Step-deviation Method

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 m₁, m₂, m₃, m₄, ……, m_n are the class marks of the class intervals and f₁, f₂, f₃, f₄, …….., 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.

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.

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.

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

9th Grade Math

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