RD Sharma Solutions For Class 7 Maths Exercise 23.2 Chapter 23 Data Handling - II (Central Values)

RD Sharma Class 7 Solutions Chapter 23 Ex 23.2 PDF Free Download

Students can view and download the PDF of RD Sharma Solutions for Class 7 Maths Exercise 23.2 of Chapter 23 Data Handling – II available here. The BYJU’S experts in Maths formulate the solutions for questions present in this exercise. This exercise explains exterior and interior angles of a triangle. RD Sharma Solutions for Class 7 includes answers to all questions present in this exercise. This exercise includes the arithmetic mean of grouped data.

Download the PDF of RD Sharma Solutions For Class 7 Chapter 23 – Data Handling – II (Central Values) Exercise 23.2

 

rd sharma class 7 maths solution ch 23 ex 2
rd sharma class 7 maths solution ch 23 ex 2
rd sharma class 7 maths solution ch 23 ex 2
rd sharma class 7 maths solution ch 23 ex 2
rd sharma class 7 maths solution ch 23 ex 2
rd sharma class 7 maths solution ch 23 ex 2
rd sharma class 7 maths solution ch 23 ex 2

 

Access answers to Maths RD Sharma Solutions For Class 7 Chapter 23 – Data Handling – II (Central values) Exercise 23.2

1. A die was thrown 20 times and the following scores were recorded:

5, 2, 1, 3, 4, 4, 5, 6, 2, 2, 4, 5, 5, 6, 2, 2, 4, 5, 5, 1

Prepare the frequency table of the scores on the upper face of the die and find the mean score.

Solution:

The frequency table for the given data is as follows:

x:

1

2

3

4

5

6

f:

2

5

1

4

6

2

To compute arithmetic mean we have to prepare the following table:

Scores (xi)

Frequency (fi)

xi fi

1

2

2

2

5

10

3

1

3

4

4

16

5

6

30

6

2

12

Total

Σ fi = 20

Σ fi xi

Mean score = Σ fi xi/ Σ fi

= 73/20

= 3.65

2. The daily wages (in Rs) of 15 workers in a factory are given below:

200, 180, 150, 150, 130, 180, 180, 200, 150, 130, 180, 180, 200, 150, 180

Prepare the frequency table and find the mean wage.

Solution:

Wages (xi)

130

150

180

200

Number of workers (fi)

2

4

6

3

To compute arithmetic mean we have to prepare the following table:

xi

fi

xi fi

130

2

260

150

4

600

180

6

1080

200

3

600

Total

Σ fi = N = 15

Σ fi xi = 2540

Mean score = Σ fi xi/ Σ fi

= 2540/15

= 169.33

3. The following table shows the weights (in kg) of 15 workers in a factory:

Weight (in Kg)

60

63

66

72

75

Number of workers

4

5

3

1

2

Calculate the mean weight.

Solution:

Calculation of mean:

xi

fi

xi fi

60

4

240

63

5

315

66

3

198

72

1

72

75

2

150

Total

Σ fi = N = 15

Σ fi xi = 975

Mean score = Σ fi xi/ Σ fi

= 975/15

= 65 kg

4. The ages (in years) of 50 students of a class in a school are given below:

Age (in years)

14

15

16

17

18

Number of students

15

14

10

8

3

Find the mean age.

Solution:

Calculation of mean:

xi

fi

xi fi

14

15

210

15

14

210

16

10

160

17

8

136

18

3

54

Total

Σ fi = N = 50

Σ fi xi = 770

Mean score = Σ fi xi/ Σ fi

= 770/50

= 15.4 years

5. Calculate the mean for the following distribution:

x:

5

6

7

8

9

f:

4

8

14

11

3

Solution:

xi

fi

xi fi

5

4

20

6

8

48

7

14

98

8

11

88

9

3

27

Total

Σ fi = N = 40

Σ fi xi = 281

Mean score = Σ fi xi/ Σ fi

= 281/40

= 7.025

6. Find the mean of the following data:

x:

19

21

23

25

27

29

31

f:

13

15

16

18

16

15

13

Solution:

xi

fi

xi fi

19

13

247

21

15

315

23

16

368

25

18

450

27

16

432

29

15

435

31

13

403

Total

Σ fi = N = 106

Σ fi xi = 2650

Mean score = Σ fi xi/ Σ fi

= 2650/106

= 25

7. The mean of the following data is 20.6. Find the value of p.

x:

10

15

p

25

31

f:

3

10

25

7

5

Solution:

xi

fi

xi fi

10

3

30

15

10

150

P

25

25p

25

7

175

31

5

175

Total

Σ fi = N = 50

Σ fi xi = 530 + 25p

Mean score = Σ fi xi/ Σ fi

20.6 = 530 + 25p/50

530 + 25 p = 20.6 x 50

25 p = 1030 – 530

p = 500/25

p = 20

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

x:

5

10

15

20

25

f:

6

p

6

10

5

Solution:

xi

fi

xi fi

5

6

30

10

P

10p

15

6

90

20

10

200

25

5

125

Total

Σ fi = 27 + p

Σ fi xi = 445 + 10p

Mean score = Σ fi xi/ Σ fi

15 = 445 + 10p/27 + p

445 + 10 p = 405 + 15p

5 p = 445 – 405

p = 40/5

p = 8

9. 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:

xi

fi

xi fi

8

12

96

12

16

192

15

20

300

P

24

24p

20

16

320

25

8

200

30

4

120

Total

Σ fi = N = 100

Σ fi xi = 1228 + 24p

Mean score = Σ fi xi/ Σ fi

16.6 = 1228 + 24p/100

1228 + 24 p = 16.6 x 100

24 p = 1660 – 1228

p = 432/24

p = 18

10. 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:

xi

fi

xi fi

5

2

10

8

5

40

10

8

80

12

22

264

P

7

7p

20

4

80

25

2

50

Total

Σ fi = N = 50

Σ fi xi = 524 + 7p

Mean score = Σ fi xi/ Σ fi

12.58 = 524 + 7p/50

524 + 7 p = 12.58 x 50

7 p = 629 – 524

p = 105/7

p = 15

11. 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:

xi

fi

xi fi

3

6

18

5

8

40

7

15

105

9

P

9p

11

8

88

13

4

52

Total

Σ fi = N = 41 + p

Σ fi xi = 303 + 9p

Mean score = Σ fi xi/ Σ fi

7.68 = 303 + 9p/41 + p

303 + 9 p = 314.88 + 7.68p

1.32 p = 314.88 – 303

p = 11.88/1.32

p = 9

12. Find the value of p, if the mean of the following distribution is 20

x:

15

17

19

20 + p

23

f:

2

3

4

5p

6

Solution:

xi

fi

xi fi

15

2

30

17

3

51

19

4

76

20 + p

5P

(20 + p) 5p

23

6

138

Total

Σ fi = 15 + 5p

Σ fi xi = 295 + (20 +p) 5p

Mean score = Σ fi xi/ Σ fi

20 = [(295 + (20 + p) 5p)]/ 15 + 5p

295 + 100 p + 5p2 = 300 + 100p

5p2 = 300 – 295

5p2= 5

p2 = 1

p = 1


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