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Question
Find the median from the following data:
| Marks | No of students |
| Below 10 | 12 |
| Below 20 | 32 |
| Below 30 | 57 |
| Below 40 | 80 |
| Below 50 | 92 |
| Below 60 | 116 |
| Below 70 | 164 |
| Below 80 | 200 |
Solution
| Class | frequency (f) | cumulative frequency (f) |
| 0 – 10 | 12 | 12 |
| 10 – 20 | 32 | (32 - 12) = 20 |
| 20 – 30 | 57 | (57 - 32) = 25 |
| 30 – 40 | 80 | (80 - 57) = 23 |
| 40 – 50 | 92 | (92 - 80) =12 |
| 50 – 60 | 116 | (116 - 92) = 24 |
| 60 – 70 | 164 | (164 - 116) = 48 |
| 70 – 80 | 200 | (200 - 164) = 36 |
| N = Σ𝑓 = 200 |
Now, N = 200
`⇒ N/2 = 100`.
The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 – 60.
Thus, the median class is 50 – 60.
∴ l = 50, h = 10, f = 24, cf = c.f. of preceding class = 92 and `N/2` = 100.
∴ Median, `M = l + {h×((N/2−cf)/f)}`
`= 50 + {10× ((100 − 92)/24)}`
= 50 + 3.33
= 53.33
Hence, median = 53.33.
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