Class 10 Exam > Class 10 Notes > Mathematics (Maths) Class 10 > RD Sharma Solutions: Statistics (Exercise 14A)
Statistics (Exercise 14A) 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. If the mean of 5 observation , 2, 4, 6 x x x x and 8 x , find the value of x.
Sol:
Mean of given observations =
11 =
55 = 5x + 20
5x = 55 20
5x = 35
x =
x = 7
Hence, the value of x is 7.
2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be
new mean?
Sol:
Mean of given observations =
Mean of 25 observations = 27
Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 (25 × 7)
= 675 175
= 500
Then, new mean = = 20
Thus, the new mean will be 20.
3. Compute the mean for following data:
Class 1 3 3 5 5 - 7 7 9
Frequency 12 22 27 19
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
1 3 12 2 24
3 5 22 4 88
5 7 27 6 162
7 9 19 8 152
Total
fi = 80 fi xi = 426
Page 2
1. If the mean of 5 observation , 2, 4, 6 x x x x and 8 x , find the value of x.
Sol:
Mean of given observations =
11 =
55 = 5x + 20
5x = 55 20
5x = 35
x =
x = 7
Hence, the value of x is 7.
2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be
new mean?
Sol:
Mean of given observations =
Mean of 25 observations = 27
Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 (25 × 7)
= 675 175
= 500
Then, new mean = = 20
Thus, the new mean will be 20.
3. Compute the mean for following data:
Class 1 3 3 5 5 - 7 7 9
Frequency 12 22 27 19
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
1 3 12 2 24
3 5 22 4 88
5 7 27 6 162
7 9 19 8 152
Total
fi = 80 fi xi = 426
The mean of given data is given by
=
i i
i
i
i
f x
f
=
= 5.325
Thus, the mean of the following data is 5.325.
4. Find the mean using direct method:
Class 0 10 10 20 20 30 30 40 40 50 50 60
Frequency 7 5 6 12 8 2
Sol:
Class Frequency (fi) Mid values (x i)
fi × xi
0 10 7 5 35
10 20 5 15 75
20 30 6 25 150
30 40 12 35 420
40 50 8 45 360
50 60 2 55 110
fi = 40 i i
f x = 1150
Mean, =
i i
i
f x
f
1150
40
= 28.75
= 28.75
5. Find the mean of the following data, using direct method:
Class 25 35 35 45 45 55 55 65 65 75
Frequency 6 10 8 12 4
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
25 35 6 30 180
35 45 10 40 400
45 55 8 50 400
55 65 12 60 720
65 75 4 70 280
fi = 40 i i
f x = 1980
Page 3
1. If the mean of 5 observation , 2, 4, 6 x x x x and 8 x , find the value of x.
Sol:
Mean of given observations =
11 =
55 = 5x + 20
5x = 55 20
5x = 35
x =
x = 7
Hence, the value of x is 7.
2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be
new mean?
Sol:
Mean of given observations =
Mean of 25 observations = 27
Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 (25 × 7)
= 675 175
= 500
Then, new mean = = 20
Thus, the new mean will be 20.
3. Compute the mean for following data:
Class 1 3 3 5 5 - 7 7 9
Frequency 12 22 27 19
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
1 3 12 2 24
3 5 22 4 88
5 7 27 6 162
7 9 19 8 152
Total
fi = 80 fi xi = 426
The mean of given data is given by
=
i i
i
i
i
f x
f
=
= 5.325
Thus, the mean of the following data is 5.325.
4. Find the mean using direct method:
Class 0 10 10 20 20 30 30 40 40 50 50 60
Frequency 7 5 6 12 8 2
Sol:
Class Frequency (fi) Mid values (x i)
fi × xi
0 10 7 5 35
10 20 5 15 75
20 30 6 25 150
30 40 12 35 420
40 50 8 45 360
50 60 2 55 110
fi = 40 i i
f x = 1150
Mean, =
i i
i
f x
f
1150
40
= 28.75
= 28.75
5. Find the mean of the following data, using direct method:
Class 25 35 35 45 45 55 55 65 65 75
Frequency 6 10 8 12 4
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
25 35 6 30 180
35 45 10 40 400
45 55 8 50 400
55 65 12 60 720
65 75 4 70 280
fi = 40 i i
f x = 1980
Mean, =
i i
i
f x
f
=
= 49.5
= 49.5
6. Find the mean of the following data, using direct method:
Class 0 100 100 200 200 300 300 400 400 500
Frequency 6 9 15 12 8
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
0 - 100 6 50 300
100 200 9 150 1350
200 300 15 250 3750
300 400 12 350 4200
400 500 8 450 3600
fi = 50 i i
f x = 13200
Mean, =
i i
i
f x
f
=
= 264
= 264
7. Using an appropriate method, find the mean of the following frequency distribution:
Class 84 90 90 96 96 102 102 108 108 114 114 120
Frequency 8 10 16 23 12 11
Which method did you use, and why?
Sol:
Class Frequency (fi) Mid values (x i) (fi xi)
84 90 8 87 696
90 96 10 93 930
96 102 16 99 1584
102 108 23 105 2415
108 114 12 111 1332
114 120 11 117 1287
Total
fi = 80 fi xi = 8244
The mean of the data is given by,
Page 4
1. If the mean of 5 observation , 2, 4, 6 x x x x and 8 x , find the value of x.
Sol:
Mean of given observations =
11 =
55 = 5x + 20
5x = 55 20
5x = 35
x =
x = 7
Hence, the value of x is 7.
2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be
new mean?
Sol:
Mean of given observations =
Mean of 25 observations = 27
Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 (25 × 7)
= 675 175
= 500
Then, new mean = = 20
Thus, the new mean will be 20.
3. Compute the mean for following data:
Class 1 3 3 5 5 - 7 7 9
Frequency 12 22 27 19
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
1 3 12 2 24
3 5 22 4 88
5 7 27 6 162
7 9 19 8 152
Total
fi = 80 fi xi = 426
The mean of given data is given by
=
i i
i
i
i
f x
f
=
= 5.325
Thus, the mean of the following data is 5.325.
4. Find the mean using direct method:
Class 0 10 10 20 20 30 30 40 40 50 50 60
Frequency 7 5 6 12 8 2
Sol:
Class Frequency (fi) Mid values (x i)
fi × xi
0 10 7 5 35
10 20 5 15 75
20 30 6 25 150
30 40 12 35 420
40 50 8 45 360
50 60 2 55 110
fi = 40 i i
f x = 1150
Mean, =
i i
i
f x
f
1150
40
= 28.75
= 28.75
5. Find the mean of the following data, using direct method:
Class 25 35 35 45 45 55 55 65 65 75
Frequency 6 10 8 12 4
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
25 35 6 30 180
35 45 10 40 400
45 55 8 50 400
55 65 12 60 720
65 75 4 70 280
fi = 40 i i
f x = 1980
Mean, =
i i
i
f x
f
=
= 49.5
= 49.5
6. Find the mean of the following data, using direct method:
Class 0 100 100 200 200 300 300 400 400 500
Frequency 6 9 15 12 8
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
0 - 100 6 50 300
100 200 9 150 1350
200 300 15 250 3750
300 400 12 350 4200
400 500 8 450 3600
fi = 50 i i
f x = 13200
Mean, =
i i
i
f x
f
=
= 264
= 264
7. Using an appropriate method, find the mean of the following frequency distribution:
Class 84 90 90 96 96 102 102 108 108 114 114 120
Frequency 8 10 16 23 12 11
Which method did you use, and why?
Sol:
Class Frequency (fi) Mid values (x i) (fi xi)
84 90 8 87 696
90 96 10 93 930
96 102 16 99 1584
102 108 23 105 2415
108 114 12 111 1332
114 120 11 117 1287
Total
fi = 80 fi xi = 8244
The mean of the data is given by,
=
i i
i
i
i
f x
f
=
= 103.05
Thus, the mean of the following data is 103.05.
8. If the mean of the following frequency distribution is 24, find the value of p.
Class 0 10 10 20 20 30 30 40 40 - 50
Frequency 3 4 P 3 2
Sol:
The given data is shown as follows:
Class Frequency (fi) Mid values (x i) (fi xi)
0 10 3 5 15
10 20 4 15 60
20 30 p 25 25p
30 40 3 35 105
40 50 2 45 90
Total
fi = 12 + p fi xi = 270 + 25p
The mean of the given data is given by,
=
i i
i
i
i
f x
f
24 =
24 (12 + p) = 270 + 25p
288 + 24p = 270 + 25p
25p 24p = 288 270
p = 18
Hence, the value of p is 18.
9. The following distribution shows the daily pocket allowance of children of a locality. If the
mean pocket allowance is 18 , find the missing frequency f.
Daily
pocket
allowance
11 13 13 15 15 17 17 19 19 - 21 21 23 23 25
Page 5
1. If the mean of 5 observation , 2, 4, 6 x x x x and 8 x , find the value of x.
Sol:
Mean of given observations =
11 =
55 = 5x + 20
5x = 55 20
5x = 35
x =
x = 7
Hence, the value of x is 7.
2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be
new mean?
Sol:
Mean of given observations =
Mean of 25 observations = 27
Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 (25 × 7)
= 675 175
= 500
Then, new mean = = 20
Thus, the new mean will be 20.
3. Compute the mean for following data:
Class 1 3 3 5 5 - 7 7 9
Frequency 12 22 27 19
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
1 3 12 2 24
3 5 22 4 88
5 7 27 6 162
7 9 19 8 152
Total
fi = 80 fi xi = 426
The mean of given data is given by
=
i i
i
i
i
f x
f
=
= 5.325
Thus, the mean of the following data is 5.325.
4. Find the mean using direct method:
Class 0 10 10 20 20 30 30 40 40 50 50 60
Frequency 7 5 6 12 8 2
Sol:
Class Frequency (fi) Mid values (x i)
fi × xi
0 10 7 5 35
10 20 5 15 75
20 30 6 25 150
30 40 12 35 420
40 50 8 45 360
50 60 2 55 110
fi = 40 i i
f x = 1150
Mean, =
i i
i
f x
f
1150
40
= 28.75
= 28.75
5. Find the mean of the following data, using direct method:
Class 25 35 35 45 45 55 55 65 65 75
Frequency 6 10 8 12 4
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
25 35 6 30 180
35 45 10 40 400
45 55 8 50 400
55 65 12 60 720
65 75 4 70 280
fi = 40 i i
f x = 1980
Mean, =
i i
i
f x
f
=
= 49.5
= 49.5
6. Find the mean of the following data, using direct method:
Class 0 100 100 200 200 300 300 400 400 500
Frequency 6 9 15 12 8
Sol:
Class Frequency (fi) Mid values (x i)
(fi × xi)
0 - 100 6 50 300
100 200 9 150 1350
200 300 15 250 3750
300 400 12 350 4200
400 500 8 450 3600
fi = 50 i i
f x = 13200
Mean, =
i i
i
f x
f
=
= 264
= 264
7. Using an appropriate method, find the mean of the following frequency distribution:
Class 84 90 90 96 96 102 102 108 108 114 114 120
Frequency 8 10 16 23 12 11
Which method did you use, and why?
Sol:
Class Frequency (fi) Mid values (x i) (fi xi)
84 90 8 87 696
90 96 10 93 930
96 102 16 99 1584
102 108 23 105 2415
108 114 12 111 1332
114 120 11 117 1287
Total
fi = 80 fi xi = 8244
The mean of the data is given by,
=
i i
i
i
i
f x
f
=
= 103.05
Thus, the mean of the following data is 103.05.
8. If the mean of the following frequency distribution is 24, find the value of p.
Class 0 10 10 20 20 30 30 40 40 - 50
Frequency 3 4 P 3 2
Sol:
The given data is shown as follows:
Class Frequency (fi) Mid values (x i) (fi xi)
0 10 3 5 15
10 20 4 15 60
20 30 p 25 25p
30 40 3 35 105
40 50 2 45 90
Total
fi = 12 + p fi xi = 270 + 25p
The mean of the given data is given by,
=
i i
i
i
i
f x
f
24 =
24 (12 + p) = 270 + 25p
288 + 24p = 270 + 25p
25p 24p = 288 270
p = 18
Hence, the value of p is 18.
9. The following distribution shows the daily pocket allowance of children of a locality. If the
mean pocket allowance is 18 , find the missing frequency f.
Daily
pocket
allowance
11 13 13 15 15 17 17 19 19 - 21 21 23 23 25
Number of
children
7 6 9 13 f 5 4
Sol:
The given data is shown as follows:
Daily pocket Number of
children (fi)
Class mark (xi) fi xi
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
Total
fi = 44 + f fi xi = 752 + 20f
The mean of the given data is given by,
=
i i
i
i
i
f x
f
18 =
18 (44 + f) = 752 + 20f
792 + 18f = 752 + 20f
20f 18f = 792 752
2f = 40
f = 20
Hence, the value of f is 20.
10. The mean of following frequency distribution is 54. Find the value of p.
Class 0 20 20 40 40 60 60 80 80 100
Frequency 7 p 10 9 13
Sol:
The given data is shown as follows:
Class Frequency (fi) Class mark (x i) fi xi
0 20 7 10 70
20 40 p 30 30p
40 60 10 50 500
60 80 9 70 630
80 100 13 90 1170
Total
fi = 39 + p fi xi = 2370 + 30p
Read More
|
124 videos|457 docs|77 tests
|
Related Exams
About this Document
|
4.66/5 Rating |
|
Nov 08, 2024 Last updated |
Document Description: RD Sharma Solutions: Statistics (Exercise 14A) for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation.
The notes and questions for RD Sharma Solutions: Statistics (Exercise 14A) have been prepared according to the Class 10 exam syllabus. Information about RD Sharma Solutions: Statistics (Exercise 14A) covers topics
like and RD Sharma Solutions: Statistics (Exercise 14A) Example, for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Statistics (Exercise 14A).
Introduction of RD Sharma Solutions: Statistics (Exercise 14A) in English is available as part of our Mathematics (Maths) Class 10
for Class 10 & RD Sharma Solutions: Statistics (Exercise 14A) 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 14A) RD Sharma Solutions | Mathematics (Maths) Class 10
Description
Full syllabus notes, lecture & questions for Statistics (Exercise 14A) 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 14A)
In this doc you can find the meaning of RD Sharma Solutions: Statistics (Exercise 14A) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Statistics (Exercise 14A) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Statistics (Exercise 14A) tests, examples and also practice Class 10
tests
|
Explore Courses for Class 10 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Summary
,Semester Notes
,Important questions
,practice quizzes
,Sample Paper
,video lectures
,Viva Questions
,ppt
,mock tests for examination
,MCQs
,Exam
,Free
,Statistics (Exercise 14A) RD Sharma Solutions | Mathematics (Maths) Class 10
,Statistics (Exercise 14A) RD Sharma Solutions | Mathematics (Maths) Class 10
,Extra Questions
,past year papers
,Objective type Questions
,shortcuts and tricks
,Statistics (Exercise 14A) RD Sharma Solutions | Mathematics (Maths) Class 10
,study material
,Previous Year Questions with Solutions
,
Additional Information about RD Sharma Solutions: Statistics (Exercise 14A) for Class 10 Preparation
RD Sharma Solutions: Statistics (Exercise 14A) Free PDF Download
The RD Sharma Solutions: Statistics (Exercise 14A) 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 14A) now and kickstart your journey towards success in the Class 10 exam.
Importance of RD Sharma Solutions: Statistics (Exercise 14A)
The importance of RD Sharma Solutions: Statistics (Exercise 14A) 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 14A) Notes
RD Sharma Solutions: Statistics (Exercise 14A) 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 14A).
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 14A) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Statistics (Exercise 14A) Class 10 Questions
The "RD Sharma Solutions: Statistics (Exercise 14A) 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 14A) on the App
Students of Class 10 can study RD Sharma Solutions: Statistics (Exercise 14A) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Statistics (Exercise 14A),
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 14A) is prepared as per the latest Class 10 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup