#### Standard Deviation

Standard Deviation

Standard deviation is the position square root of the arithmetic mean of the squares of the deviations of the given observation from their arithmetic mean.

Amongst all the methods of finding out dispersion, the standard deviation is regarded as the best. It is free from those defects with which the earlier methods (range, quartile deviation and mean deviation) suffer. Its value is based upon each and every item of the series and it also take into account algebraic signs. Standard deviation is also known as 'Root-Mean-Square Deviation' because it is the square root of the arithmetic mean of the square of the deviations. It is denoted by the Greek letter $$\sigma$$ (known as Sigma).

1. While computing mean deviation, algebraic signs are ignored, whereas while calculating standard deviation, algebraic signs are taken into account.
2. The mean deviation can be computed either from mean, median or mode, whereas standard deviation is always computed from arithmetic mean.

Calculation of Standard Deviation

Individual Series

(1) Direct Method or Actual Mean Method:

Let X1, X2, X3, .......................... Xn be the N-variate and $$\overline{X}$$ be the arithmetic mean.

Define: x = X - $$\overline{X}$$ = deviation taken from actual mean.

Then, standard deviation is given by the following formula:

\begin{align*} \sigma = \sqrt {\frac {\sum{x^2}}N}\\ \end{align*}

(2)Short-cut Method of Assumed Mean Method:

Let X1, X2, X3, ..........................Xn be the N-variate values and A be the assumed mean.

Define: d = X - A = deviation taken from assumed mean.

Then, standard deviation is defined by the following formula:

\begin{align*} \sigma &= \sqrt {\frac {\sum{d^2}}N - (\frac {\sum {d}}N)^2}\\ \end{align*}

This method is suitable when the actual mean is in fraction (or decimal).

Discrete Series

Let X1, X2, X3, ........................ Xn be the variable values and f1, f2, f3, ....................... fn is their frequencies.

Let, $$\overline{X}$$ and A be the actual mean and assumed mean respectively. Then the standard deviation is defined by the following formulae:

(1) Direct Method or Actual Mean Method:

If x = X - $$\overline{X}$$, then $$\sigma$$ = $$\sqrt {\frac {\sum {fx^2}}N}$$

(2) Short cut Method or Assumed Mean Method:

If d = X - A, then $$\sigma$$ = $$\sqrt {\frac {\sum{fd^2}}N - (\frac {\sum {fd}}N)^2}$$

(3) Step-Deviation Method:

If d' = $$\frac {X - A}h$$, where h is common factor, then $$\sigma$$ = $$\sqrt {\frac {\sum {fd'^2}}N - (\frac {\sum {fd'}}N)^2}$$× h

Continuous Series

Let X1, X2, X3, ............................. Xn be the mid-values of the classes and f1, f2, f3, ....................... fn be their frequencies.

Let $$\overline{X}$$ and A be the actual mean and assumed mean. Then, the standard deviation is defined by the following formulae:

(1) Direct Method:

If x = X - $$\overline{X}$$, then $$\sigma$$ = $$\sqrt {\frac {\sum {fx^2}}N}$$

(2) Short-cut Method:

If d = X - A, then $$\sigma$$ = $$\sqrt {\frac {\sum {fd^2}}N - ({\frac {\sum {fd}}{N}})^2}$$

(3) Step Deviation Method:

If d' = $$\frac {X - A}h$$, h = class size or common factor, then $$\sigma$$ = $$\sqrt {\frac {\sum {fd'^2}}N - (\frac {\sum {fd'}}N)^2}$$× h

Coefficient of standard deviation and coefficient of variation

Standard deviation is the absolute measure of dispersion based on the standard deviation is known as the coefficient of standard deviation. Thus,

Coefficient of S.D. = $$\frac {S.D.}{Mean}$$

Also, the coefficient of standard deviation multiplied by 100 is known as the coefficient of variation (C.V.). Thus,

Coefficient of Variation (C.V.) = $$\frac {S.D.}{Mean}$$× 100

Merits and Demerits of Standard Deviation

Merits

1. The value of standard deviation is based on each and every item of the data.
2. It is free from those defects with which other methods like range, quartile deviation, mean deviation suffer.
3. It is less affected by fluctuations of sampling.

Demerits

1. As compared to the measure, it is somewhat more difficult to understand and compute.
2. Its value is unduly affected by extreme observations.

• The smaller value of the coefficient of variation indicates the data is consistency or uniform.
• the greater value of the coefficient of variation indicates the data is not consistency or uniform.
• In the comparative study of two data, the data with the smaller value of a coefficient of variation is considered as good or uniform.
• The maximum value of a coefficient of variation is 1 or 100%.

Calculating Standard Deviation

 x d = x - $$\overline{X}$$ d2 12 -22 484 25 -9 81 29 -5 25 37 3 9 41 7 49 45 11 121 49 15 225 $$\sum x$$ = 238 $$\sum {d^2}$$ = 994

Here,

$$\sum x$$ = 238

N = 7

Mean ($$\overline{X})$$ = $$\frac {\sum{x}}N$$ = $$\frac {238}7$$ = 34

\begin{align*} ∴ Standard\;Deviation\;(σ) &= \sqrt {\frac {\sum{d^2}}N}\\ &= \sqrt {\frac {994}7}\\ &= \sqrt {142}\\ &= 11.92_{Ans}\\ \end{align*}

Given Data in ascending order is:

11, 14, 15, 17, 18

Let: Assumed mean (A) = 15

Calculating Standard Deviation

 Number (x) d = X - A d2 11 -4 16 14 -1 1 15 0 0 17 2 4 18 3 9 $$\sum d$$ = 0 $$\sum {d^2}$$ = 30

\begin{align*}{\therefore}\;Standard \: Deviation \:(\sigma)&=\sqrt{\frac{\sum d^2}{N}- \left( \frac{\sum d}{N}\right)^2}\\&= \sqrt {\frac {30}{5}-(\frac {0}{5})^2}\\ &= \sqrt {6}\\ &= 2.45_{Ans}\end{align*}

Calculating Standard Deviation

 X f fx d = X - $$\overline{X}$$ d2 fd2 12 2 24 -2.2 4.84 9.68 13 3 39 -1.2 1.44 4.32 14 6 84 -0.2 0.04 0.24 15 4 60 0.8 0.64 2.56 16 2 32 1.8 3.24 6.56 17 1 17 2.8 7.84 7.84 N = 18 $$\sum {fx}$$ = 256 $$\sum {fd^2}$$ = 31.12

\begin{align*} Mean\;(\overline{X}) &= \frac {\sum {fx}}N\\ &= \frac {256}{18}\\ &= 14.2\\ \end{align*}

\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum {fd^2}}N}\\ &= \sqrt {\frac {31.12}{18}}\\ &= \sqrt {1.72}\\ &= 1.31_{Ans}\\ \end{align*}

\begin{align*} Coefficient\;of\;variation &= \frac {\sigma}{\overline{X}}\;×\;100\%\\ &= \frac {1.31}{14.2}\;×\;100\%\\ &= 9.23\%_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Score (x) Frequency (f) fx d = x - $$\overline {X}$$ d2 fd2 18 4 72 6 36 144 20 2 40 8 64 128 14 18 252 2 4 72 16 9 144 4 16 144 10 25 250 -2 4 100 12 27 324 0 0 0 8 14 112 -4 16 224 6 1 6 -6 36 36 N = 100 $$\sum {fx}$$= 1200 $$\sum {fd^2}$$ = 848

\begin{align*} Mean\;(\overline {X}) &= \frac {\sum {fx}}N\\ &= \frac {1200}{100}\\ &= 12\\ \end{align*}

\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum {fd^2}}N}\\ &= \sqrt {\frac {848}{100}}\\ &= \sqrt {8.48}\\ &= 2.91_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Marks (X) No. of students (f) fx d = x - $$\overline {X}$$ d2 fd2 5 2 10 -12 144 288 10 3 30 -7 49 147 15 5 75 -2 4 20 20 6 120 3 9 54 25 3 75 8 64 192 30 1 30 13 169 169 N = 20 $$\sum {fx}$$ = 340 $$\sum {fd^2}$$ =870

\begin{align*} Mean\;(\overline {X})\; &= \frac {\sum {fx}}N\\ &= \frac {340}{20}\\ &= 17\\ \end{align*}

\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum {fd^2}}N}\\ &= \sqrt {\frac {870}{20}}\\ &= \sqrt {43.5}\\ &= 6.59_{Ans}\\ \end{align*}

Calculating Standard Deviation

 X f fx fx2 12 2 24 288 13 3 39 507 14 6 84 1176 15 4 60 900 16 2 32 512 17 1 17 289 N = 18 $$\sum {fx}$$ = 256 $$\sum {fx^2}$$ = 3672

\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fx^2}}N - (\frac {\sum {fx}}N)^2}\\ &= \sqrt {\frac {3672}{18} - (\frac {256}{18})^2}\\ &= \sqrt {204 - 202.27}\\ &= \sqrt {1.73}\\ &= 1.32_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Height No. of Plants (f) mid-value (m) fm d = m - $$\overline{X}$$ fd2 0-8 6 4 24 -17.4 1816.56 8-16 7 12 84 -9.4 618.52 16-24 10 20 200 -1.4 19.6 24-32 8 28 224 6.6 348.48 32-40 9 36 324 14.6 1918.44 N = 40 $$\sum{fm}$$ = 856 $$\sum{fd^2}$$ = 4721.6

\begin{align*} Mean\;({\overline {X}})\; &= \frac {\sum {fm}}N\\ &= \frac {856}{40}\\ &= 21.4\\ \end{align*}

\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fd^2}}N}\\ &= \sqrt {\frac {4721.6}{40}}\\ &= 10.87_{Ans}\\ \end{align*}

\begin{align*} Coefficient\;of\;Variation\;(C.V.) &= \frac {\sigma}{\overline {X}} \times {100}\%\\ &= \frac {10.87}{21.4} \times {100}\%\\ &= 50.79\%_{Ans}\\ \end{align*}

Let: Assumed Mean (A) = 45

Calculation of the Standard Deviation

 Class Interval (x) Mid-value (m) Frequency (f) d = m - A d2 fd fd2 20-30 25 2 -20 400 -40 800 30-40 35 3 -10 100 -30 300 40-50 45 4 0 0 0 0 50-60 55 5 10 100 50 500 60-70 65 4 20 400 80 1600 70-80 75 2 30 900 60 1800 N = 20 $$\sum{fd}$$ = 120 $$\sum{fd^2}$$ = 5000

\begin{align*} Mean\;(\overline{X}) &= A + \frac {\sum {fd}}N\\ &= 45 + \frac {120}{20}\\ &= 45 + 6\\ &= 51\\ \end{align*}

\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fx^2}}N - (\frac {\sum {fx}}N)^2}\\ &= \sqrt {\frac {5000}{20} - (\frac {120}{20})^2}\\ &= \sqrt {250 - 36}\\ &= \sqrt {214}\\ &= 14.63_{Ans}\\ \end{align*}

\begin{align*} Coefficient\;of\;Variation\;(C.V.) &= \frac {\sigma}{\overline {X}} \times {100}\%\\ &= \frac {14.63}{51} \times {100}\%\\ &= 29\%_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Marks obtained Mid-value No. of students (f) fx d= X - $$\overline{X}$$ fd d2 fd2 10-20 15 4 60 -17.5 -70 306.25 1225 20-30 25 6 150 -7.5 -45 56.25 337.5 30-40 35 10 350 2.5 25 6.25 62.5 40-50 45 3 135 12.5 37.5 156.25 468.75 50-60 55 2 110 22.5 45 506.25 1012.5 N = 25 $$\sum{fx}$$ = 805 $$\sum{fd}$$ = 7.5 $$\sum{fd^2}$$ = 3106.25

\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fx}}{N}\\ &= \frac {805}{25}\\ &= 32.5\\ \end{align*}

\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fx^2}}N - (\frac {\sum {fx}}N)^2}\\ &= \sqrt {\frac {3106.25}{25} - (\frac {-7.5}{25})^2}\\ &= \sqrt {124.25 - 0.09}\\ &= \sqrt {124.16}\\ &= 11.14_{Ans}\\ \end{align*}

Let: Assumed Mean (A) = 35

Calculating Standard Deviation

 Daily Sales Mid-value(m) Frequency (f) d = $$\frac {x - A}{10}$$ d2 fd fd2 10-20 15 4 -2 4 -8 16 20-30 25 10 -1 1 -10 10 30-40 35 12 0 0 0 0 40-50 45 8 1 1 8 8 50-60 55 6 2 4 12 24 N = 40 $$\sum {fd}$$ = 2 $$\sum{fd^2}$$ = 58

\begin{align*}\therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fx^2}}N - (\frac {\sum {fx}}N)^2}\;\times i\\ &= \sqrt {\frac {58}{40} - (\frac {2}{40})^2}\;\times {10}\\ &= \sqrt {1.45 - 0.0025}\;\times {10}\\ &= \sqrt {1.4475}\;\times {10}\\ &= 1.203\;\times{10}\\&= 12.03_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Class Interval Mid-value (m) Frequency (f) fm fm2 0-4 2 7 14 28 4-8 6 7 42 252 8-12 10 10 100 1000 12-16 14 15 210 2940 16-20 18 7 126 2268 20-24 22 6 132 2904 N = 52 $$\sum{fm}$$ = 624 $$\sum{fm^2}$$ = 9392

\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fm^2}}N - (\frac {\sum {fm}}N)^2}\\ &= \sqrt {\frac {9392}{52} - (\frac {624}{52})^2}\\ &= \sqrt {180.61 - 144}\\ &= \sqrt {36.61}\\ &= 6.05_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Marks secured Mid-value (m) No. of Students (f) fm d = m - $$\overline{X}$$ d2 fd2 10-20 15 8 120 -18.6 345.96 2767.68 20-30 25 12 300 8.6 73.96 887.52 30-40 35 15 525 1.4 1.96 29.4 40-50 45 9 405 11.4 129.96 1169.64 50-60 55 6 330 21.4 457.96 2747.76 N = 50 $$\sum {fm}$$ = 1680 $$\sum {fd^2}$$ = 7602

\begin{align*} Mean\;(\overline {X}) &= \frac {\sum {fm}}N\\ &= \frac {1680}{50}\\ &= 33.6\\ \end{align*}

\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum {fd^2}}N}\\ &= \sqrt {\frac {7602}{50}}\\ &= \sqrt {152.04}\\ &= 12.33_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Marks Secured Mid-value (m) No. of students (f) fm d= m - $$\overline {X}$$ d2 fd2 0-20 10 2 20 -43 1849 3698 20-40 30 8 240 -23 529 4232 40-60 50 16 800 -3 9 144 60-80 70 10 700 17 289 2890 80-100 90 4 360 37 1369 5476 N = 40 $$\sum {fm}$$ = 2120 $$\sum{fd^2}$$ = 16440

\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fm}}N\\ &= \frac {2120}{40}\\ &= 53\\ \end{align*}

\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fd^2}}N}\\ &= \sqrt {\frac {16440}{40}}\\ &= \sqrt {411}\\ &= 20.27_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Weight in kg. No. of boys (f) Mid-value (m) fm d = m - $$\overline{X}$$ d2 fd2 10-20 2 15 30 -22 484 968 20-30 5 25 125 -12 144 720 30-40 6 35 210 -2 4 24 40-50 3 45 135 8 64 192 50-60 2 55 110 18 324 648 60-70 2 65 130 28 784 1568 N = 20 $$\sum{fm}$$ = 740 $$\sum{fd^2}$$ = 4120

\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fm}}N\\ &= \frac {740}{20}\\ &= 37\\ \end{align*}

\begin{align*} Standard\;Deviation\;(\sigma) &= \sqrt {\sum{fd^2}N}\\ &= \sqrt {\frac {4120}{20}}\\ &= \sqrt {206}\\ &= 14.35_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Daily Sales Mid-value (x) No. of shops (f) fx d = x - $$\overline{X}$$ d2 fd2 0-10 5 2 10 -18.6 345.96 691.92 10-20 15 9 135 -8.6 73.96 665.64 20-30 25 10 250 1.4 1.96 19.6 30-40 35 7 245 11.4 129.96 909.72 40-50 45 1 45 21.4 457.96 457.96 N = 29 $$\sum {fx}$$ = 685 $$\sum {fd^2}$$ = 2744.84

\begin{align*} Mean\;(\overline{X}) &= \frac {\sum{fx}}N\\ &= \frac {685}{29}\\ &= 23.6\\ \end{align*}

\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fd^2}}N}\\ &= \sqrt {\frac {2744.84}{29}}\\ &= \sqrt {94.65}\\ &= 9.73_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Marks Obtained No. of Students (f) Mid-value (m) fm fm2 0-10 4 5 20 100 10-20 7 - 4 = 3 15 45 675 20-30 14 - 7 = 7 25 175 4375 30-40 16 - 14 = 2 35 70 2450 40-50 22 - 16 = 6 45 270 12150 N = 22 $$\sum{fm}$$ = 580 $$\sum{fm^2}$$ = 19750

\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fm^2}}N - (\frac {\sum{fm}}N)^2}\\ &= \sqrt {\frac {19750}{22} - (\frac {580}{22})^2}\\ &= \sqrt {897.72 - 695.04}\\ &= \sqrt {202.69}\\ &= 14.24_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Marks Obtained (x) No. of Students (f) fx d d2 fd2 5 5 25 -25 625 3125 15 8 120 -15 225 1800 25 10 250 -5 25 250 35 15 525 5 25 375 45 8 360 15 225 1800 55 4 220 25 625 2500 N = 50 $$\sum{fx}$$ = 1500 $$\sum {fd^2}$$ = 9850

\begin{align*} Mean\;(\overline{X}) &= \frac {\sum {fx}}N\\ &= \frac {1500}{50}\\ &= 30\\ \end{align*}

\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fd^2}}N}\\ &= \sqrt {\frac {9850}{50}}\\ &= \sqrt {197}\\ &= 14.04_{Ans}\\ \end{align*}

Calculating Standard Deviation

 Marks Obtained Mid-value (m) d = m - 35 f fd fd2 10-20 15 -20 4 -80 1600 20-30 25 -10 6 -60 600 30-40 35 0 10 0 0 40-50 45 10 3 30 300 50-60 55 20 2 40 800 N = 25 $$\sum{fd}$$ = -70 $$\sum{fd^2}$$ = 3300

\begin{align*} \therefore\;Standard\;Deviation\;(\sigma) &= \sqrt {\frac {\sum{fd^2}}N - (\frac {\sum{fd}}N)^2}\\ &= \sqrt {\frac {3300}{25} - (\frac {-70}{25})^2}\\ &= \sqrt {132 - 7.84}\\ &= \sqrt {124.16}\\ &= 11.14_{Ans}\\ \end{align*}

0%

6.05

6.5

3.15

7.5

4.29, 0.039

6.15, 0.056

3.95, 0.027

2.23, 0.096

6.73

7.63

6.26

8.25

4.369

6.652

7.325

5.395

5.6

5.9

6.6

7.5

2.01

3.02

4.05

5.52

41.56

42.22

45.55

47.32

14.2

12.1

13.3

15.7

9.13

8.25

6.63

7.70

9.93

11.33

13.25

12.29

11.03

12.02

10.36

13.09

11.33

10.05

13.39

12.69

11.23

10.05

9.05

8.32

12.36

10.02

11.25

13.39

10.32

11.75

13.75

12.75