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प्रश्न
Calculate the median from the following frequency distribution table:
| Class | 5 – 10 | 10 – 15 | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 |
| Frequency | 5 | 6 | 15 | 10 | 5 | 4 | 2 | 2 |
उत्तर
| Class | Frequency (f) | Cumulative Frequency (cf) |
| 5 – 10 | 5 | 5 |
| 10 – 15 | 6 | 11 |
| 15 – 20 | 15 | 26 |
| 20 – 25 | 10 | 36 |
| 25 – 30 | 5 | 41 |
| 30 – 35 | 4 | 45 |
| 35 – 40 | 2 | 47 |
| 40 – 45 | 2 | 49 |
| N = Σ𝑓 = 49 |
Now, N = 49
`⇒ N/2` = 24.5.
The cumulative frequency just greater than 24.5 is 26 and the corresponding class is 15 - 20.
Thus, the median class is 15 – 20.
∴ l = 15, h = 5, f = 15, cf = c.f. of preceding class = 11 and `N/2` = 24.5.
Now,
Median, M = `l + {h×((N/2−cf)/f)}`
`= 15 + {5 × ((24.5 − 11)/15)}`
= 15 + 4.5
= 19.5
Hence, the median = 19.5.
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संबंधित प्रश्न
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| Length (in mm) | Number of leaves |
| 118 − 126 | 3 |
| 127 – 135 | 5 |
| 136 − 144 | 9 |
| 145 – 153 | 12 |
| 154 – 162 | 5 |
| 163 – 171 | 4 |
| 172 – 180 | 2 |
Find the median length of the leaves.
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