In probability theory and statistics, the standard deviation of a statistical population, a data set, or a probability distribution is the square root of its variance. Standard deviation is a widely used measure of the variability or dispersion, being algebraically more tractable though practically less robust than the expected deviation or average absolute deviation.
Source – Wikipedia.
Solving Sample Standard Deviation – Examples:
The following example problems show sample standard deviation.
 Solving the sample standard deviation for the values 8, 15, 17 and 16.
(i) We can find the mean and deviation.
X = 8, 15, 17, 16
M = `(8 + 15 + 17 + 16)/4`
= `56/4`
= 14
(ii) Then we can find the sum of (X – M)^{ 2}
X  XM  (XM)^{2} 
815
17 16 
814 = 61514 = 1
1714 = 3 1614 = 2

361
9 4 
Total  50 
N = 4, the total number of values.
Then N1 = 4 – 1
= 3
(iii) The Standard Deviation can be located by the method.
`(sqrt50)/(sqrt3)` = `7.07/1.73`
= 4.08
 Solving the sample standard deviation for the values 20, 14, 6, 15, 17 and 12.
Solution:
(i) We can find the mean and deviation.
X = 20, 14, 6, 15, 17, 12
M = `(20 + 14 + 6 + 15 + 17 + 12)/6`
= `84/6`
= 14
(ii) Then we can find the sum of (X – M)^{ 2}
X  XM  (XM)^{2} 
2014
6 15 17 12 
2014 = 61414 = 0
614 = 8 1514 = 1 1714 = 3 1214 = 2 
360
64 1 9 4 
Total  110 
N = 6, the total number of values.
Then N1 = 6 – 1
= 5
(iii) The Standard Deviation can be located by the method.
`(sqrt110)/(sqrt5)` = `10.48/2.23`
= 4.70
 Solving the sample standard deviation for the values 17, 14, 12, 9 and 13.
(i) We can find the mean and deviation.
X = 17, 14, 12, 9, 13
M = `(17 + 14 + 12 + 9 + 13)/5`
= `65/5`
= 13
(ii) Then we can find the sum of (X – M)^{ 2}
X  XM  (XM)^{2} 
1714
12 9 13 
1713 = 41413 = 1
1213 = 1 913 = 4 1313 = 0 
161
1 16 0 
Total  34 
N = 5, the total number of values.
Then N1 = 5 – 1
= 4
(iii) The Standard Deviation can be located by the method.
`(sqrt34)/(sqrt4)` = `5.83/2`
= 2.91
 Solving the sample standard deviation for the values 6, 15, 17 and 10.
(i) We can find the mean and deviation.
X = 6, 15, 17, 10
M = `(6 + 15 + 17 + 10)/4`
= `48/4`
= 12
(ii) Then we can find the sum of (X – M)^{ 2}
X  XM  (XM)^{2} 
615
17 10 
612 = 61512 = 3
1712 = 5 1012 = 2 
369
25 4 
Total  74 
N = 4, the total number of values.
Then N1 = 4 – 1
= 3
(iii) The Standard Deviation can be located by the method.
`(sqrt74)/(sqrt3)` = `8.60/1.73`
= 4.97
Solving Sample Standard Deviation – Practice Problems:
The following practice problems show sample standard deviation.
 Solving the sample standard deviation for the values 15, 8, 12 and 17.
= 3.92
 Solving the sample standard deviation for the values 6, 12, 20, 15, 17 and 8.
Solution:
= 5.36
 Solving the sample standard deviation for the values 17, 13, 7, 14 and 9.
= 4.
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