Measure Of Dispersion Of Grouped And Ungrouped Data

a) For ungrouped data

The range is the difference between the highest and the lowest values.

Example 6.26

Find the range of:

a) 65, 66, 67, 68, 71, 73, 74, 77

$\text{Range} = 77 - 65 = 12$

b) 42, 54, 58, 62, 67, 77, 85, 93, 100

$\text{Range} = 100 - 42 = 58$

b) For grouped data

Range is the difference between the upper class boundary of the highest interval and the lower class boundary of the lowest class interval.

Example 6.27

Class intervals	Frequencies
21-25	4
26-30	9
31-35	16
36-40	10
41-45	10
46-50	1

$\text{Range} = 50.5 - 20.5 = 30$

Variance is the sum of squared deviations divided by the number of values or the average of squares of deviations. Deviation is the distance from the mean ($\bar{x}$), indicating how far a value $x$ is from the mean. It determines the consistency of a variable and the spread of data.

a) For ungrouped data

Let $x_i$ represent a variable with values $x_1, x_2, x_3, \ldots, x_n$ whose mean is $\bar{x}$.

$\text{Variance} = \frac{\text{Sum of squares of deviations}}{\text{Number of observations}}$

$\mathrm{Var}(x) = \frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{N}$

or

$\mathrm{Var}(x) = \frac{\sum_{i=1}^{n} x_i^2}{N} - \left(\frac{\sum_{i=1}^{n} x_i}{N}\right)^2$

where $N$ is the total number of observations.

Other formulae for calculating variance are:

i) $\mathrm{Var}(x) = \frac{1}{N}\sum_{i=1}^{n} d_i^2 - \left(\frac{1}{N}\sum_{i=1}^{n} d_i\right)^2$ where $d_i = x_i - A$ (assumed mean method).

ii) $\mathrm{Var}(x) = c^2\left[\frac{1}{N}\sum_{i=1}^{n} u_i^2 - \left(\frac{1}{N}\sum_{i=1}^{n} u_i\right)^2\right]$ where $u_i = d_i/c = (x_i - A)/c$ (coding method).

b) For grouped data

Let $x_i$ represent class marks for grouped data with values $x_1, x_2, x_3, \ldots, x_n$ whose mean is $\bar{x}$ and $f_i$ are the frequencies.

The formulae for computing variance are:

i) $\mathrm{Var}(x) = \frac{\sum_{i=1}^{n} f_i(x_i - \bar{x})^2}{\sum_{i=1}^{n} f_i}$

or

$\mathrm{Var}(x) = \frac{\sum_{i=1}^{n} f_i x_i^2}{\sum_{i=1}^{n} f_i} - \left(\frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i}\right)^2$

where $\sum_{i=1}^{n} f_i = N$

Remark: The formula

$\mathrm{Var}(x) = \frac{\sum_{i=1}^{n} f_i(x_i - \bar{x})^2}{\sum_{i=1}^{n} f_i}$

can also be used for ungrouped data with values $x_1, x_2, x_3, \ldots, x_n$ which occur with frequencies $f_1, f_2, f_3, \ldots f_n$, respectively.

ii) $\mathrm{Var}(x) = \frac{1}{N}\sum_{i=1}^{n} f_i d_i^2 - \left(\frac{1}{N}\sum_{i=1}^{n} f_i d_i\right)^2$ where $d_i = x_i - A$ (assumed mean method).

iii) $\mathrm{Var}(x) = c^2\left[\frac{1}{N}\sum_{i=1}^{n} f_i u_i^2 - \left(\frac{1}{N}\sum_{i=1}^{n} f_i u_i\right)^2\right]$ where $u_i = d_i/c = (x_i - A)/c$ (coding method).

Example 6.27.

Given the data set: 32, 41, 47, 53, 57. Find the variance.

$\bar{x} = \frac{\sum x}{N} = \frac{32 + 41 + 47 + 53 + 57}{5} = \frac{230}{5} = 46$

$x$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
32	-14	196
41	-5	25
47	1	1
53	7	49
57	11	121

$\mathrm{Var}(x) = \frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{N}$

$\mathrm{Var}(x) = \frac{196 + 25 + 1 + 49 + 121}{5} = \frac{392}{5} = 78.4$

$\therefore \mathrm{Var}(x) = 78.4$

Standard deviation is a useful measure of dispersion based on the mean. It portrays how values are spread around the mean. It is the average deviation between all observed values and the sample mean and is calculated as the square root of the variance, denoted by $\sigma$.

$\text{Standard deviation} = \sqrt{\text{Variance}}$

For a variable $x$, $\sigma_x = \sqrt{\mathrm{Var}(x)}$

a) For ungrouped data

$\sigma_x = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{N}}$

or

$\sigma_x = \sqrt{\mathrm{Var}(x)} = \sqrt{\frac{\sum_{i=1}^{n} x_i^2}{N} - \left(\frac{\sum_{i=1}^{n} x_i}{N}\right)^2}$

where $N$ represents total observations and $\bar{x}$ represents the mean.

Other formulae for finding standard deviation are:

i) $\sigma_x = \sqrt{\frac{1}{N}\sum_{i=1}^{n} d_i^2 - \left(\frac{1}{N}\sum_{i=1}^{n} d_i\right)^2}$ where $d_i = x_i - A$ (assumed mean method).

ii) $\sigma_x = c\sqrt{\frac{1}{N}\sum_{i=1}^{n} u_i^2 - \left(\frac{1}{N}\sum_{i=1}^{n} u_i\right)^2}$ where $u_i = d_i/c = (x_i - A)/c$ (coding method).

b) For grouped data

Let $x_i$ represent class marks with values $x_1, x_2, x_3, \ldots, x_n$ whose mean is $\bar{x}$ and $f_i$ are the frequencies.

Formulae for computing standard deviation are:

i) $\sigma_x = \sqrt{\mathrm{Var}(x)} = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2 f_i}{N}} = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2 f_i}{\sum_{i=1}^{n} f_i}}$

Remark: The formula

$\sigma_x = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2 f_i}{\sum_{i=1}^{n} f_i}}$

can also be used for ungrouped data values $x_1, x_2, x_3, \ldots, x_n$ which occur with frequencies $f_1, f_2, f_3, \ldots f_n$ respectively.

ii) $\sigma_x = \sqrt{\frac{1}{N}\sum_{i=1}^{n} f_i d_i^2 - \left(\frac{1}{N}\sum_{i=1}^{n} f_i d_i\right)^2}$ where $d_i = x_i - A$ (assumed mean method).

iii) $\sigma_x = c\sqrt{\frac{1}{N}\sum_{i=1}^{n} f_i u_i^2 - \left(\frac{1}{N}\sum_{i=1}^{n} f_i u_i\right)^2}$ where $u_i = d_i/c = (x_i - A)/c$ (coding method).

Example 6.28

Class intervals	Frequencies
60 – 62	5
63 – 65	18
66 – 68	42
69 – 71	27
72 – 74	8

Calculate:

a) the Variance

b) the Standard deviation

Solution

a) Using:

$\bar{x} = \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i} = \frac{6745}{100} = 67.45$

Class Intervals	$f_i$	$x_i$	$f_i x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$	$f_i(x_i - \bar{x})^2$
60 – 62	5	61	305	-6.45	41.6025	208.0125
63 – 65	18	64	1152	-3.45	11.9025	214.245
66 – 68	42	67	2814	-0.45	0.2025	8.505
69 – 71	27	70	1890	2.55	6.5025	175.5675
72 – 74	8	73	584	5.55	30.8025	246.42
$\sum f_i = 100$				$\bar{x} = 67.45$		$\sum f_i(x_i - \bar{x})^2 = 852.75$

$\mathrm{Var}(x) = \frac{\sum(x_i - \bar{x})^2 f_i}{\sum f_i} = \frac{852.75}{100} = 8.5275$

$\therefore \mathrm{Var}(x) = 8.5275$

b) Using

$\sigma_x = \sqrt{\frac{\sum(x_i - \bar{x})^2 f_i}{\sum f_i}}$

$\sigma_x = \sqrt{\mathrm{Var}(x)}$

$\sigma_x = \sqrt{8.5275}$

$\sigma_x = 2.9202$

Standard deviation is 2.9202

Alternatively: By using assumed mean method and coding method: Let assumed mean, $A = 67$

Class intervals	$f_i$	$x_i$	$d_i = x_i - A$	$f_i d_i$	$f_i d_i^2$	$u_i = (x_i - A)/c$	$f_i u_i$	$f_i u_i^2$
60 – 62	5	61	-6	-30	180	-2	-10	20
63 – 65	18	64	-3	-54	162	-1	-18	18
66 – 68	42	67	0	0	0	0	0	0
69 – 71	27	70	3	81	243	1	27	27
72 – 74	8	73	6	48	288	2	16	32
$\sum f_i = 100$				$\sum f_i d_i = 45$		$\sum f_i d_i^2 = 873$

$\sum f_i u_i = 15$

$\sum f_i u_i^2 = 97$

a)

i) $\mathrm{Var}(x) = \frac{1}{N}\sum f_i d_i^2 - \left(\frac{1}{N}\sum f_i d_i\right)^2$

$= \frac{1}{100}(873) - \left(\frac{45}{100}\right)^2$

$= 8.73 - 0.2025$

$= 8.5275$

ii) $\mathrm{Var}(x) = c^2\left[\frac{1}{N}\sum f_i u_i^2 - \left(\frac{1}{N}\sum f_i u_i\right)^2\right]$

$= 3^2\left[\frac{97}{100} - \left(\frac{15}{100}\right)^2\right]$

$= 9[0.97 - 0.0225]$

$= 9[0.9475]$

$= 8.5275$

b)

i) $\sigma_x = \sqrt{\frac{1}{N}\sum f_i d_i^2 - \left(\frac{1}{N}\sum f_i d_i\right)^2}$

$= \sqrt{8.73 - \left(\frac{45}{100}\right)^2}$

$= \sqrt{8.5275}$

$= 2.9202$

ii) $\sigma_x = c\sqrt{\frac{1}{N}\sum f_i u_i^2 - \left(\frac{1}{N}\sum f_i u_i\right)^2}$

$= 3\sqrt{\frac{97}{100} - \left(\frac{15}{100}\right)^2}$

$= 3\sqrt{0.9475}$

$= 3 \times 0.9734$

$= 2.9202$