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

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 CoolGyan’S experts in Maths formulate the solutions for questions present in the textbook based on the understanding abilities of students. Exercise 23.2 mainly explains about the exterior and interior angles of a triangle. RD Sharma Solutions for Class 7 PDF can be downloaded by the students and referred to while solving the textbook problems. This exercise also includes the arithmetic mean of grouped data.

Download the PDF of RD Sharma Solutions For Class 7 Maths 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 35
f: 3 10 25 7 5

Solution:

xi fi xi fi
10 3 30
15 10 150
P 25 25p
25 7 175
35 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

Me
an 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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