1. Calculate Mode 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)`	Mid value `(x)` `(3)`	`f*x` `(4)=(2)xx(3)`	`cf` `(6)`
1	3	1	3	3
2	4	2	8	7
5	10	5	50	17
6 - 10	23	8	184	40
10 - 20	20	15	300	60
20 - 30	20	25	500	80
30 - 50	15	40	600	95
50 - 70	3	60	180	98
70 - 100	2	85	170	100
---	---	---	---	---
	`n = 100`	-----	`sum f*x=1995`	-----

Mean `bar x = (sum fx)/n`

`=1995/100`

`=19.95`

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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`

---

Mode :
The given data is uni-model.
Hence, we find the mode with the help of the formula.
`Z = 3M - 2 bar x`

`=3 * 15 - 2 * 19.95`

`=45 - 39.9`

`=5.1`

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2. Calculate Mode 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)`	Mid value `(x)` `(3)`	`f*x` `(4)=(2)xx(3)`	`cf` `(6)`
2	1	2	2	1
3	2	3	6	3
4	2	4	8	5
5 - 9	8	7	56	13
10 - 14	15	12	180	28
15 - 19	8	17	136	36
20 - 29	4	24.5	98	40
---	---	---	---	---
	`n = 40`	-----	`sum f*x=486`	-----

Mean `bar x = (sum fx)/n`

`=486/40`

`=12.15`

---

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`

---

Mode :
The given data is uni-model.
Hence, we find the mode with the help of the formula.
`Z = 3M - 2 bar x`

`=3 * 11.8333 - 2 * 12.15`

`=35.5 - 24.3`

`=11.2`

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