Telanagana SCERT Solution Class VIII (8) Math Chapter 7 Frequency Distribution Tables and Graphs Exercise 7.2
(1) Max. value = 63
Minimum value = 5
Range 63 – 5 = 58
Therefore, length of interval = 58/6 = 9.6 = 10
Then
Class | Interval frequency |
0 – 9 | 5 |
10 – 19 | 8 |
20 – 29 | 10 |
30 – 39 | 7 |
40 – 49 | 7 |
50 – 59 | 5 |
60 – 69 | 3 |
(2) Max number = 40
Min number = 15
Range = (40 – 15) + 1
Class number = 26/4 = 6.5 = 7
Class | Frequency |
15 – 22 | 12 |
23 – 30 | 11 |
31 – 38 | 6 |
39 – 46 | 1 |
(3) Given distribution are
4 – 11, 12 – 19, 20 – 27, 28 – 35, 36 – 43
Next two distribution are 44 – 51, 52 – 59.
(i) Length of each class interval 11 – 4 = 7
(ii) Class boundaries 11, 19, 27, 35, 43
(iii)
Class Interval | Class mark |
4 – 11 | 7.5 |
12 – 19 | 15.5 |
20 – 27 | 23.5 |
28 – 35 | 31.5 |
36 – 43 | 39.5 |
(4)
(i) Class intervals of the data 22 – 10 = 12.
(ii) Class intervals are 6- 16, 16 – 28, 28 – 40, 40 – 51, 52 – 64, 64 – 76
Class | Intervals |
<16 | 6 |
<28 | 6 + 14 = 20 |
<40 | 20 + 20 = 40 |
<52 | 40 + 21 = 61 |
<64 | 61 + 9 = 70 |
<76 | 70 + 5 = 75 |
(iii)
Class | Intervals |
> 6 | 6 + 69 = 75 |
> 16 | 14 + 55 = 69 |
> 28 | 20 + 35 = 55 |
> 40 | 21 + 14 = 35 |
> 52 | 9 + 5 = 14 |
> 64 | 5 |
(5)
Class | Distribution frequency |
0 – 10 | 2 |
10 – 20 | 10 |
20 – 30 | 4 |
30 – 40 | 7 |
40 – 50 | 12 |
(6)
Class | Frequency | Cumulative class than | Cumulative greater than |
1 – 3 | 10 | 10 | 59 |
4 – 6 | 12 | 10 + 12 = 22 | 49 |
7 – 9 | 15 | 15 + 22 = 27 | 37 |
10 – 12 | 13 | 37 + 13 = 50 | 22 |
13 – 15 | 9 | 50 + 9 = 59 | 9 |
(7)
Runs | No. of Students | Frequency |
0 – 10 | 3 | 3 |
10 – 20 | 8 | 8 – 3 = 5 |
20 – 30 | 19 | 19 – 8 = 11 |
30 – 40 | 25 | 25 – 19 = 6 |
40 – 50 | 30 | 30 – 25 = 5 |
(8)
No of Books | Greater than cumulative frequency | Frequency | Less than cumulative frequency |
1 – 10 | 42 | 42 – 36 = 6 | 6 |
11 – 20 | 36 | 36 – 23 = 13 | 6 + 13 = 19 |
21 – 30 | 23 | 23 – 14 = 9 | 19 + 9 = 28 |
31 – 40 | 14 | 14 – 6 = 8 | 28 + 8 = 36 |
41 – 50 | 6 | 6 | 36 + 6 = 42 |