What is step deviation method? Explain with an example.
gopika43:
hii
Answers
Answered by
0
Step Deviation Method (Mean)
Step Deviation : Sometimes, during the application of the short-cut method for finding the mean, the deviation d, are divisible by a common number ‘h’ .In this case the di = xi – A is reduced to a great extent as di becomes di / h. So the formula of mean by this is :
Where ui = ( xi – A) / h ; h = class width and N = Σ fi
Finding mean by using this formula is known as the Step Deviation Method.
Some solved examples
1) Apply Step - Deviation method to find arithmetic mean of the following frequency distribution.
variate
5
10
15
20
25
30
Frequency
20
43
75
67
72
45
Solution:
Let the assumed mean be A = 20 and h = 5.
Variate (xi)
Frequency (fi)
Deviation= di = xi - 20
ui = (xi - 20 )/ 5
fi ui
5
20
-15
-3
-60
10
43
-10
-2
-86
15
75
-5
-1
-75
20
67
0
0
0
25
72
5
1
72
30
45
10
2
90
N = Σ fi = 322
-59
N = 322, A = 20 , h = 5 and Σ fi ui = - 59
⇒ Mean = 20 + 5 ( - 59 / 322)
⇒ Mean = 20 – 0.91
∴ Mean = 19.09
_____________________________________________________________
2) Find the mean of following frequency distribution:
Class interval
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
Number of workers
7
10
15
8
10
Solution :
Class intervals Mid values (xi) Frequency (fi) di = xi - 25 ui = (xi - 25) / 10 fi ui
0-10
5
7
-20
-2
-14
10-20
15
10
-10
-1
-10
20-30
25
15
0
0
0
30-40
35
8
10
1
8
40-50
45
10
20
2
20
N = Σ fi = 50 4
A = 25 , h = 10 , N = 50 and Σ fi ui = 4
⇒ Mean = 25 + 10 x ( 4 / 50)
⇒ Mean = 25 + 0.8
∴ Mean = 25.8
Step Deviation : Sometimes, during the application of the short-cut method for finding the mean, the deviation d, are divisible by a common number ‘h’ .In this case the di = xi – A is reduced to a great extent as di becomes di / h. So the formula of mean by this is :
Where ui = ( xi – A) / h ; h = class width and N = Σ fi
Finding mean by using this formula is known as the Step Deviation Method.
Some solved examples
1) Apply Step - Deviation method to find arithmetic mean of the following frequency distribution.
variate
5
10
15
20
25
30
Frequency
20
43
75
67
72
45
Solution:
Let the assumed mean be A = 20 and h = 5.
Variate (xi)
Frequency (fi)
Deviation= di = xi - 20
ui = (xi - 20 )/ 5
fi ui
5
20
-15
-3
-60
10
43
-10
-2
-86
15
75
-5
-1
-75
20
67
0
0
0
25
72
5
1
72
30
45
10
2
90
N = Σ fi = 322
-59
N = 322, A = 20 , h = 5 and Σ fi ui = - 59
⇒ Mean = 20 + 5 ( - 59 / 322)
⇒ Mean = 20 – 0.91
∴ Mean = 19.09
_____________________________________________________________
2) Find the mean of following frequency distribution:
Class interval
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
Number of workers
7
10
15
8
10
Solution :
Class intervals Mid values (xi) Frequency (fi) di = xi - 25 ui = (xi - 25) / 10 fi ui
0-10
5
7
-20
-2
-14
10-20
15
10
-10
-1
-10
20-30
25
15
0
0
0
30-40
35
8
10
1
8
40-50
45
10
20
2
20
N = Σ fi = 50 4
A = 25 , h = 10 , N = 50 and Σ fi ui = 4
⇒ Mean = 25 + 10 x ( 4 / 50)
⇒ Mean = 25 + 0.8
∴ Mean = 25.8
Attachments:
Answered by
1
Hope this helps you
Attachments:
Similar questions