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Question
Calculate the median for the following data:
| Class | 19 – 25 | 26 – 32 | 33 – 39 | 40 – 46 | 47 – 53 | 54 - 60 |
| Frequency | 35 | 96 | 68 | 102 | 35 | 4 |
Solution
First, we will convert the data into exclusive form.
| Class | Frequency (f) | Cumulative Frequency (cf) |
| 18.5 – 25.5 | 35 | 35 |
| 25.5 – 32.5 | 96 | 131 |
| 32.5 – 39.5 | 68 | 199 |
| 39.5 – 46.5 | 102 | 301 |
| 46.5 – 53.5 | 35 | 336 |
| 53.5 – 60.5 | 4 | 340 |
| N = Σ𝑓 = 340 |
Now, N = 340
`⇒ N/2 = 170`.
The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5 – 39.5.
Thus, the median class is 32.5 – 39.5.
∴ l = 32.5, h = 7, f = 68, cf = c.f. of preceding class = 131 and `N/2` = 170.
∴ Median, `M = l + {h×((N/2−cf)/f)}`
`= 32.5 + {7 × ((170 − 131)/68)}`
= 32.5 + 4.01
= 36.51
Hence, the median = 36.51.
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