Using the information given, it is found that the class width for this frequency distribution table is of 1.
In this problem, these following classes are given:
0 – 1 14
2 – 3 1
4 – 5 8
6 – 7 12
8 – 9 12
The classes not given, which are 1 - 2, 3 - 4 and 5 - 6, have values of 0.
The <u>difference between the bounds of the classes is of 1</u>, thus, the class width is of 1.
A similar problem is given at brainly.com/question/24701109
I think it would have to be 1.46.
Explanation: if you subtract 3.56 from 2.1 you’ll get 1.46