Solving Sample Standard Deviation

 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 X-M (X-M)2
815

17

16

8-14 = -615-14 = 1

17-14 = 3

16-14 = 2

 

361

9

4

Total 50

N = 4, the total number of values.

Then N-1 = 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 X-M (X-M)2
2014

6

15

17

12

20-14 = 614-14 = 0

6-14 = -8

15-14 = 1

17-14 = 3

12-14 = -2

360

64

1

9

4

Total 110

N = 6, the total number of values.

Then N-1 = 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 X-M (X-M)2
1714

12

9

13

17-13 = 414-13 = 1

12-13 = -1

9-13 = -4

13-13 = 0

161

1

16

0

Total 34

N = 5, the total number of values.

Then N-1 = 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 X-M (X-M)2
615

17

10

6-12 = -615-12 = 3

17-12 = 5

10-12 = -2

369

25

4

Total 74

N = 4, the total number of values.

Then N-1 = 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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