1. Calculate Median from the following mixed data

Class	Frequency
1	3
2	4
5	10
6 - 10	23
10 - 20	20
20 - 30	20
30 - 50	15
50 - 70	3
70 - 100	2

Solution:

Class `(1)`	Frequency `(f)` `(2)`	`cf` `(6)`
1	3	3
2	4	7
5	10	17
6 - 10	23	40
10 - 20	20	60
20 - 30	20	80
30 - 50	15	95
50 - 70	3	98
70 - 100	2	100
---	---	---
	`n = 100`	-----

To find Median Class
= value of `(n/2)^(th)` observation

= value of `(100/2)^(th)` observation

= value of `50^(th)` observation

From the column of cumulative frequency `cf`, we find that the `50^(th)` observation lies in the class `10 - 20`.

`:.` The median class is `10 - 20`.

Now,
`:. L = `lower boundary point of median class `=10`

`:. n = `Total frequency `=100`

`:. cf = `Cumulative frequency of the class preceding the median class `=40`

`:. f = `Frequency of the median class `=20`

`:. c = `class length of median class `=10`

Median `M = L + (n/2 - cf)/f * c`

`=10 + (50 - 40)/20 * 10`

`=10 + (10)/20 * 10`

`=10 + 5`

`=15`

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2. Calculate Median from the following mixed data

Class	Frequency
2	1
3	2
4	2
5 - 9	8
10 - 14	15
15 - 19	8
20 - 29	4

Solution:

Class `(1)`	Frequency `(f)` `(2)`	`cf` `(6)`
2	1	1
3	2	3
4	2	5
5 - 9	8	13
10 - 14	15	28
15 - 19	8	36
20 - 29	4	40
---	---	---
	`n = 40`	-----

To find Median Class
= value of `(n/2)^(th)` observation

= value of `(40/2)^(th)` observation

= value of `20^(th)` observation

From the column of cumulative frequency `cf`, we find that the `20^(th)` observation lies in the class `10 - 14`.

`:.` The median class is `9.5 - 14.5`.

Now,
`:. L = `lower boundary point of median class `=9.5`

`:. n = `Total frequency `=40`

`:. cf = `Cumulative frequency of the class preceding the median class `=13`

`:. f = `Frequency of the median class `=15`

`:. c = `class length of median class `=5`

Median `M = L + (n/2 - cf)/f * c`

`=9.5 + (20 - 13)/15 * 5`

`=9.5 + (7)/15 * 5`

`=9.5 + 2.3333`

`=11.8333`

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