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.

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