## 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,

$$\mathbf{Standard \,Deviation\, = \,σ}$$ $$\mathbf{Variance \,=\, σ^{2}}$$ $$\mathbf{Coefficient\,of\,SD\,=\,{{{\frac{σ}{\overline{X}}} }}}$$ $$\mathbf{Coefficient\,of\,variation\,=\,{{{\frac{σ}{\overline{X}}} ×\,100 }}}$$

### 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:

1. Actual Mean Method

$$\mathbf{σ\, = \,\sqrt{\frac {Σd^{2}}{N}}}$$, where d = X - x̄

2. Assumed Mean Method

$$\mathbf{σ\, = \,\sqrt{\frac{Σd^{2}}{N}\,-\,{\Bigl(\frac{Σd}{N}\Bigr)}^{2}} }$$

3. Direct Method

$$\mathbf{σ\, = \,\sqrt{\frac{Σx^{2}}{N}\,-\,{{\overline{X}}}^{2}} }$$ or $$\mathbf{σ\, = \,\sqrt{\frac{Σx^{2}}{N}\,-\,{\Bigl(\frac{Σx}{N}\Bigr)}^{2}} }$$

4. Step Deviation Method

$$\mathbf{σ\, = \,\sqrt{\frac{Σd'^{2}}{N}\,-\,{\Bigl(\frac{Σd'}{N}\Bigr)}^{2}} \,×\,c }$$

### 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:

1. Actual Mean Method

$$\mathbf{σ\, = \,\sqrt{\frac {Σfx^{2}}{Σf}}}$$ or $$\mathbf{σ\, = \,\sqrt{\frac {Σfd^{2}}{Σf}}}$$,

where d = X - X

2. Assumed Mean Method

$$\mathbf{σ\, = \,\sqrt{\frac{Σfd^{2}}{Σf}\,-\,{\Bigl(\frac{Σfd}{Σf}\Bigr)}^{2}} }$$

where d = X - A

3. Direct Method

$$\mathbf{σ\, = \,\sqrt{\frac{Σfx^{2}}{Σf}\,-\,{\Bigl(\frac{Σfx}{Σf}\Bigr)}^{2}} }$$

4. Step Deviation Method

$$\mathbf{σ\, = \,\sqrt{\frac{Σfd'^{2}}{Σf}\,-\,{\Bigl(\frac{Σfd'}{Σf}\Bigr)}^{2}} \,×\,c }$$

where d = X - A

$$d'\, = \, {{{\frac{(x-A)}{C}} }}$$

### 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.

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

Table 6.27
x f fx (X - X)(X - 9) X x2 fx2
6 3 18 -3 9 27
7 6 42 -2 4 24
8 9 72 -1 1 9
9 13 117 0 0 0
10 8 80 1 1 8
11 5 55 2 4 20
12 4 48 3 9 36
Σf = 48 Σfx = 432 Σfx2 = 124

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

### 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.

Table 6.29
x f (d = X - A)(A = 10) d2 fd fd2
6 3 -4 16 -12 48
7 6 -3 9 -18 54
8 9 -2 4 -18 36
9 13 -1 1 -13 13
10 8 0 0 0 0
11 5 1 1 5 5
12 4 2 4 8 16
Σf = 48 Σfd = -48 Σfd2 = 172

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

### 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.

Table 6.31
x f fx x2 fx2
6 3 18 36 108
7 6 42 49 294
8 9 72 64 576
9 13 117 81 1053
10 8 80 100 800
11 5 55 121 605
12 4 48 144 576
Σf = 48 Σfx = 432 Σfx2 = 4012

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

### 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.

Table 6.33
x f d = X - A (A = 25) $$\mathbf{d'\, = \, {{{\frac{(X-A)}{C}} }}}$$ C = 5 fd' d'2 fd'2
10 2 -15 -3 -6 9 18
15 8 -10 -2 -16 4 32
20 10 10 -1 -10 1 10
25 15 0 0 0 0 0
30 3 5 1 3 1 3
35 2 10 2 4 4 8
Σf = 40 Σfd' = -25 Σf'd2 = 71

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

### 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:

1. Actual Mean Method

$$\mathbf{σ\, = \,\sqrt{\frac {Σfx^{2}}{Σf}}}$$

2. Assumed Mean Method

$$\mathbf{σ\, = \,\sqrt{\frac{Σfd^{2}}{Σf}\,-\,{\Bigl(\frac{Σfd}{Σf}\Bigr)}^{2}} }$$

where d = X - A

3. Direct Method

$$\mathbf{σ\, = \,\sqrt{\frac{Σfm^{2}}{Σf}\,-\,{\Bigl(\frac{Σfm}{Σf}\Bigr)}^{2}} }$$

4. Step Deviation Method

Deviation d can be converted into d' by multiplying it with the class interval, C.

$$\mathbf{σ\, = \,\sqrt{\frac{Σfd'^{2}}{Σf}\,-\,{\Bigl(\frac{Σfd'}{Σf}\Bigr)}^{2}} \,×\,c }$$

where d = X - A

$$d'\, = \, {{{\frac{d}{C}} }}$$

### Actual Mean Method

• Let us find standard deviation for the following data by actual mean method.
• Table 6.34
x f
40 - 50 2
50 - 60 5
60 - 70 12
70 - 80 18
80 - 90 8
90 - 100 5

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

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

Table 6.35
x f m fm x (m - X) fx x2 fx2
40 - 50 2 45 90 -28 -56 784 1568
50 - 60 5 55 275 -18 -90 324 1620
60 - 70 12 65 780 -8 -96 64 768
70 - 80 18 75 1350 2 36 4 72
80 - 90 8 85 680 12 96 144 1152
90 - 100 5 95 475 22 110 484 2420
Σf = 50 Σfm = 3650 0 Σfx2 = 7600

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

### Assumed Mean Method

• Let us find standard deviation for the following data by assumed mean method.
• Table 6.36
x f
40 - 50 2
50 - 60 5
60 - 70 12
70 - 80 18
80 - 90 8
90 - 100 5

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

Table 6.37
x f m d (x - 75) d2 fd fd2
40 - 50 2 45 -30 900 -60 1800
50 - 60 5 55 -20 400 -100 2000
60 - 70 12 65 -10 100 -120 1200
70 - 80 18 75 0 0 0 0
80 - 90 8 85 10 100 80 800
90 - 100 5 95 20 400 100 200
Σf = 50 Σfd = -100 Σfd2 = 7800

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

### Direct Method

• Let us find standard deviation for the following data by direct method.
• Table 6.38
x f
40 - 50 2
50 - 60 5
60 - 70 12
70 - 80 18
80 - 90 8
90 - 100 5

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

Table 6.39
x f m fm fm2
40 - 50 2 45 -90 4050
50 - 60 5 55 -275 15125
60 - 70 12 65 -780 50700
70 - 80 18 75 1350 101250
80 - 90 8 85 680 57800
90 - 100 5 95 475 45125
Σf = 50 Σfm = 3650 Σfm2 = 274050

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

### Step Deviation Method

• Let us find standard deviation for the following data by step deviation method.
• Table 6.40
x f
40 - 50 2
50 - 60 5
60 - 70 12
70 - 80 18
80 - 90 8
90 - 100 5

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

Table 6.41
x f m $$\mathbf{d'\, = \, {{{\frac{(m-75)}{10}} }}}$$ fd' d'2 fd'2
40 - 50 2 45 -3 -6 9 18
50 - 60 5 55 -2 -10 4 20
60 - 70 12 65 -1 -12 1 12
70 - 80 18 75 0 0 0 0
80 - 90 8 85 1 8 1 8
90 - 100 5 95 2 10 4 20
Σf = 50 Σfd' = -10 Σfd'2 = 78

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

### Properties of SD

1. SD is calculated from AM because; the sum of the squares of the deviations taken from the AM is least.

2. SD is independent of the change of origin. That is, if a constant A is added or subtracted from each of the items of series, then SD remains unchanged.

3. SD is affected by change of scale. That is, if each item of series is multiplied or divided by a constant, say, c, then the SD is also affected by the same constant c.

### MERITS OF STANDARD DEVIATION

• Rigidly defined.
• Its value is always definite.
• Based on all items.
• It is capable of further algebraic treatment.
• It possesses many mathematical properties.
• It is less affected by sampling fluctuations.

### DEMERITS OF STANDARD DEVIATION

• Calculation is not easy.
• It is not understood by a layman.
• Much affected by extreme values.
• Gives much importance to extreme values than values near the mean (this happens because of taking square of the deviations).

### 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).

Table 6.42
Absolue Measure Relative Measure
$$\mathbf{Range\, = \, L\,-\,S}$$ $$\mathbf{ Coefficient \,of \,Range \,= \,{{\frac{L - S }{L + S}}}}$$
$$\mathbf{ Quartile \,Deviation\, =\, {{{\frac{Q_3 - Q_1}{2}} }}}$$ $$\mathbf{ Coefficient\, of\, Quartile\, Deviation\, =\, {{{\frac{Q_3 - Q_1}{Q_3 + Q_1}} }}}$$
$$\mathbf{Mean\, Deviation\, =\, {{{\frac{Σ|D|}{N}} }} }$$ $$\mathbf{Coefficient\, of\, MD\, =\, {{{\frac{MD}{{Mean\,/\,Median\,/\,Mode}}} }}}$$
$$\mathbf{Standard \, Deviation \, = \,\sqrt{\frac {Σx^{2}}{Σf}}}$$; $$\mathbf{\sqrt{\frac{Σd^{2}}{Σf}\,-\,{\Bigl(\frac{Σd}{Σf}\Bigr)}^{2}} }$$; $$\mathbf{\sqrt{\frac{Σfd^{2}}{Σf}\,-\,{\Bigl(\frac{Σfd}{Σf}\Bigr)}^{2}} }$$ $$\mathbf{Coefficient\, of\, SD\, =\, {{{\frac{σ}{\overline{X}}} }}} × 100$$