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Question
Assertion (A): The difference in class marks of the modal class and the median class of the following frequency distribution table is 0.
| Class interval |
20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
| Frequency | 1 | 3 | 2 | 6 | 4 |
Reason (R): Modal class and median class are always the same for a given frequency distribution.
Options
Both A and R are correct and R is the correct explanation for A.
Both A and R are correct and R is not the correct explanation for A.
A is true, but R is false.
Both A and R are true.
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Solution
A is true, but R is false.
Explanation:
Total frequency n = 1 + 3 + 2 + 6 + 4 = 16.
The median position is `n/2` = 8th observation.
Cumulative frequencies are 1, 4, 6, 12, 16.
So, the 8th observation lies in class 50 – 60.
Thus, median class = 50 – 60.
Modal class = Class with highest frequency = 50 – 60 (frequency 6).
Both modal and median classes are 50 – 60, their class-marks (midpoints) are 55, so the difference = 0.
Hence, Assertion A is true.
Reason R is false because modal and median classes need not always coincide for every frequency distribution they coincide only when the modal class happens to contain the median position.
For example, classes 10 – 20, 20 – 30, 30 – 40 with frequencies 4, 1, 4 have modal classes 10 – 20 and 30 – 40 (tie) while the median class is 20 – 30 so they differ.
