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

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       ### 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