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Question
If the median of the following frequency distribution is 32.5, find the values of `f_1 and f_2`.
| Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 -40 | 40 – 50 | 50 – 60 | 60 – 70 | Total |
| Frequency | `f_1` |
5 |
9 | 12 | `f_2` | 3 | 2 | 40 |
Solution
| Class | Frequency (f) | Cumulative Frequency (cf) |
| 0 – 10 | `f_1` | `f_1` |
| 10 – 20 | 5 | `f_1`+5 |
| 20 – 30 | 9 | `f_1`+14 |
| 30 – 40 | 12 | `f_1`+26 |
| 40 – 50 | `f_2` | `f_1`+`f_2`+26 |
| 50 – 60 | 3 | `f_1`+`f_2`+29 |
| 60 – 70 | 2 | `f_1`+`f_2`+31 |
| N = Σ𝑓 = 40 |
Now,` f_1 + f_2 + 31 = 40`
⇒` f_1 + f_2 = 9`
⇒ `f_2 = 9 - f_1`
The median is 32.5 which lies in 30 – 40.
Hence, median class = 30 – 40
Here, `l = 30, N/2 = 40/2 = 20, f = 12 and cf = 14 + f_1`
Now, median = 32.5
`⇒ l + ((N/2−Cf)/f) × h = 32.5`
`⇒ 30 + ((20 −(14 + f1))/ 12)× 10 = 32.5`
`⇒ (6 − f_1)/ 12` × 10 = 2.5
`⇒ (60 − 10f_1)/12 = 2.5`
`⇒ 60 – 10f_1 = 30`
`⇒ 10f_1 = 30`
`⇒ f_1 = 3`
From equation (i), we have:
`f_2 = 9 – 3`
`⇒ f_2 = 6`
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