + 
= 150 + 3.64
= 153.64
Hence, the median = 153.64.
7. Calculate the missing frequency from the following distribution, it being given that the 
median of distribution is 24.
Sol: 
Class Frequency (fi) Cumulative Frequency (cf)
0 10 5 5
10 20 25 30
20 30 x x + 30
30 40 18 x + 48
40 50 7 x + 55
Median is 24 which lies in 20 30
Median class = 20 30
Let the unknown frequency be x.
Class 0 10 10 20 20 30 30 40 40 - 50
Frequency 5 25 ? 18 7

	Mathematics (Maths) Class 10 123 videos\|457 docs\|77 tests

Mathematics (Maths) Class 10

123 videos|457 docs|77 tests

Join Course for Free

Top Courses for Class 10

View all

Related Exams

Class 10

About this Document

	4.95/5 Rating
	Nov 23, 2024 Last updated

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

Introduction of RD Sharma Solutions: Statistics (Exercise 14B) in English is available as part of our Mathematics (Maths) Class 10 for Class 10 & RD Sharma Solutions: Statistics (Exercise 14B) 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 14B) RD Sharma Solutions | Mathematics (Maths) Class 10

Description

Full syllabus notes, lecture & questions for Statistics (Exercise 14B) 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 14B)

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

	Mathematics (Maths) Class 10 123 videos\|457 docs\|77 tests

Mathematics (Maths) Class 10

123 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

study material

past year papers

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

Objective type Questions

Semester Notes

Previous Year Questions with Solutions

Extra Questions

Free

MCQs

Exam

practice quizzes

ppt

Summary

Viva Questions

mock tests for examination

pdf

shortcuts and tricks

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

Important questions

Sample Paper

video lectures

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

;

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

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

The RD Sharma Solutions: Statistics (Exercise 14B) 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 14B) now and kickstart your journey towards success in the Class 10 exam.

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

The importance of RD Sharma Solutions: Statistics (Exercise 14B) 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 14B) Notes

RD Sharma Solutions: Statistics (Exercise 14B) 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 14B). 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 14B) Notes on EduRev are your ultimate resource for success.

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

The "RD Sharma Solutions: Statistics (Exercise 14B) 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 14B) on the App

Students of Class 10 can study RD Sharma Solutions: Statistics (Exercise 14B) alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Statistics (Exercise 14B), 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 14B) is prepared as per the latest Class 10 syllabus.

Education Revolution