Standard Deviation

The technique of the calculation of mean deviation is mathematically illogical as in its calculation the algebraic signs are ignored. This drawback is removed in the calculation of standard deviation. One of the easiest ways of doing away with algebraic signs is to square the figures and this process is adopted in the calculation of standard deviation. In the calculation of SD, first the AM is calculated and the deviations of. various items from the AM are squared. The squared deviations are summed up and the sum is divided by the number of items, The positive square root of the number will give SD. That is, SD is the positive square root of the mean of squared deviations from mean.

The concept of standard deviation was first used by Karl Pearson in the year 1893. It is the most commonly used measure of dispersion. It satisfies most of the properties laid down for an ideal measure of dispersion. Note that SD is calculated from AM only. Just as mean is the best measure of central tendency, standard deviation is the best measure of dispersion. Standard deviation is calculated on the basis of mean only.

" Standard deviation is defined as the square root of the arithmetic average of the squares of deviations taken from the arithmetic average of a series. "

It is also known as the root-mean-square deviation for the reason that it is the square root of the mean of the squared deviations from AM.

Standard deviation is denoted by the Greek letter σ (small letter ‘sigma’).

The term variance is used to describe the square of the standard deviation. The term was first used by R. A. Fisher in 1913.

Standard deviation is an absolute measure of dispersion. The corresponding relative measure is called coefficient of SD. Coefficient of variation is also a relative measure. A series with more coefficient of variation is regarded as less consistent or less stable than a series with less coefficient of variation.

Symbolically,

Individual Series

Different methods are used to calculate standard deviation of individual series. All these methods result in the same value of standard deviation. These are given below:

Actual Mean Method

Let us find standard deviation for the following data by actual mean method.

Height: 160, 160, 161, 162, 163, 163, 163, 164, 164, 170.

We need to find d and d², it is shown in the below given table.

$$ {\overline{X}}\,= \, {{{\frac{ΣX}{N}} }} $$ $$ = \, {{{\frac{1630}{10}} }} $$ $$ = \, 163 $$

Table 6.22
x	d = (X - x̅)=(X - 163)	d2
160	-3	9
160	-3	9
161	-2	4
162	-1	1
163	0	0
163	0	0
163	0	0
164	1	1
164	1	1
170	7	49
ΣX = 1630, N = 10		Σd2 = 74

$$ \mathbf{σ\, = \,\sqrt{\frac {Σd^{2}}{N}}} $$ $$ \mathbf{ = \,\sqrt{\frac {74}{10}}} $$ $$ \mathbf{ = \,\sqrt{7.4}} $$ $$ \mathbf{ = \,2.72} $$

Assumed Mean Method

Let us find standard deviation for the following data by assumed mean method.

Height: 160, 160, 161, 162, 163, 163, 163, 164, 164, 170.

We need to find d and d², it is shown in the below given table.

$$ Assumed \,Mean \,=\, 162 $$

Table 6.23
x	d = (X - 162)	d2
160	-2	4
160	-2	4
161	-1	1
162	0	0
163	1	1
163	1	1
163	1	1
164	2	2
164	2	2
170	8	64
	Σd = 10	Σd2 = 84

$$ \mathbf{σ\, = \,\sqrt{\frac{Σd^{2}}{N}\,-\,{\Bigl(\frac{Σd}{N}\Bigr)}^{2}} } $$ $$ \mathbf{σ\, = \,\sqrt{\frac{84}{10}\,-\,{\Bigl(\frac{10}{10}\Bigr)}^{2}} } $$ $$ \mathbf{ = \,\sqrt{8.4\,-\,1}} $$ $$ \mathbf{ = \,\sqrt{7.4}} $$ $$ \mathbf{ = \,2.72} $$

Direct Method

Let us find standard deviation for the following data by direct method.

Height: 160, 160, 161, 162, 163, 163, 163, 164, 164, 170.

We need to find x², it is shown in the below given table.

Table 6.24
x	x2
160	25600
160	25600
161	25921
162	26244
163	26569
163	26569
163	26569
164	26896
164	26896
170	28900
Σx = 1630	Σx2 = 265764

$$ \mathbf{σ\, = \,\sqrt{\frac{Σx^{2}}{N}\,-\,{\Bigl(\frac{Σx}{N}\Bigr)}^{2}} } $$ $$ \mathbf{σ\, = \,\sqrt{\frac{265764}{10}\,-\,{\Bigl(\frac{1630}{10}\Bigr)}^{2}} } $$ $$ \mathbf{ = \,\sqrt{26576.4\,-\,26569}} $$ $$ \mathbf{ = \,\sqrt{7.4}} $$ $$ \mathbf{ = \,2.72} $$

Step Deviation Method

Let us find standard deviation for the following data by step deviation method.

5, 10, 25, 30, 50.

We need to find d, d', and d'². Deviations taken from 25 and common factor 5, it is shown in the below given table.

Table 6.25
x	d = (X - 25)	\( \mathbf{d'\, = \, {{{\frac{(x-25)}{5}} }}} \)	d'2
5	-20	-4	16
10	-15	-3	9
25	0	0	0
30	5	1	1
50	25	5	25
		Σd' = -1	Σd'2 = 51

$$ \mathbf{σ\, = \,\sqrt{\frac{Σd'^{2}}{N}\,-\,{\Bigl(\frac{Σd'}{N}\Bigr)}^{2}} \,×\,c } $$ $$ \mathbf{ = \,\sqrt{\frac{51}{5}\,-\,{\Bigl(\frac{-1}{5}\Bigr)}^{2}} \,×\,5 } $$ $$ \mathbf{ = \,\sqrt{10.2\,-\,(.04)}\,×\,5} $$ $$ \mathbf{ = \,\sqrt{10.16}\,×\,5} $$ $$ \mathbf{ = \,3.187\,×5} $$ $$ \mathbf{ = \,15.936} $$

Discrete Series

Standard deviation can be calculated in four ways:

Actual Mean Method

Let us find standard deviation for the following data by actual mean method.

Table 6.26
x	f
6	3
7	6
8	9
9	13
10	8
11	5
12	4

We need to find fx, x, x² and fx², this is shown in the below given table.

$$ {\overline{X}}\,= \, {{{\frac{ΣfX}{Σf}} }} $$ $$ = \, {{{\frac{432}{48}} }} $$ $$ = \, 9 $$

$$ \mathbf{σ\, = \,\sqrt{\frac {Σfx^{2}}{Σf}}} $$ $$ \mathbf{ = \,\sqrt{\frac {124}{48}}} $$ $$ \mathbf{ = \,\sqrt{2.58}} $$ $$ \mathbf{ = \,1.6} $$

Assumed Mean Method

Let us find standard deviation for the following data by assumed mean method.

Table 6.28
x	f
6	3
7	6
8	9
9	13
10	8
11	5
12	4

We need to find d, d², x², fd and fd², this is shown in the below given table.

$$ \mathbf{σ\, = \,\sqrt{\frac{Σfd^{2}}{Σf}\,-\,{\Bigl(\frac{Σfd}{Σf}\Bigr)}^{2}} } $$ $$ \mathbf{ = \,\sqrt{\frac{172}{48}\,-\,{\Bigl(\frac{-48}{48}\Bigr)}^{2}} } $$ $$ \mathbf{ = \,\sqrt{3.58\,-\,1} } $$ $$ \mathbf{ = \,\sqrt{2.58} } $$ $$ \mathbf{ = \,1.6 } $$

Direct Method

Let us find standard deviation for the following data by direct method.

Table 6.30
x	f
6	3
7	6
8	9
9	13
10	8
11	5
12	4

We need to find fx, x² and fx², this is shown in the below given table.

$$ \mathbf{σ\, = \,\sqrt{\frac{Σfx^{2}}{Σf}\,-\,{\Bigl(\frac{Σfx}{Σf}\Bigr)}^{2}} } $$ $$ \mathbf{ = \,\sqrt{\frac{4012}{48}\,-\,{\Bigl(\frac{432}{48}\Bigr)}^{2}} } $$ $$ \mathbf{ = \,\sqrt{83.58\,-\,81} } $$ $$ \mathbf{ = \,\sqrt{2.58} } $$ $$ \mathbf{ = \,1.61 } $$

Step Deviation Method

Let us find standard deviation for the following data by step deviation method.

Table 6.32
x	f
10	2
15	8
20	10
25	15
30	3
35	2

We need to find d, d', fd', d'² and fd'², this is shown in the below given table.

$$ \mathbf{σ\, = \,\sqrt{\frac{Σfd'^{2}}{Σf}\,-\,{\Bigl(\frac{Σfd'}{Σf}\Bigr)}^{2}} \,×\,c } $$ $$ \mathbf{ = \,\sqrt{\frac{71}{40}\,-\,{\Bigl(\frac{-25}{40}\Bigr)}^{2}} \,×\,5 } $$ $$ \mathbf{ = \,\sqrt{1.775\,-\,.391} \,×\,5 } $$ $$ \mathbf{ = \,\sqrt{1.384} \,×\,5 } $$ $$ \mathbf{ = \,1.1764 \,×\,5 } $$ $$ \mathbf{ = \,5.88} $$

Continuous Series

In continuous series we have class intervals for the variable. So we have to find out the mid-point for the various classes. Then the problem becomes similar to those of discrete series.

Standard deviation can be calculated in four ways:

Actual Mean Method

Let us find standard deviation for the following data by actual mean method.

We need to find m, fm, x (m - X, fx, x² and fx², this is shown in the below given table.

$$ {\overline{X}}\,= \, {{{\frac{Σfm}{Σf}} }} $$ $$ = \, {{{\frac{3650}{50}} }} $$ $$ = \, 73 $$

$$ \mathbf{σ\, = \,\sqrt{\frac {Σfx^{2}}{Σf}}} $$ $$ \mathbf{ = \,\sqrt{\frac {7600}{50}}} $$ $$ \mathbf{ = \,\sqrt{152}} $$ $$ \mathbf{ = \, {12.33}} $$

Assumed Mean Method

Let us find standard deviation for the following data by assumed mean method.

We need to find m, d (x - 75), d², fd and fd², this is shown in the below given table.

$$ \mathbf{σ\, = \,\sqrt{\frac{Σfd^{2}}{Σf}\,-\,{\Bigl(\frac{Σfd}{Σf}\Bigr)}^{2}} } $$ $$ \mathbf{= \,\sqrt{\frac{7800}{50}\,-\,{\Bigl(\frac{-100}{50}\Bigr)}^{2}} } $$ $$ \mathbf{= \,\sqrt{156\,-\,(-4)} } $$ $$ \mathbf{= \,\sqrt{156\,-\,4} } $$ $$ \mathbf{= \,\sqrt{152 }} $$ $$ \mathbf{= \,12.33} $$

Direct Method

Let us find standard deviation for the following data by direct method.

We need to find m, fm, and fm², this is shown in the below given table.

$$ \mathbf{σ\, = \,\sqrt{\frac{Σfm^{2}}{Σf}\,-\,{\Bigl(\frac{Σfm}{Σf}\Bigr)}^{2}} } $$ $$ \mathbf{ = \,\sqrt{\frac{274050}{50}\,-\,{\Bigl(\frac{3650}{50}\Bigr)}^{2}} } $$ $$ \mathbf{ = \,\sqrt{5481\,-\,5329} } $$ $$ \mathbf{ = \,\sqrt{152} } $$ $$ \mathbf{ = \,12.33 } $$

Step Deviation Method

Let us find standard deviation for the following data by step deviation method.

We need to find m, d', fd', d'² and fd'², this is shown in the below given table.

$$ \mathbf{σ\, = \,\sqrt{\frac{Σfd'^{2}}{Σf}\,-\,{\Bigl(\frac{Σfd'}{Σf}\Bigr)}^{2}} \,×\,c } $$ $$ \mathbf{ = \,\sqrt{\frac{78}{50}\,-\,{\Bigl(\frac{-10}{50}\Bigr)}^{2}} \,×\,10 } $$ $$ \mathbf{ = \,\sqrt{1.56 \,-\, .04} \,×\,10 } $$ $$ \mathbf{ = \,\sqrt{1.52} \,×\,10 } $$ $$ \mathbf{ = \,1.233 \,×\,10 } $$ $$ \mathbf{ = \,12.33 } $$

Properties of SD

Absolute and Relative Measures of Dispersion

Absolute measures of dispersion are expressed in the same statistical unit in which the original data are given such as rupees, tonnes, centimeters, etc. In case two sets of data are expressed in different units, absolute measures of dispersion are not comparable. In such cases, measures of relative dispersion should be used.

A measure of relative dispersion is the ratio of measure of absolute dispersion to an appropriate average. It is sometimes called a coefficient of dispersion because coefficient means a pure number that is independent of the unit of measurement. Greater the value of coefficient of dispersion more is the variability in a distribution (less consistency).