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notes
To calculate the mean of grouped data, the first step is to determine the midpoint (also called a class mark) of each interval, or class. These midpoints must then be multiplied by the frequencies of the corresponding classes. The sum of the products divided by the total number of values will be the value of the mean.
Average of any observation is known as mean. In this chapter there are three methods to find mean
1) Direct method
2) Assumed mean method
3) Step deviation method
We will take a common example to understand these three methods
1) Direct method
Class interval 
1025 
2540 
4055 
5570 
7085 
85100 
No. of students 
2 
3 
7 
6 
6 
6 
CI class interval
fi Number of students
xi= class mark= `"Upper class limit+ Lower class limit"/2`
∑ means summation, which means the total
CI 
fi 
xi 
fixi 
1025 
2 
(10+25)/2= 17.5 
35 
2540 
3 
32.5 
97.5 
4055 
7 
47.5 
332.5 
5570 
6 
62.5 
375 
7085 
6 
77.5 
465 
85100 
6 
92.5 
555 

∑ fi= 30 

∑ fixi= 1860 
Mean through direct method= `bar(x)`= `(sum "fixi")/ (sum "fi")`
Mean through direct method= `1860/30= 62`
2) Assumed mean method
Mean through assumed mean method= `bar(x)`= ` a+(sum "fidi")/(sum "fi")`
where a= assumed mean i.e any value of xi
di= deviation= xia
CI 
fi 
xi 
fixi 
di=xia 
fidi 
1025 
2 
17.5 
35 
30 
60 
2540 
3 
32.5 
97.5 
15 
45 
4055 
7 
47.5 
332.5 
0 
0 
5570 
6 
62.5 
375 
15 
90 
7085 
6 
77.5 
465 
30 
180 
85100 
6 
92.5 
555 
45 
270 

∑ fi= 30 

∑ fixi= 1860 

∑ fidi= 435 
Let a= 47.5
Mean through assumed mean method= `bar(x)`= ` a+(sum "fidi")/(sum "fi")`
= 47.5+ 435/30
= 47.5+ 14.5
Mean through assumed mean method= 62
3) Step deviation method
Mean through step deviation method= `bar(x)`= `a+ (sum "fiui")/(sum "fi") xx h`
where, ui= modified class mark= `(di)/h`
h= Class size
CI 
fi 
xi 
fixi 
di=xia 
fidi 
ui=di/h 
fiui 
1025 
2 
17.5 
35 
30 
60 
2 
4 
2540 
3 
32.5 
97.5 
15 
45 
1 
3 
4055 
7 
47.5 
332.5 
0 
0 
0 
0 
5570 
6 
62.5 
375 
15 
90 
1 
6 
7085 
6 
77.5 
465 
30 
180 
2 
12 
85100 
6 
92.5 
555 
45 
270 
3 
18 

∑fi= 30 

∑ fixi= 1860 

∑ fidi= 435 

∑fiui= 29 
Mean through step divation method= `bar(x)`= `a+ (sum "fiui")/(sum "fi") xx h`
=`47.5+ (29/30) xx 5`
=` 47.5+ 14.5`
Mean through step deviation= `bar(x)`= 62
As you can see, the mean obtained is same i.e 62 from any of the method.
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Related QuestionsVIEW ALL [15]
The following table gives the frequency distribution of trees planted by different Housing Societies in a particular locality:
No. of Trees  No. of Housing Societies 
1015  2 
1520  7 
2025  9 
2530  8 
3035  6 
3540  4 
Find the mean number of trees planted by Housing Societies by using ‘Assumed Means Method’
The measurements (in mm) of the diameters of the head of the screws are given below:
Diameter (in mm)  No. of Screws 
33 — 35  10 
36 — 38  19 
39 — 41  23 
42 — 44  21 
45 — 47  27 
Calculate mean diameter of head of a screw by ‘Assumed Mean Method’.
There are three dealers A, B and C in Maharashtra. Suppose, the trade of each of them in september 2018 was as shown in the following table.
The rate of GST on each transaction was 5%.
Read the table and answer the questions below it.
Dealer  GST collected on the sale 
GST paid at the time of purchase 
ITC  Tax paid to the Govt. 
Taxbalance with the Govt. 
A  Rs.5000  Rs. 6000  Rs. 5000  Rs. 0  Rs. 1000 
B  Rs 5000  Rs. 4000  Rs. 4000  Rs. 1000  Rs. 0 
C  Rs.5000  Rs. 5000  Rs. 5000  Rs. 0  Rs. 0 
(i) How much amount did the dealer A get by sale ?
(ii) For how much amount did the dealer B buy the articles ?
(iii) How much is the balance of CGST and SGST left with the government that was paid by A ?
Frequency distribution of daily commission received by 100 salesmen is given below :
Daily Commission (in Rs.) 
No. of Salesmen 
100120  20 
120140  45 
140160  22 
160180  09 
180200  04 
Find mean daily commission received by salesmen, by the assumed mean method.
The following frequency distribution table shows the amount of aid given to 50 flood affected families. Find the mean of the amount of aid.
Amount of aid
(Thosand rupees)

50  60  60  70  70  80  80  90  90  100 
No. of families  7  13  20  6  4 