Question

The width of each of nine classes in a frequency distribution is 2.5 and the lower class boundary of the lowest class 10.6. Then the upper class boundary of the highest class is
A. 35.6
B. 33.1
C. 30.6
D. 28.1

Answer 1

The required upper class boundary  is 33.1.

Width of each of nine classes = 2.5 (Given)

Lower class boundary of the lowest class = 10.6 (Given)

Thus,

Total width of 9 classes = 9 × 2.5 = 22.5

Now,

Upper class boundary of highest the class = Lower class boundary of the lowest class + Total width

= 10.6 + 22.5

= 33.1

Therefore, the required upper class boundary  is 33.1.

Answer 2

Answer:

33.1 is you answer hope it helps you