Formula

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

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

3. Mode `Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`

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Examples
1. Calculate Mean, Median, Mode from the following grouped data

Class	Frequency
2 - 4	3
4 - 6	4
6 - 8	2
8 - 10	1

Solution:

Class `(1)`	Frequency `(f)` `(2)`	Mid value `(x)` `(3)`	`f*x` `(4)=(2)xx(3)`	`cf` `(6)`
2-4	3	3 `3=(2+4)/2`	9 `9=3xx3` `(4)=(2)xx(3)`	3 `3=0+3` `(6)=`Previous `(6)+(2)`
4-6	4	5 `5=(4+6)/2`	20 `20=4xx5` `(4)=(2)xx(3)`	7 `7=3+4` `(6)=`Previous `(6)+(2)`
6-8	2	7 `7=(6+8)/2`	14 `14=2xx7` `(4)=(2)xx(3)`	9 `9=7+2` `(6)=`Previous `(6)+(2)`
8-10	1	9 `9=(8+10)/2`	9 `9=1xx9` `(4)=(2)xx(3)`	10 `10=9+1` `(6)=`Previous `(6)+(2)`
---	---	---	---	---
--	`n = 10`	--	`sum f*x=52`	--

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

`=52/10`

`=5.2`

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

= value of `(10/2)^(th)` observation

= value of `5^(th)` observation

From the column of cumulative frequency `cf`, we find that the `5^(th)` observation lies in the class `4 - 6`.

`:.` The median class is `4 - 6`.

Now,
`:. L = `lower boundary point of median class `=4`

`:. n = `Total frequency `=10`

`:. cf = `Cumulative frequency of the class preceding the median class `=3`

`:. f = `Frequency of the median class `=4`

`:. c = `class length of median class `=2`

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

`=4 + (5 - 3)/4 * 2`

`=4 + (2)/4 * 2`

`=4 + 1`

`=5`

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To find Mode Class
Here, maximum frequency is `4`.

`:.` The mode class is `4 - 6`.

`:. L = `lower boundary point of mode class `=4`

`:. f_1 = ` frequency of the mode class `=4`

`:. f_0 = ` frequency of the preceding class `=3`

`:. f_2 = ` frequency of the succedding class `=2`

`:. c = ` class length of mode class `=2`

`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`

`=4 + ((4 - 3)/(2*4 - 3 - 2)) * 2`

`=4 + (1/3) * 2`

`=4 + 0.6667`

`=4.6667`

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2. Calculate Mean, Median, Mode from the following grouped data

X	Frequency
0	1
1	5
2	10
3	6
4	3

Solution:

`x` `(1)`	Frequency `(f)` `(2)`	`f*x` `(3)=(2)xx(1)`	`cf` `(5)`
0	1	0 `0=1xx0` `(3)=(2)xx(1)`	1 `1=0+1` `(5)=`Previous `(5)+(2)`
1	5	5 `5=5xx1` `(3)=(2)xx(1)`	6 `6=1+5` `(5)=`Previous `(5)+(2)`
2	10	20 `20=10xx2` `(3)=(2)xx(1)`	16 `16=6+10` `(5)=`Previous `(5)+(2)`
3	6	18 `18=6xx3` `(3)=(2)xx(1)`	22 `22=16+6` `(5)=`Previous `(5)+(2)`
4	3	12 `12=3xx4` `(3)=(2)xx(1)`	25 `25=22+3` `(5)=`Previous `(5)+(2)`
---	---	---	---
	`n=25`	`sum f*x=55`	--

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

`=55/25`

`=2.2`

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Median :
M = value of `(n/2)^(th)` observation

= value of `(25/2)^(th)` observation

= value of `12^(th)` observation

From the column of cumulative frequency `cf`, we find that the `12^(th)` observation is `2`.

Hence, the median of the data is `2`.

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Mode :
the frequency of observation `2` is maximum (`10`)

`:. Z = 2`

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