Question

Class intervals in a frequency distribution are $(3-6),(7-10),(11-14),(15-18)$ etc. The actual upper limit of the class interval $(7-10)$ is

• A. $10+\frac{1}{2}(11-10)$
• B. $10-\frac{1}{2}(11-10)$
• C. $10+\frac{1}{2}(11+10)$
• D. $10-\frac{1}{2}(11+10)$

Answer 1

The correct option is A $10+\frac{1}{2}(11-10)$

We change the lower and upper limits of each class intervals in such a way so as to make those limits continuous.
Thus, the new limits for the class $7-10$ look like $7-\frac{1}{2}\times (7-6)=6.5$ and $10+\frac{1}{2}\times (11-10)=10.5$
Also, the lower limit of class $11-14$ becomes $11-\frac{1}{2}\times (15-14)=10.5$ and thus the limits are continuous.