RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.1

RD Sharma Class 10 Solutions Chapter 7 Ex 7.1 PDF Free Download

Statistics is purely about data analysis. So, with the given data students will have to find the mean by using different methods. In this exercise, students will have to find the mean of grouped data by using the direct method. The RD Sharma Solutions Class 10 prepared by subject experts at BYJU’S is a very useful resource for students to refer and prepare well for their exams. Download RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics Exercise 7.1, PDF provided below for detailed solutions to problems.

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.1 Download PDF

RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.1 01
RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.1 02
RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.1 03
RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.1 04

Access RD Sharma Solutions for Class 10 Chapter 7 Statistics Exercise 7.1

1. Calculate the mean for the following distribution:

x:

5

6

7

8

9

f:

4

8

14

11

3

Solution:

x

f

fx

5

4

20

6

8

48

7

14

98

8

11

88

9

3

27

N = 40

Σ fx = 281

Mean = Σ fx/ N = 281/40

∴ Mean = 7.025

2. Find the mean of the following data:

x:

19

21

23

25

27

29

31

f:

13

15

16

18

16

15

13

Solution:

x

f

fx

19

13

247

21

15

315

23

16

368

25

18

450

27

16

432

29

15

435

31

13

403

N = 106

Σ fx = 2620

Mean = Σ fx/ N = 2620/106

∴ Mean = 25

3. If the mean of the following data is 20.6. Find the value of p.

x:

10

15

p

25

35

f:

3

10

25

7

5

Solution:

x

f

fx

10

3

30

15

10

150

p

25

25p

25

7

175

35

5

175

N = 50

Σ fx = 530 + 25p

We know that,

Mean = Σ fx/ N = (2620 + 25p)/ 50

Given,

Mean = 20.6

⇒ 20.6 = (530 + 25p)/ 50

(20.6 x 50) – 530 = 25p

p = 500/ 25

∴ p = 50

4. If the mean of the following data is 15, find p.

x:

5

10

15

20

25

f:

6

p

6

10

5

Solution:

x

f

fx

5

6

30

10

p

10p

15

6

90

20

10

200

25

5

125

N = p + 27

Σ fx = 445 + 10p

We know that,

Mean = Σ fx/ N = (445 + 10p)/ (p + 27)

Given,

Mean = 15

⇒ 15 = (445 + 10p)/ (p + 27)

15p + 405 = 445 + 10p

5p = 40

∴ p = 8

5. Find the value of p for the following distribution whose mean is 16.6

x:

8

12

15

p

20

25

30

f:

12

16

20

24

16

8

4

Solution:

x

f

fx

8

12

96

12

16

192

15

20

300

P

24

24p

20

16

320

25

8

200

30

4

120

N = 100

Σ fx = 1228 + 24p

We know that,

Mean = Σ fx/ N = (1228 + 24p)/ 100

Given,

Mean = 16.6

⇒ 16.6 = (1228 + 24p)/ 100

1660 = 1228 + 24p

24p = 432

∴ p = 18

6. Find the missing value of p for the following distribution whose mean is 12.58

x:

5

8

10

12

p

20

25

f:

2

5

8

22

7

4

2

Solution:

x

f

fx

5

2

10

8

5

40

10

8

80

12

22

264

P

7

7p

20

4

80

25

2

50

N = 50

Σ fx = 524 + 7p

We know that,

Mean = Σ fx/ N = (524 + 7p)/ 50

Given,

Mean = 12.58

⇒ 12.58 = (524 + 7p)/ 50

629 = 524 + 7p

7p = 629 – 524 = 105

∴ p = 15

7. Find the missing frequency (p) for the following distribution whose mean is 7.68

x:

3

5

7

9

11

13

f:

6

8

15

p

8

4

Solution:

x

f

fx

3

6

18

5

8

40

7

15

105

9

p

9p

11

8

88

13

4

52

N = 41 + p

Σ fx = 303 + 9p

We know that,

Mean = Σ fx/ N = (303 + 9p)/ (41 + p)

Given,

Mean = 7.68

⇒ 7.68 = (303 + 9p)/ (41 + p)

7.68(41 + p) = 303 + 9p

7.68p + 314.88 = 303 + 9p

1.32p = 11.88

∴ p = 11.88/1.32 = 9

Also, access other exercise solutions of RD Sharma Class 10 Maths Chapter 7 Statistics

Exercise 7.2 Solutions

Exercise 7.3 Solutions

Exercise 7.4 Solutions

Exercise 7.5 Solutions

Exercise 7.6 Solutions

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