Answer :
Let's go through the problem step-by-step:
(a) Number of classes:
The number of classes can be determined by counting the distinct groups or intervals given in the frequency distribution table.
From the table:
1. 10 - 13.9
2. 14 - 17.9
3. 18 - 21.9
4. 22 - 25.9
5. 26 - 29.9
6. 30 - 33.9
There are 6 classes.
Answer: 6 classes
(b) Lower class limit for the first class:
The lower class limit is the smallest value in the first class interval.
For the first class interval (10 - 13.9):
The lower class limit is 10.0
Answer: 10.0
(c) Class width:
The class width can be calculated by taking the difference between the lower class limits of any two consecutive classes.
For the first and second class intervals:
- The lower limit of the first class is 10.0
- The lower limit of the second class is 14.0
Class width = Lower limit of second class - Lower limit of first class
Class width = 14.0 - 10.0
Class width = 4.0
Answer: 4.0
(a) Number of classes:
The number of classes can be determined by counting the distinct groups or intervals given in the frequency distribution table.
From the table:
1. 10 - 13.9
2. 14 - 17.9
3. 18 - 21.9
4. 22 - 25.9
5. 26 - 29.9
6. 30 - 33.9
There are 6 classes.
Answer: 6 classes
(b) Lower class limit for the first class:
The lower class limit is the smallest value in the first class interval.
For the first class interval (10 - 13.9):
The lower class limit is 10.0
Answer: 10.0
(c) Class width:
The class width can be calculated by taking the difference between the lower class limits of any two consecutive classes.
For the first and second class intervals:
- The lower limit of the first class is 10.0
- The lower limit of the second class is 14.0
Class width = Lower limit of second class - Lower limit of first class
Class width = 14.0 - 10.0
Class width = 4.0
Answer: 4.0
