Measures of Dispersion - Practice Questions & MCQ

Measures of the Spread 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

The 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 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 $|\mathrm{x}-\mathrm{a}|$
To find the mean deviation about mean or median or any other value M of ungrouped data, the 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 ${ }^{\prime}{ }^{\prime}$, i.e.,

Mean deviation about 'a', $\quad$ M.D. $(a)=\frac{1}{n} \sum_{i=1}^n\left|x_i-a\right|$
Mean deviation about mean, $\quad$ 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 $n$ distinct values $x_1, x_2, \ldots, x_n$ occurring with frequencies $f_1, f_2, \ldots, f_n$ respectively.

$
\begin{array}{ccc}
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.