How to find Mean of Grouped Data:

(ii) Method 2: The Assumed Mean Method

Mean of deviations is given as:
		$d̄$ = ${Σf_i d_i }/{Σf_i}$
		
Since $d_i=(x_i-a)$

So, $d̄$ = ${Σf_i (x_i-a)}/{Σf_i}$
	   = ${Σf_i x_i}/{Σf_i}$ -   ${aΣf_i}/{Σf_i}$
	 $d̄$  = $x̄$ -   $a$
	 $x̄$ =   $a$ + $d̄$
	 $x̄$ =   $a$ +${Σf_i d_i }/{Σf_i}$

Example:Conside the previos example.
Let the assumed mean of $x̄_i$  be $a$= 17.5 or 22.5.

Case 1: Let $a$ = 17.5

 
Time taken (in seconds)
Frequency ($f_i$)
Midpoint($x_i$)
$d_i=x_i - a$
$f_id_i$
 
5 ≤ t < 10
1
7.5
-10
-10
 
10 ≤ t < 15
4
12.5
-5
-20
 
15 ≤ t < 20
6
17.5
0
0
 
20 ≤ t < 25
4
22.5
5
20
 
25 ≤ t < 30
2
27.5
10
20
 
30 ≤ t < 35
3
32.5
15
45
 
Total 
20
55
 

  $x̄$ =   $a$ +${Σf_i d_i }/{Σf_i}$
             = 17.5 + $55/20$
	    = $20.25$


Case 2: Let $a$=22.5

 
Time taken (in seconds)
Frequency ($f_i$)
Midpoint($x_i$)
$d_i=x_i - a$
$f_id_i$
 
5 ≤ t < 10
1
7.5
-15
-15
 
10 ≤ t < 15
4
12.5
-10
-40
 
15 ≤ t < 20
6
17.5
-5
-30
 
20 ≤ t < 25
4
22.5
0
0
 
25 ≤ t < 30
2
27.5
5
10
 
30 ≤ t < 35
3
32.5
10
30
 
Total 
20
-45
 

  $x̄$ =   $a$ +${Σf_i d_i }/{Σf_i}$
             = 22.5 + $(-45)/20$
             = 22.5 - $2.25$
	    = $20.25$