By Team Aakash Byju's | 28th October 2022
A.
B.
C.
D.
The correct answer is option B. 2(Mid-value of the class) – (Upper-class limit of the class).
Brief Explanation
For a continuous frequency distribution of a class, let a be the lower-class limit, b be the upper-class limit and m be the mid-value of the class.
We know that the mid-value of the class is defined as
Since we need to find the lower-class limit of the class, we shall isolate the term b in the previous equation.
Therefore, 2m = a + b or b = 2m– a i.e.,, the lower-class limit = 2(mid-value of the class) – (upper-class limit of the class).