Stepdeviation Method
Here we will learn the stepdeviation method for finding the mean of classified data.
We know that the direct method of finding the mean of
classified data gives
Mean A = \(\frac{\sum m_{i}f_{i}}{\sum f_{i}}\)
where m_{1}, m_{2}, m_{3}, m_{4}, ……, m_{n} are the class marks of the class intervals and f_{1}, f_{2}, f_{3}, f_{4}, …….., f_{n} are the frequencies of the corresponding classes.
For class intervals of equal size of width l, let the assumed mean be a.
Taking d_{i} = m_{i} – a, where m_{i }= class mark of the ith class interval and d_{i}’ = \(\frac{d_{1}}{l}\) = \(\frac{m_{1}  a}{l}\), the above formula for mean becomes
A = \(\frac{\sum m_{i}f_{i}}{\sum f_{i}}\)
= \(\frac{\sum (l{d_{i}}' + a)f_{i}}{\sum f_{i}}\)
= \(\frac{\sum (l{d_{i}}'f_{i} + af_{i}) }{\sum f_{i}}\)
= \(\frac{\sum l{d_{i}}'f_{i} + \sum af_{i}}{\sum f_{i}}\)
= \(\frac{\sum af_{i}f_{i}}{\sum f_{i}}\) + \(\frac{\sum l{d_{i}}'f_{i}}{\sum f_{i}}\)
= \(\frac{a\sum f_{i}}{\sum f_{i}}\) + \(\frac{l\sum {d_{i}}'f_{i}}{\sum f_{i}}\)
Therefore, A = a + l ∙ \(\frac{\sum {d_{i}}'f_{i}}{\sum f_{i}}\)
This is the formula for determining the mean by stepdeviation method.
Solved Examples on Finding Mean Using Step Deviation Method:
Find the mean of the following distribution using the stepdeviation method.
Solution:
Here, the intervals are of equal size. So we can apply the stepdeviation method, in which
A = a + l ∙ \(\frac{\sum {d_{i}}'f_{i}}{\sum f_{i}}\)
where a = assumed mean,
l = common size of class intervals
f_{i} = frequency of the ith class interval
d_{i}’ = \(\frac{m_{1}  a}{l}\), mi being the calss mark of the ith class interval.
Putting the calculated values in a table, we have the following.
Class Intervals

Class Mark (m_{i})

Frequency (f_{i})

d_{i }= m_{i}  a = m_{i}  28

d_{i}’= \(\frac{d_{i}}{l}\) = \(\frac{d_{i}}{8}\)

d_{i}’f_{i}

0  8

4

10

24

3

30

8  16

12

20

16

2

40

16  24

20

14

8

1

14

24  32

28

16

0

0

0

32  40

36

18

8

1

18

40  48

44

22

16

2

44



\(\sum f_{i}\) = 100



\(\sum {d_{i}}'f_{i}\) = 22

Therefore, A = a + l ∙ \(\frac{\sum {d_{i}}'f_{i}}{\sum f_{i}}\)
= 28 + 8 ∙ \(\frac{22}{100}\)
= 28 – 1.76
= 26.24.
2. Find the mean percentage of the work completed for a
project in a country from the following frequency distribution by using the
stepdeviation method.
Solution:
The frequency table with overlapping class intervals is as
follows.
Taking the assumed mean a = 50, the calculations will be as
below.
Class Intervals

Class Mark (m_{i})

d_{i} = m_{i}  a = m_{i}  50

d_{i}’ = \(\frac{m_{1}  a}{l}\) = \(\frac{m_{1}  50}{20}\)

Frequency (f_{i})

d_{i}’f_{i}

0  20

10

40

2

15

30

20  40

30

20

1

45

45

40  60

50

0

0

15

0

60  80

70

20

1

17

17

80  100

90

40

2

8

16

\(\sum f_{i}\) = 100
\(\sum {d_{i}}'f_{i}\) = 42
Therefore, mean = a + l ∙ \(\frac{\sum {d_{i}}'f_{i}}{\sum f_{i}}\)
= 50 + 20 × \(\frac{42}{100}\)
= 50  \(\frac{42}{5}\)
= 50 – 8.4
= 41.6.
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