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# Mean Deviation For Ungrouped Data

The basic difference between grouped data and ungrouped data is that in the case of latter, the data is unorganized and is in random form. This type of data is also known as raw data, whereas in the case of grouped data, it is organized in the form of groups or which has been categorized in terms of the frequency distribution. These groups are known as class intervals.

For example, Marks of 10 students (out of 100) are given as:
45, 60, 65, 78, 91, 38, 67, 81, 12, 55

This form of data is ungrouped in nature.

This can be represented in grouped form as:

 Marks Frequency 0-9 0 10-19 1 20-29 0 30-39 1 40-49 1 50-59 1 60-69 3 70-79 1 80-89 1 90-99 1

## Steps to Calculate the Mean Deviation for Ungrouped data

To calculate the mean deviation for ungrouped data, the following steps are followed:

Let the set of data consist of  observations

$$\begin{array}{l} x_1,x_2,x_3………..x_n\end{array}$$
.

Step i) The measure of central tendency about which mean deviation is to be found out is calculated. Let this measure be a.

Step ii) Calculate the absolute deviation of each observation from the measure of central tendency calculated in step (i) i.e.,

$$\begin{array}{l} |x_1-a|,|x_2-a|,|x_3-a|………..|x_n-a| \end{array}$$

Step iii) Evaluate the mean of all the absolute deviations. This gives the mean absolute deviation (M.A.D) about ‘a‘ for ungrouped data i.e.,
M.A.D (a) =

$$\begin{array}{l} \large \frac {\sum^n_{i=1} |x_i – a|}{n}\end{array}$$

In case the measure of central tendency is mean the above equation can be rewritten as:

M.A.D(

$$\begin{array}{l}\bar x \end{array}$$
)=
$$\begin{array}{l}\large \frac {\sum^n_{i=1} |x_i – \bar x|}{n}\end{array}$$

Where
$$\begin{array}{l} \bar x \end{array}$$
=Mean

Similarly around median,
M.A.D(a) =

$$\begin{array}{l}\large \frac {\sum^n_{i=1} |x_i – M|}{n}\end{array}$$

Where M=Median

Let us go through an example to have a better insight of the topic.

### Examples

Example 1: In a mango eating competition the number of mangoes eaten by six contestants in an hour is as follows:
12, 18, 21, 26, 17, 20
Find the mean deviation about the mean for the given data.

Solution 1: Let us follow the steps mentioned to calculate M.A.D about mean.
Step i) The mean of the following data can be given by,

$$\begin{array}{l}\large \bar x \end{array}$$
=
$$\begin{array}{l}\large \frac {\sum^n_{i=1} x_i}{n}\end{array}$$

$$\begin{array}{l}\large \bar x \end{array}$$
=
$$\begin{array}{l}\large \frac {12+18+21+26+17+20}{6}\end{array}$$
= 19

Step ii) Let us now calculate the absolute deviation around each observation,

$$\begin{array}{l}~~~~~~~~~~~~~~~\end{array}$$
|
$$\begin{array}{l}x_1 – \bar x \end{array}$$
|= |12-19|=7
$$\begin{array}{l}~~~~~~~~~~~~~~~\end{array}$$
|
$$\begin{array}{l}x_2 – \bar x \end{array}$$
|=|18-19|=1
$$\begin{array}{l}~~~~~~~~~~~~~~~\end{array}$$
|
$$\begin{array}{l}x_3 – \bar x \end{array}$$
|=|21-19|=2
$$\begin{array}{l}~~~~~~~~~~~~~~~\end{array}$$
|
$$\begin{array}{l}x_4 – \bar x \end{array}$$
|=|26-19|=7
$$\begin{array}{l}~~~~~~~~~~~~~~~\end{array}$$
|
$$\begin{array}{l}x_5 – \bar x \end{array}$$
|=|17-19|=2
$$\begin{array}{l}~~~~~~~~~~~~~~~\end{array}$$
|
$$\begin{array}{l}x_6 – \bar x \end{array}$$
|=|19-20|=1

Step iii) For calculating the mean deviation for ungrouped data:

$$\begin{array}{l}~~~~~~~~~~~~~~\end{array}$$
M.A.D(x)=
$$\begin{array}{l}\large \frac {\sum^n_{i=1}|x_i-a|}{n}\end{array}$$

$$\begin{array}{l}~~~~~~~~~~~~~~\end{array}$$
⇒M.A.D(x)=
$$\begin{array}{l}\large \frac {7+1+2+7+2+1}{6}= \frac{20}{6} \end{array}$$
= 3.333

Example 2: The runs scored by Sachin in 7 different matches are given as:

$$\begin{array}{l}~~~~~~~~~~~~~~~~\end{array}$$
89,91,54,66,13,97,06

Calculate the mean deviation about the median for this data.

Solution 2: Let us arrange the following data in ascending order to find out the median i.e.,

$$\begin{array}{l}~~~~~~~~~~~~~~~~~~\end{array}$$
06,13,54,66,89,91,97

As the number of observations is odd the median is given by

$$\begin{array}{l} \left( \frac {n+1}{2}\right)^{th}\end{array}$$
observation i.e.
$$\begin{array}{l}4^{th}\end{array}$$
observation which is 66.

Absolute deviation around each observation will be

$$\begin{array}{l}|x_i-M| \end{array}$$
i.e.,

$$\begin{array}{l}~~~~~~~~~~~~~~~~~~~~\end{array}$$
$$\begin{array}{l}|x_1-M|\end{array}$$
=|6-66|=60
$$\begin{array}{l}~~~~~~~~~~~~~~~~~~~~\end{array}$$
$$\begin{array}{l}|x_2-M|\end{array}$$
=|13-66|=53
$$\begin{array}{l}~~~~~~~~~~~~~~~~~~~~\end{array}$$
$$\begin{array}{l}|x_3-M|\end{array}$$
=|54-66|=12
$$\begin{array}{l}~~~~~~~~~~~~~~~~~~~~\end{array}$$
$$\begin{array}{l}|x_4-M|\end{array}$$
=|66-66|=0
$$\begin{array}{l}~~~~~~~~~~~~~~~~~~~~\end{array}$$
$$\begin{array}{l}|x_5-M|\end{array}$$
=|89-66|=23
$$\begin{array}{l}~~~~~~~~~~~~~~~~~~~~\end{array}$$
$$\begin{array}{l}|x_6-M|\end{array}$$
=|91-66|=25
$$\begin{array}{l}~~~~~~~~~~~~~~~~~~~~\end{array}$$
$$\begin{array}{l}|x_7-M|\end{array}$$
=|97-66|=31

The mean absolute deviation for the ungrouped data is given by:

$$\begin{array}{l}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\end{array}$$
M.A.D(M) =
$$\begin{array}{l}\large \frac {\sum^n_{i=1}|x_i – M|}{n}\end{array}$$

$$\begin{array}{l}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\end{array}$$
⇒M.A.D(M)=
$$\begin{array}{l}\large \frac {60+53+12+0+23+25+31}{7}\end{array}$$
=29.142