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Step Deviation Method

Step Deviation Method is one of the strategies to find the mean for grouped data.

The mean is the mathematical average of a set of two or more numbers. It is an arithmetic average of the given set of data, which is determined by dividing the sum of all the data points by the no. of data points in the data set.

\(\mbox{Mean} = \frac{\mbox{Sum of observations}}{\mbox{Number of observations}}\)

Depending on the size of the data, there are three strategies to find the mean for grouped data, they are,

  • Direct Method
  • Assumed Mean Method
  • Step-deviation Method

Step Deviation Method Formula

When we have large data values, the step-deviation method is used to find the mean. 

The formula is given as follows:

\(\mbox{Mean} (\overline{x}) = a + h \frac{\sum f_i u_i}{\sum f_i}\)

Here,
\(a\) = Assumed mean
\(m_i = \frac{\mbox{upper limit } + \mbox{ lower limit}}{2}\) (used for grouped frequencies)
\(h\) = Size of class interval
\(f_i\) = Frequency of \(i^{th}\) class
\(x_i\) = Class mark = \(\frac{\mbox{upper class limit } + \mbox{ lower class limit}}{2}\)
\(x_i – a\) = Deviation of \(i^{th}\) class
\(u_i = \frac{(x_i – a)}{h}\)
\(\sum f_i\) = \(N\) = Total number of observations

Solved Examples

Question 1. Use step deviation method to find the mean of the following table.

No. of letters120240360480600
Receivers20406080100

Solution. As we know step deviation method formula is given by,

\(\mbox{Mean} (\overline{x}) = a + h \frac{\sum f_i u_i}{\sum f_i}\)

Now, rearranging the table we get;

No. of lettersReceivers
12020
24040
36060
48080
600100
\(\sum f_i = 300\)

Let \(a\) = 360
\(h\) = 120, class interval 

(Since 240 – 120 = 360 – 240 = 480 – 360 = 600 – 480 = 120)

Now, we will add a column for \(u_i\);

No. of lettersReceivers (\(f_i\))\(u_i = \frac{x_i – a}{h}\)
12020-2
24040-1
360 (\(a\))600
480801
6001002
\(\sum f_i = 300\)

Now, we find \(\sum f_i u_i\);

No. of lettersReceivers (\(f_i\))\(u_i = \frac{x_i – a}{h}\)\(f_i u_i\)
12020-2-40
24040-1-40
360 (\(a\))6000
48080180
6001002200
\(\sum f_i = 300\)\(\sum f_i u_i = 200\)

Using the data we got here in the \(\overline x\) formula;

Therefore, \(\overline x = a + h \frac{\sum f_i u_i}{\sum f_i}\)

\(\overline x = 360 + 120 \frac{200}{300} = 360 + 80 = 440 \)

Hence, the mean for this distribution, \(\overline x = 440\) 

Question 2. Find the mean of the following distribution using the step deviation method.

Class Interval0 – 88 – 1616 – 2424 – 3232 – 40
Frequency820141622

Solution. As we know step deviation method formula is given by,

\(\mbox{Mean} (\overline{x}) = a + h \frac{\sum f_i u_i}{\sum f_i}\)

Now, rearranging the table we get;

Class IntervalClass Mark (\(x_i \,or \,m_i\)) = \(\frac{lower\,limit + upper\,limit}{2}\)Frequency (\(f_i\))
0 – 848
8 – 161220
16 – 242014
24 – 322816
32 – 403622
\(\sum f_i = 80\)

Let \(a\) = 20
\(h\) = 8, class interval 

Now we will add a column for \(u_i\) and \(f_i u_i\);

Class IntervalClass Mark (\(x_i\))Frequency (\(f_i\))\(u_i\)\(f_i u_i\)
0 – 848-2-16
8 – 161220-1-20
16 – 2420 (\(a\)1400
24 – 322816116
32 – 403622222
\(\sum f_i = 80\)\(\sum f_i u_i = 2\)

Now, putting values in the step deviation method formula;

\(\overline x = a + h \frac{\sum f_i u_i}{\sum f_i}\)

\(\overline x = 20 + 8\frac{2}{80} = 20.2\)

Hence, the mean for this distribution, \(\overline x = 20.2\)

Read More on Statistics

FAQs

How do we find assumed mean in Step deviation method?

Assumed mean is to be taken the middle value of the given data set.
Assumed mean(a), if data available is in;
Odd = \(\frac{n+1}{2}\)
Even = \(\frac{n}{2}\) & \(\frac{n}{2} + 1\)
In case of even, we can choose any one of the two values.

What is I in step deviation method?

\(I\) is the Class interval, sometimes it is also denoted using \(h\).
Similarly \(u_i\) is also sometimes denoted using \(d_i\) & \(f_i\) as \(R_i\).

How do we find class interval (I or h)?

We can get the class interval by subtracting one class with previous class. Class interval is uniform throughout given data.

Can Mean & assumed mean be same?

Yes, sometimes Mean & assumed mean(a) can be same.
But if the same question is repeated with taking different assumed mean(a) the final mean will not change.

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