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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 [196]
Find the mean from the following frequency distribution of marks at a test in statistics:
Marks(x) | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 |
No. of students (f) | 15 | 50 | 80 | 76 | 72 | 45 | 39 | 9 | 8 | 6 |
Find the mean of each of the following frequency distributions
Class interval | 0 - 8 | 8 - 16 | 16 - 24 | 24 - 32 | 32 - 40 |
Frequency | 5 | 9 | 10 | 8 | 8 |
The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students
Age (in years) | 15 | 16 | 17 | 18 | 19 | 20 |
No. of students | 3 | 8 | 10 | 10 | 5 | 4 |
Find the mean of the following data, using direct method:
Class | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 |
Frequency | 6 | 9 | 15 | 12 | 8 |
Find the mean of the following data, using direct method:
Class | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
Frequency | 6 | 10 | 8 | 12 | 4 |