Here we will learn class boundaries or actual class limits.
For overlapping class intervals, the class limits are also called class boundaries or actual class limits.
In the case of nonoverlapping class intervals, the class limits are different from class boundaries.
Let the nonoverlapping class intervals for a grouped data be 1 – 10, 11 – 20, 21 – 30, etc. The gap between any two consecutive intervals is 1 (11 – 10 = 1, 21 – 20 = 1, etc.). Change the class intervals into class intervals (1  \(\frac{1}{2}\))  (10 + \(\frac{1}{2}\)), (11  \(\frac{1}{2}\))  (20 + \(\frac{1}{2}\)), (21  \(\frac{1}{2}\))  (30 + \(\frac{1}{2}\)), etc., i.e., 0.5 – 10.5, 10.5 – 20.5, 20.5 – 30.5, etc.
These class intervals are now overlapping. As there is no value of the variable between 10 and 11 or 20 and 21, etc., the frequencies of the intervals do not change. The class limits of 0.5 – 10.5 are 0.5 (lower limit) and 10.5 (upper limit). 0.5 and 10.5 are the class boundaries (actual class limits) of the class interval 1 – 10 in the nonoverlapping case.
Thus in the case of nonoevrlapping class intervals,
The actual lower limit = lower limit  \(\frac{1}{2}\) × (gap)
The actual upper limit = upper limit + \(\frac{1}{2}\) × (gap)
Solved Example on Class Boundaries or Actual Class Limits:
If the class marks of two consecutive overlapping intervals of equal size in a distribution are 94 and 104 then find the corresponding intervals.
Solution:
The difference between 104 and 94 = 104  94 = 10.
Therefore, the class intervals are (94  \(\frac{10}{2}\))  (94 + \(\frac{10}{2}\)) and (104  \(\frac{10}{2}\))  (104 + \(\frac{10}{2}\)),
i.e., 89  99, 99  109.
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