Find missing frequency from the following data
| Class | Frequency |
| 10 - 15 | 24 |
| 15 - 20 | a |
| 20 - 25 | 90 |
| 25 - 30 | 122 |
| 30 - 35 | b |
| 35 - 40 | 56 |
| 40 - 45 | 20 |
| 45 - 50 | 33 |
Total Frequency (N) = 460 and q-1 = 21.5
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` | `cf`
`(6)` |
| 10 - 15 | 24 | 24 |
| 15 - 20 | a | 24+a |
| 20 - 25 | 90 | 114+a |
| 25 - 30 | 122 | 236+a |
| 30 - 35 | b | 236+a+b |
| 35 - 40 | 56 | 292+a+b |
| 40 - 45 | 20 | 312+a+b |
| 45 - 50 | 33 | 345+a+b |
| --- | --- | |
| `n=345+a+b` | |
`n = 460`
`345 + a+b= 460`
`a+b=115 ->(1)`
To find `Q_1` class
Here, `Q_1` is `21.5`.
`:.` The `Q_1` class is `20 - 25`.
Now,
`:. L = `lower boundary point of `Q_1` class `=20`
`:. n = `Total frequency `=460`
`:. cf = `Cumulative frequency of the class preceding the `Q_1` class `=24 + a`
`:. f = `Frequency of the `Q_1` class `=90`
`:. c = `class length of `Q_1` class `=5`
`Q_1` ` = L + (( n)/4 - cf)/f * c`
`21.5=20 + (( 460)/4 - (24 + a))/90 * 5`
`21.5 - 20=(115 - (24 + a))/90 * 5`
`1.5=(-a+91)/90 * 5`
`1.5*90=(-a+91)*5`
`135=-5a+455`
`5a-320 = 0`
`5a = 320`
`a = 320/5`
`a = 64 ->(2)`
Now solving this 2 equations using substitution method
`a+b=115`
and `a=64`
`a+b=115 ->(1)`
`a=64 ->(2)`
Putting `a=64` in equation `(1)`, we get
`=>a+b=115`
`=>64+b=115`
`=>b=115-64`
`=>b=51`
`:.a=64" and "b=51`
Thus, the missing frequencies are `64 and 51` respectively.
This material is intended as a summary. Use your textbook for detail explanation.
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