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प्रश्न
If the median of the following frequency distribution is 32.5. Find the values of f1 and f2.
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | Total |
| Frequency | f1 | 5 | 9 | 12 | f2 | 3 | 2 | 40 |
उत्तर
Given: Median = 32.5
We prepare the cumulative frequency table, as given below.
| Class interval: | Frequency: (fi) | Cumulative frequency (c.f.) |
| 0-10 | f1 | f1 |
| 10-20 | 5 | 5 + f1 |
| 20-30 | 9 | 14 + f1 |
| 30-40 | 12 | 26 +f1 |
| 40-50 | f2 | 26 +f1 + f2 |
| 50-60 | 3 | 29 + f1 + f2 |
| 60-70 | 2 | 31 + f1 + f2 |
| N = 40 = 31 + f1 +f2 |
Now, we have
N = 40
31 + f1 + f2 = 40
f2 = 9 - f1 ........(1)
Also, `("N")/(2) = 20`
Since median = 32.5 so the median class is 30 - 40.
Here, l = 30, f = 12 , F = 14 + f1 and h = 10
We know that
Median = `"l" + {{("N")/(2) -"F")/("f")} xx "h"`
`32.5 = 30 + {(20-(14+"f"_1))/(12)} xx 10`
`2.5 = ((6 - "f"_1) xx 10)/(12)`
2.5 x 12 = 60 - 10`"f"_1`
`"f"_1 = (30)/(10)`
= 3
Putting the value of `f_1` in (1), we get
`"f"_2` = 9 - 3
= 6
Hence, the missing frequencies are 3 and 6.
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