Consider the data:
| Class | 65–85 | 85–105 | 105–125 | 125–145 | 145–165 | 165–185 | 185–205 |
|---|---|---|---|---|---|---|---|
| Frequency | 4 | 5 | 13 | 20 | 14 | 7 | 4 |
The difference of the upper limit of the median class and the lower limit of the modal class is
0
19
20
38
Step 1: Find total frequency (N).
Add all frequencies: \(4 + 5 + 13 + 20 + 14 + 7 + 4 = 67\). So, \(N = 67\).
Step 2: Find median class.
Median is at position \(N/2 = 67/2 = 33.5\).
Now write cumulative frequencies: 65–85 → 4, 85–105 → 9, 105–125 → 22, 125–145 → 42, 145–165 → 56, 165–185 → 63, 185–205 → 67.
Since 33.5 lies between 22 and 42, the median class = 125–145.
Upper limit of median class = 145.
Step 3: Find modal class.
The class with the highest frequency is 125–145 (frequency = 20).
So, modal class = 125–145.
Lower limit of modal class = 125.
Step 4: Find the required difference.
Difference = (Upper limit of median class) − (Lower limit of modal class)
= \(145 − 125 = 20\).
Final Answer: Option C (20)