Here we will learn the step-deviation 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 step-deviation method.
Solved Examples on Finding Mean Using Step Deviation Method:
Find the mean of the following distribution using the step-deviation method.
Class Interval
0 - 8
8 - 16
16 - 24
24 - 32
32 - 40
40 - 48
Frequency
10
20
14
16
18
22
Solution:
Here, the intervals are of equal size. So we can apply the step-deviation 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 step-deviation 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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