Question:
If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is
(a) 2 m+l
(b) 2 m-l
(c) m-l
(d) m-2l
Solution:
(b) Let x and y be the lower and upper class limit of a continuous frequency distribution.
Now, mid-point of a class = (x + y)/2 = m [given]
=> x + y = 2 m =x + l = 2m
[∴ y = l = upper class limit (given)]
=> x = 2 m-l
Hence, the lower class limit of the class is 2m – l.