Standard Deviation

Standard Deviation

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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

formulas
formulas

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*}

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  • Find the standard deviation of the given data:

    20, 25, 28, 30, 40, 35

    6.05


    6.5


    3.15


    7.5


  • Calculate standard deviation and its coefficient:

     

    X 140 142 150 147 145
    f 4 8 9 3 2

    4.29, 0.039


    6.15, 0.056


    3.95, 0.027


    2.23, 0.096


  • Find the standard deviation from the following data:

    Mid Value 2.5 7.5 12.5 17.5 22.5 27.5
    f 3 2 7 9 1 6

    6.73


    7.63


    6.26


    8.25


  • Find the standard deviation of the data:

    X 92 93 97 98 102 104 109
    f 3 2 3 2 6 3 3

    4.369


    6.652


    7.325


    5.395


  • Find the standard deviation:

    X 15 25 30 25 40 45
    f 1 5 10 12 8 4

    5.6


    5.9


    6.6


    7.5


  • Find the standard deviation:

    X 12 14 16 18 20 25
    f 6 7 10 15 10 2

    2.01


    3.02


    4.05


    5.52


  • Find the standard deviation:

    x = 20, 25, 28, 30, 40, 35.

    41.56


    42.22


    45.55


    47.32


  • Calculate the standard deviation:

    Marks 0-10 10-20 20-30 30-40 40-50
    No. of student 15 17 12 9 12

    14.2


    12.1


    13.3


    15.7


  • Find the standard deviation:

    Marks 0-6 0-12 0-18 0-24 0-30
    no. of student 4 7 12 18 20

    9.13


    8.25


    6.63


    7.70


  • Find the standard deviation:

    Marks  30-40 40-50 50-60 60-70 70-80
    No. of student 2 3 6 5 4

    9.93


    11.33


    13.25


    12.29


  • Find the standard deviation:

    X 0-10 10-20 20-30 30-40 40-50 50-60
    f 1 5 8 10 4 2

    11.03


    12.02


    10.36


    13.09


  • Find the standard deviation:

    X 0-10 10-20 20-30 30-40 40-50
    f 5 4 4 6 1

    11.33


    10.05


    13.39


    12.69


  • Find the standard deviation:

    Marks 10 20 30 40 50
    No. of student 8 12 15 9 6

    11.23


    10.05


    9.05


    8.32


  • Find the standard deviation and its coefficient:

    C . I. 0-10 10-20 20-30 30-40 40-50
    Frequency 4 6 5 3 2

    12.36


    10.02


    11.25


    13.39


  • Find the standard deviation and its coefficient:

    Class 10 20 30 40 50
    Frequency 2 4 3 7 4

    10.32


    11.75


    13.75


    12.75


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smriti khanal

if the median of the following data is 24,find the standard deviationmarks obtained:- 0-10 10-20 20-30 30-40 40-50no of students:- 9 ,21,x,15,10