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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 |
10-25 |
25-40 |
40-55 |
55-70 |
70-85 |
85-100 |
|
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 |
|
10-25 |
2 |
(10+25)/2= 17.5 |
35 |
|
25-40 |
3 |
32.5 |
97.5 |
|
40-55 |
7 |
47.5 |
332.5 |
|
55-70 |
6 |
62.5 |
375 |
|
70-85 |
6 |
77.5 |
465 |
|
85-100 |
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= xi-a
|
CI |
fi |
xi |
fixi |
di=xi-a |
fidi |
|
10-25 |
2 |
17.5 |
35 |
-30 |
-60 |
|
25-40 |
3 |
32.5 |
97.5 |
-15 |
-45 |
|
40-55 |
7 |
47.5 |
332.5 |
0 |
0 |
|
55-70 |
6 |
62.5 |
375 |
15 |
90 |
|
70-85 |
6 |
77.5 |
465 |
30 |
180 |
|
85-100 |
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=xi-a |
fidi |
ui=di/h |
fiui |
|
10-25 |
2 |
17.5 |
35 |
-30 |
-60 |
-2 |
-4 |
|
25-40 |
3 |
32.5 |
97.5 |
-15 |
-45 |
-1 |
-3 |
|
40-55 |
7 |
47.5 |
332.5 |
0 |
0 |
0 |
0 |
|
55-70 |
6 |
62.5 |
375 |
15 |
90 |
1 |
6 |
|
70-85 |
6 |
77.5 |
465 |
30 |
180 |
2 |
12 |
|
85-100 |
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.
Video Tutorials
Shaalaa.com | Statistics part 4 (Direct Method for mean)
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 |
| 10-15 | 2 |
| 15-20 | 7 |
| 20-25 | 9 |
| 25-30 | 8 |
| 30-35 | 6 |
| 35-40 | 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 |
| 100-120 | 20 |
| 120-140 | 45 |
| 140-160 | 22 |
| 160-180 | 09 |
| 180-200 | 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 |
