Measures of Dispersion: Definition, Formulas and Examples

Measures of the Dispersion of the Data

An important characteristic of any set of data is the variation in the data. The degree to which the numerical data tends to vary about an average value is called the dispersion or scatteredness of the data.

The following are the measures of dispersion:

1. Range

2. Mean Deviation

3. Standard deviation and Variance

Range

Range is the difference between the highest and the lowest value in a set of observations.

The range of data gives us a rough idea of variability or scatter but does not tell about the dispersion of the data from a measure of central tendency.

Mean Deviation

Mean deviation measures the deviation of the average mean to the given set of data.

Mean deviation for ungrouped data

Let n observations are $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \ldots ., \mathrm{x}_{\mathrm{n}}$.
If $x$ is a number, then its deviation from any given value $a$ is $|x-a|$
To find the mean deviation about mean or median or any other value M of ungrouped data, following steps are involved:
1. Calculate the measure of central tendency about which we need to find the mean deviation. Let it be ' $a$ '
2. Find the deviation of each $x_i$ from $a$, i.e., $\left|x_1-a\right|,\left|x_2-a\right|,\left|x_3-a\right|, \ldots,\left|x_n-a\right|$
3. Find the mean of these deviations. This mean is the mean deviation about ' $a$ ', i.e.,

Mean deviation about 'a', M.D. $(a)=\frac{1}{n} \sum_{i=1}^n\left|x_i-a\right|$
Mean deviation about mean, M.D. $(\bar{x})=\frac{1}{n} \sum_{i=1}^n\left|x_i-\bar{x}\right|$
Mean deviation about median, M.D.(Median $) \left.=\frac{1}{n} \sum_{i=1}^n \right\rvert\, x_i-$ Median $\mid$

Mean deviation for ungrouped frequency distribution

Let the given data consist of $\underline{n}$ distinct values $\underline{x_1}, \underline{x_2}, \ldots, x_n$ occurring with frequencies $\underline{f_1}, \underline{f_2}, \ldots, f_n$ respectively.

$
\begin{array}{lll}
x: x_1 & x_2 & x_3 \ldots x_n \\
f: f_1 & f_2 & f_3 \ldots f_n
\end{array}
$

1. Mean Deviation About Mean

First find the mean, i.e.

$
\bar{x}=\frac{\sum_{i=1}^n x_i f_i}{\sum_{i=1}^n f_i}=\frac{1}{\mathrm{~N}} \sum_{i=1}^n x_i f_i
$

N is the sum of all frequencies
Then, find the deviations of observations $x_i$ from the mean $\bar{x}$ and take their absolute values, i.e., $\left|x_i-\bar{x}\right|$ for all $i=1,2, \ldots, n$
After this, find the mean of the absolute values of the deviations
$\operatorname{M.D.}(\bar{x})=\frac{\sum_{i=1}^n f_i\left|x_i-\bar{x}\right|}{\sum_{i=1}^n f_i}=\frac{1}{N} \sum_{i=1}^n f_i\left|x_i-\bar{x}\right|$

2. Mean Deviation About any value 'a'

$
\text { M.D.(a) }=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left|x_i-\mathrm{a}\right|
$

Mean deviation for grouped frequency distribution

The formula for mean deviation is the same as in the case of ungrouped frequency distribution. Here, $x_i$ is the midpoint of each class.
Note
The mean deviation about the median is the lowest as compared to the mean deviation about any other value.

Standard Deviation

The standard deviation is a number that measures how far data values are from their mean.
The positive square root of the variance is called the standard deviation. The standard deviation is usually denoted by $\sigma$ and it is given by

$
\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2}
$

Variance

The mean of the squares of the deviations from the mean is called the variance and is denoted by $\sigma^2$ (read as sigma square). Variance is a quantity that leads to a proper measure of dispersion.

The variance of $n$ observations $x_1, x_2, \ldots, x_n$ is given by

$
\sigma^2=\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2
$

Variance and Standard Deviation of a Ungrouped Frequency Distribution

The given data is

$
\begin{aligned}
& x: x_1, \quad x_2, \quad x_3, \quad \ldots \quad x_n \\
& f: f_1, f_2, f_3, \ldots f_n
\end{aligned}
$

In this case, Variance $\left(\sigma^2\right)=\frac{1}{N} \sum_{i=1}^n f_i\left(x_i-\bar{x}\right)^2$ and, Standard Deviation $(\sigma)=\sqrt{\frac{1}{N} \sum_{i=1}^n f_i\left(x_i-\bar{x}\right)^2}$ where, $\mathrm{N}=\sum_{i=1}^n f_i$

Variance and Standard deviation of a grouped frequency distribution

The formula for variance and standard deviation are the same as in the case of ungrouped frequency distribution. Here, $x_i$ is the mid point of each class.

Another formula for Standard Deviation

Variance
$\begin{aligned}
\begin{aligned}
\left(\sigma^2\right) & =\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left(x_i-\bar{x}\right)^2=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left(x_i^2+\bar{x}^2-2 \bar{x} x_i\right) \\
& =\frac{1}{N}\left[\sum_{i=1}^n f_i x_i^2+\sum_{i=1}^n \bar{x}^2 f_i-\sum_{i=1}^n 2 \bar{x} f_i x_i\right] \\
& =\frac{1}{N}\left[\sum_{i=1}^n f_i x_i^2+\bar{x}^2 \sum_{i=1}^n f_i-2 \bar{x} \sum_{i=1}^n x_i f_i\right] \\
& =\frac{1}{N}\left[\sum_{i=1}^n f_i x_i^2+\bar{x}^2 N-2 \bar{x} \cdot N \bar{x}\right] \\
& =\frac{1}{N}\left[\sum_{i=1}^n f_i x_i^2+\bar{x}^2 N-2 \bar{x} \cdot N \bar{x}\right] \\
{\left[\because \frac{1}{N} \sum_{i=1}^n x_i f_i\right.} & \left.=\bar{x} \text { or } \sum_{i=1}^n x_i f_i=\mathrm{N} \bar{x}\right] \\
& =\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i x_i^2+\bar{x}^2-2 \bar{x}^2=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i x_i^2-\bar{x}^2
\end{aligned}
\end{aligned}$

Standard Deviation

$\sigma = \sqrt{\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i x_i^2-\bar{x}^2}$

Recommended Video Based on Measures of Dispersion

Solved Example Based On Measures of Dispersion

Example 1: What is the range of the data $3,8,6,5,2,1,9,3,2$ ?
1) $9$
2) $10$
3) $8$
4) $5$

Solution
Range - The range is the difference between the smallest and largest observations. It is the simplest measure of Dispersion
Range $=9-1=8$

Hence,the answer is an option 3.

Example 2:The mean of $5$ observations is $5$ and their variance is $124$ . If three of the observations are $1,2$ and $6$ ; then the mean deviation from the mean of the data is :
1) $2.4$
2) $2.8$
3) $2.5$
4) $2.6$

Solution
Initially, we need to look at the following concepts:
Arithmetic Mean -
For the values $x_1, x_2, \ldots . x_n$ of the variant $x$ the arithmetic mean is given by

$
\bar{x}=\frac{x_1+x_2+x_3+\cdots+x_n}{n}
$

Mean Deviation -
If $x_1, x_2, \ldots x_n$ are $n$ observations then the mean deviation from the point $A$ is given by :

$
\frac{1}{n} \sum\left|x_i-A\right|
$

Variance -

In case of discrete data

$
\sigma^2=\left(\frac{\sum x_i^2}{n}\right)-\left(\frac{\sum x_i}{n}\right)^2
$
Now,

$
\begin{aligned}
& \frac{\sum x_i}{5}=5 \Rightarrow \sum x_i=25 \\
& \frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2=124 \\
& \frac{\sum x_i^2}{5}-25=124 \\
& \sum x_i^2=149 \times 5=745
\end{aligned}
$

Let the two observations be $\mathrm{a} \& \mathrm{~b}$

$
\begin{aligned}
& a+b+1+2+6=25 \\
& a+b=16 \\
& a^2+b^2+1^2+2^2+6^2=745 \\
& a^2+b^2+1+4+36=745 \\
& a^2+b^2=704
\end{aligned}
$

$
\begin{aligned}
& \text { Mean deviation }=\frac{\sum\left|x_i-5\right|}{5}=\frac{\left|x_1-5\right|+\left|x_2-5\right|+8}{5} \\
& =\frac{8+\left|x_1-5\right|+\left|11-x_1\right|}{5}=\frac{8+6}{5}=2.8
\end{aligned}
$
Hence, the answer is the option 2.

Example 3: If the mean deviation of the numbers $1,1+d, \ldots, 1+100 d$ from their mean is $255$ , then a value of $d$ is :
1) $10.1$
2) $20.2$
3) $10$
4) $5.05$

Solution
Mean Deviation -If $x_1, x_2, \ldots x_n$ are $n$ observations then the mean deviation from point $A$ is given by :

$
\frac{1}{n} \sum\left|x_i-A\right|
$

$
\text { Mean }=\frac{1+1+d+1+2 d+---\cdots----1+100 d}{101}=1+50 d
$

Mean deviation
$
\begin{aligned}
& \Rightarrow \frac{1}{101} \sum_{r=0}^{100}|(I+r d)-(I+50 d)| \\
& \Rightarrow \frac{1}{101} \times 2 d \times \frac{50 \times 51}{2}=255 \\
& d=10.1
\end{aligned}
$
Hence, the answer is the option 1.

Example 4: The mean deviation of the numbers $3,4,5,6,7$ is
1) $0$
2) $1.2$
3) $5$
4) $25$

Solution

$
\begin{aligned}
&\text { Here, the mean can be calculated as: }\\
&\bar{x}=\frac{3+4+5+6+7}{5}=5
\end{aligned}
$

x $|x-\bar{x}|$

3 2

4 1

5 0

6 1

$
\sum|x-\bar{x}|=6
$

Mean deviation from the mean

$
\begin{aligned}
& =\frac{6}{5} \\
& =1.2
\end{aligned}
$
Hence, the answer is the option (2).

Example 5: Let $\bar{X}$ and $M.D.$ be the mean and the mean deviation about $\bar{X}$ of n observations $x_i, i=1,2, \ldots \ldots \ldots \ldots, n$. If each of the observations is increased by $5$ , then the new mean and the mean deviation about the new mean, respectively, are:
1) $\bar{X}, M \cdot D$.
2) $\bar{X}+5, M \cdot D$.
3) $\bar{X}, M \cdot D \cdot+5$
4) $\bar{X}+5$, M. D. +5

Solution
Observation all increased by $5$

$
\text { New mean }=\frac{\text { new sum }}{n}=\frac{\left(x_1+5\right)+\left(x_2+5\right)+\ldots \ldots+\left(x_n+5\right)}{n}
$

$
\begin{aligned}
& =\frac{x_1+x_2+\ldots \ldots+x_n}{n}+\frac{5 n}{n} \\
& =\bar{X}+5
\end{aligned}
$

New mean deviation about the new mean:

$
\begin{aligned}
& =\frac{1}{n} \sum_{i=1}^n\left|\left(x_i+5\right)-(\bar{X}+5)\right| \\
& =\frac{1}{n} \sum_{i=1}^n\left|x_i-\bar{X}\right|
\end{aligned}
$

$=$ old mean deviation
So, the mean will be increased by 5 but there will be no change in M.D.
Hence, the answer is the option (2).