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) Direct Method or Actual Mean Method:
Let X_{1}, X_{2}, X_{3}, .......................... X_{n} 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 X_{1}, X_{2}, X_{3}, ..........................X_{n} 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).
Let X_{1}, X_{2}, X_{3}, ........................ X_{n} be the variable values and f_{1}, f_{2}, f_{3}, ....................... f_{n} 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
Let X_{1}, X_{2}, X_{3}, ............................. X_{n} be the mid-values of the classes and f_{1}, f_{2}, f_{3}, ....................... f_{n} 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
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
Demerits
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 |
3.95, 0.027
6.15, 0.056
4.29, 0.039
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 |
8.25
6.26
7.63
6.73
Find the standard deviation of the data:
X | 92 | 93 | 97 | 98 | 102 | 104 | 109 |
f | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
5.395
7.325
4.369
6.652
Find the standard deviation:
X | 15 | 25 | 30 | 25 | 40 | 45 |
f | 1 | 5 | 10 | 12 | 8 | 4 |
5.9
6.6
7.5
5.6
Find the standard deviation:
X | 12 | 14 | 16 | 18 | 20 | 25 |
f | 6 | 7 | 10 | 15 | 10 | 2 |
5.52
3.02
2.01
4.05
Find the standard deviation:
x = 20, 25, 28, 30, 40, 35.
47.32
42.22
45.55
41.56
Calculate the standard deviation:
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
No. of student | 15 | 17 | 12 | 9 | 12 |
13.3
14.2
15.7
12.1
Find the standard deviation:
Marks | 0-6 | 0-12 | 0-18 | 0-24 | 0-30 |
no. of student | 4 | 7 | 12 | 18 | 20 |
6.63
7.70
8.25
9.13
Find the standard deviation:
Marks | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
No. of student | 2 | 3 | 6 | 5 | 4 |
11.33
9.93
12.29
13.25
Find the standard deviation:
X | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
f | 1 | 5 | 8 | 10 | 4 | 2 |
12.02
11.03
13.09
10.36
Find the standard deviation:
X | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
f | 5 | 4 | 4 | 6 | 1 |
13.39
11.33
10.05
12.69
Find the standard deviation:
Marks | 10 | 20 | 30 | 40 | 50 |
No. of student | 8 | 12 | 15 | 9 | 6 |
10.05
11.23
8.32
9.05
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
13.39
10.02
11.25
Find the standard deviation and its coefficient:
Class | 10 | 20 | 30 | 40 | 50 |
Frequency | 2 | 4 | 3 | 7 | 4 |
10.32
12.75
11.75
13.75
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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
Jan 04, 2017
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