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प्रश्न
Find the median from the following data:
| Class | 1 – 5 | 6 – 10 | 11 – 15 | 16 – 20 | 21 – 25 | 26 – 30 | 31 – 35 | 35 – 40 | 40 – 45 |
| Frequency | 7 | 10 | 16 | 32 | 24 | 16 | 11 | 5 | 2 |
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उत्तर
Converting into exclusive form, we get:
| Class | Frequency (f) | Cumulative Frequency (cf) |
| 0.5 – 5.5 | 7 | 7 |
| 5.5 – 10.5 | 10 | 17 |
| 10.5 – 15.5 | 16 | 33 |
| 15.5 – 20.5 | 32 | 65 |
| 20.5 – 25.5 | 24 | 89 |
| 25.5 – 30.5 | 16 | 105 |
| 30.5 – 35.5 | 11 | 116 |
| 35.5 – 40.5 | 5 | 121 |
| 40.5 – 45.5 | 2 | 123 |
| N = Σ𝑓 = 123 |
Now, N = 123
⇒` N/2` = 61.5.
The cumulative frequency just greater than 61.5 is 65 and the corresponding class is 15.5 – 20.5.
Thus, the median class is 15.5 – 20.5.
∴ l = 15.5, h = 5, f = 32, cf = c.f. of preceding class = 33 and `N/2` = 61.5.
∴ Median,` M = l + {h×((N/2−f)/f)}`
`= 15.5 + {5 × ((61.5 − 33)/32)}`
= 15.5 + 4.45
= 19.95
Hence, median = 19.95.
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