Whenever the data values are large, and calculation is tedious, the step deviation method is applied.
The following steps are used while applying the step deviation method to calculate the arithmetic mean:
- Choose one observation from the data set and mark it as the assumed mean of the whole series. In the case of grouped data, it is not possible to pick an observation from the class intervals, so one first needs to calculate the class marks of mid-points of the intervals and mark one as the assumed mean.
- If the class intervals are not continuous, then one must first make them continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.
- The next step is to find deviations from the assumed mean (A) by deducting the mean so assumed from all the other observations. d = X - A.
- Next, we are supposed to calculate the step deviations from the deviations obtained above, by finding a common factor, denoted by c, of all the values (deviations), dividing all by this factor, and labelling the step deviations as d1
- Multiply the step deviations with the frequencies and take up the sum of the numbers so obtained.
- Apply the formula:
x̄=a+\frac{Σd_1f}{Σf}×c , where Σd1 is the sum of all the step deviations multiplied by respective frequencies and c represents the common factor. - The number so obtained is the arithmetic mean of the given data set.
Thus, the formula for the calculation of arithmetic mean by step deviation method is
x̄=a+\frac{Σd_1f}{Σf}×c
Example: Calculate the arithmetic mean for the following data set using the step deviation method:
Marks | Number of Students |
|---|---|
0 - 10 | 5 |
10 - 20 | 12 |
20 - 30 | 14 |
30 - 40 | 10 |
40 - 50 | 8 |
Solution: Class intervals are Continuous
Marks
f
m
d = m - A
A = 25
d1 = d/ c
c = 10
fd1
0 - 10
5
5
5 - 25 = −20
−2
−10
10 - 20
12
15
15 - 25 = −10
−1
−12
20 - 30
14
A = 25
25 - 25 = 0
0
0
30 - 40
10
35
35 - 25 = 10
1
10
40 - 50
8
45
45 - 25 = 20
2
16
Σf = 49
Σfd1 = 4
Mean = X̄ =
a+\frac{Σd_1f}{Σf}×c =
25+\frac{4}{49}×10 = 25 + 0.81
= 25.81
Hence, Arithmetic Mean of the given data set is 25.81
Example: Calculate the mean for the following data set using the step deviation method:
Class Interval | Frequency |
|---|---|
30 - 32 | 5 |
33 - 35 | 12 |
36 - 38 | 18 |
39 - 41 | 7 |
42 - 44 | 8 |
Solution: Class intervals are not Continuous
The given class intervals do not touch each other.
The gap between successive intervals is 1.To make them continuous, subtract 0.5 from every lower limit and add 0.5 to every upper limit.
After that, we apply the step-deviation method.
Class Interval
Continuous Interval
f
m
d = m - A
A = 37
d1 = d/ c
c = 3
fd1
30 - 32
29.5–32.5
5
31
31 - 37 = −6
−2
−10
33 - 35
32.5–35.5
12
34
34 - 37 = −3
−1
−12
36 - 38
35.5–38.5
18
A = 37
37 - 37 = 0
0
0
39 - 41
38.5–41.5
7
40
40 - 37 = 3
1
7
42 - 44
41.5–44.5
8
43
43 - 37 = 6
2
16
Σf = 50
Σfd1 = 1
Mean = X̄ =
a+\frac{Σd_1f}{Σf}×c =
37+\frac{1}{50}×3 = 37 + 0.06
= 37.06
Hence, Mean of the given data set is 37.06
Sample Questions on Calculating Mean using Step Deviation Method
Question 1. Calculate the mean using the step deviation method:
Marks | Number of students |
|---|---|
10 - 20 | 5 |
20 - 30 | 3 |
30 - 40 | 4 |
40 - 50 | 7 |
50 - 60 | 2 |
60 - 70 | 6 |
70 - 80 | 13 |
Solution:
Marks f
m
d = m - A
A = 45
d1 = d/ c
c = 10
fd1
10 - 20 5
15 −30
−3
−15
20 - 30 3
25 −20
−2
−6
30 - 40 4
35 −10
−1
−4
40 - 50 7
45 0
0
0
50 - 60 2
55 10
1
2
60 - 70 6
65 20
2
12
70 - 80 13
75 30
3
39
Σf = 40
Σfd1 = 28 Mean = X̄ =
a+\frac{Σd_1f}{Σf}×c =
45+\frac{28}{40}×10 = 45 + 7
= 52
Hence, Arithmetic Mean of the given data set is 52.
Question 2. Calculate the mean using the step deviation method:
Class Intervals | Frequency |
|---|---|
−40 to −30 | 10 |
−30 to −20 | 28 |
−20 to −10 | 30 |
−10 to 0 | 42 |
0 to 10 | 65 |
10 to 20 | 180 |
20 to 30 | 10 |
Solution:
Class Intervals
f
m
d = m - A
A = −5
d1 = d/c
c = 10
fd1
−40 to −30
10
−35
−30
−3
−30
−30 to −20
28
−25
−20
−2
−56
−20 to −10
30
−15
−10
−1
−30
−10 to 0
42
−5
0
0
0
0 to 10
65
5
10
1
65
10 to 20
180
15
20
2
360
20 to 30
10
25
30
3
30
Σf = 365 Σfd1 = 339 Mean = X̄ =
a+\frac{Σd_1f}{Σf}×c =
−5+\frac{339}{365}×10 = 4.288
Hence arithmetic mean is 4.288
Question 3. Calculate the mean using the step deviation method:
| Wages | Number of workers |
|---|---|
| 0 - 10 | 22 |
| 10 - 20 | 38 |
| 20 - 30 | 46 |
| 30 - 40 | 35 |
| 40 - 50 | 19 |
Solution:
Wages
f
m
d = m - A
A = 25
d1 = d/c
c = 10
fd1
0 - 10
22
5
−20
−2
−44
10 - 20
38
15
−10
−1
−38
20 - 30
46
25
0
0
0
30 - 40
35
35
10
1
35
40 - 50
19
45
20
2
38
Σf = 160
Σfd1 = −9
Mean = X̄ =
a+\frac{Σd_1f}{Σf}×c =
25~+~\frac{−9}{160}×10 = 24.44
Hence, arithmetic mean is 24.44
Question 4. Calculate the mean using the step deviation method:
Age | Number of People |
|---|---|
0 - 20 | 4 |
20 - 40 | 10 |
40 - 60 | 15 |
60 - 80 | 20 |
80 - 100 | 11 |
Solution:
Age
f
m
d = m - A
A = 50
d1 = d/c
c = 20
fd1
0 - 20
4
10
−40
−2
−8
20 - 40
10
30
−20
−1
−10
40 - 60
15
50
0
0
0
60 - 80
20
70
20
1
20
80 - 100
11
90
40
2
22
Σf = 60 Σfd1 = 24 Mean = X̄ =
a+\frac{Σd_1f}{Σf}×c =
50+\frac{24}{60}×20 = 50 + 8
= 58
Hence, arithmetic mean is 58.
Practice Problems: Step Deviation Method
Calculate the mean using the step deviation method.
Problem 1: Consider the following frequency distribution:
| Class Interval | Frequency (fi) |
|---|---|
| 5 - 15 | 6 |
| 15 - 25 | 9 |
| 25 - 35 | 13 |
| 35 - 45 | 10 |
| 45 - 55 | 7 |
Problem 2: Consider the following frequency distribution:
| Class Interval | Frequency (fi) |
|---|---|
| 20 - 30 | 5 |
| 30 - 40 | 8 |
| 40 - 50 | 12 |
| 50 - 60 | 15 |
| 60 - 70 | 10 |
Problem 3: Consider the following frequency distribution:
| Class Interval | Frequency (fi) |
|---|---|
| 10 - 20 | 6 |
| 20 - 30 | 11 |
| 30 - 40 | 7 |
| 40 - 50 | 15 |
| 50 - 60 | 5 |
Problem 4: Consider the following frequency distribution:
| Class Interval | Frequency (fi) |
|---|---|
| 5 - 15 | 4 |
| 15 - 25 | 7 |
| 25 - 35 | 11 |
| 35 - 45 | 15 |
| 45 - 55 | 8 |
| 55 - 65 | 5 |
