Formula

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

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

3. Mode `Z = 3M - 2 bar x`

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Examples
1. Calculate Mean, Median, 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 `1=1`	3 `3=3xx1` `(4)=(2)xx(3)`	3 `3=0+3` `(6)=`Previous `(6)+(2)`
2	4	2 `2=2`	8 `8=4xx2` `(4)=(2)xx(3)`	7 `7=3+4` `(6)=`Previous `(6)+(2)`
5	10	5 `5=5`	50 `50=10xx5` `(4)=(2)xx(3)`	17 `17=7+10` `(6)=`Previous `(6)+(2)`
6 - 10	23	8 `8=(6+10)/2`	184 `184=23xx8` `(4)=(2)xx(3)`	40 `40=17+23` `(6)=`Previous `(6)+(2)`
10 - 20	20	15 `15=(10+20)/2`	300 `300=20xx15` `(4)=(2)xx(3)`	60 `60=40+20` `(6)=`Previous `(6)+(2)`
20 - 30	20	25 `25=(20+30)/2`	500 `500=20xx25` `(4)=(2)xx(3)`	80 `80=60+20` `(6)=`Previous `(6)+(2)`
30 - 50	15	40 `40=(30+50)/2`	600 `600=15xx40` `(4)=(2)xx(3)`	95 `95=80+15` `(6)=`Previous `(6)+(2)`
50 - 70	3	60 `60=(50+70)/2`	180 `180=3xx60` `(4)=(2)xx(3)`	98 `98=95+3` `(6)=`Previous `(6)+(2)`
70 - 100	2	85 `85=(70+100)/2`	170 `170=2xx85` `(4)=(2)xx(3)`	100 `100=98+2` `(6)=`Previous `(6)+(2)`
---	---	---	---	---
	`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`

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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 Mean, Median, 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`

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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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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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