Examples of Standard Deviation

Examples of Standard Deviation:

This tutorial is about some examples of standard deviation using all methods which are discussed in the pervious tutorial.

Example:

Calculate the standard deviation for the following sample data using all methods: 2, 4, 8, 6, 10, and 12.
Solution:

Method-I: Actual Mean Method

>X
>{\left( {X - \overline X } \right)^2}
>2
>{(2 - 7)^2} = 25
>4
>{(4 - 7)^2} = 9
>8
>{(8 - 7)^2} = 1
>6
>{(6 - 7)^2} = 1
>10
>{(10 - 7)^2} = 9
>12
>{(12 - 7)^2} = 25
>\sum X = 42
>\sum {\left( {X - \overline X } \right)^2} = 70

 

\overline X = \frac{{\sum X}}{n} = \frac{{42}}{6} = 7
S = \sqrt {\frac{{\sum {{\left( {X - \overline X } \right)}^2}}}{n}}
S = \sqrt {\frac{{70}}{6}} = \sqrt {\frac{{35}}{3}} = 3.42

Method-II: Taking Assumed Mean as 6

>X
>D = \left( {X - 6} \right)
>{D^2}
>2
> - 4
>16
>4
> - 2
>4
>8
>2
>4
>6
>0
>0
>10
>4
>16
>12
>6
>36
>Total
>\sum D = 6
>\sum {D^2} = 76

 

S = \sqrt {\frac{{\sum {D^2}}}{n} - {{\left( {\frac{{\sum D}}{n}} \right)}^2}}
S = \sqrt {\frac{{76}}{6} - {{\left( {\frac{6}{6}} \right)}^2}} = \sqrt {\frac{{70}}{6}}
S = \sqrt {\frac{{35}}{3}} = 3.42

Method-III: Taking Assume Mean as Zero

>X
>{X^2}
>2
>4
>4
>16
>8
>64
>6
>36
>10
>100
>12
>144
>\sum X = 42
>\sum {X^2} = 364

S = \sqrt {\frac{{\sum {X^2}}}{n} - {{\left( {\frac{{\sum X}}{n}} \right)}^2}}
S = \sqrt {\frac{{364}}{6} - {{\left( {\frac{{42}}{6}} \right)}^2}}
S = \sqrt {\frac{{70}}{6}} = \sqrt {\frac{{35}}{3}} = 3.42

Method-IV: Taking 2as common divisor or factor

>X
>U = \left( {X - 4} \right)/2
>{U^2}
>2
> - 1
>1
>4
>0
>0
>8
>2
>4
>6
>1
>1
>10
>3
>9
>12
>4
>16
>Total
>\sum U = 9
>\sum {U^2} = 31

 

S = \sqrt {\frac{{\sum {U^2}}}{n} - {{\left( {\frac{{\sum U}}{n}} \right)}^2}} \times c
S = \sqrt {\frac{{31}}{6} - {{\left( {\frac{9}{6}} \right)}^2}} \times 2
S = \sqrt {2.92} \times 2 = 3.42

Example:
Calculate standard deviation from the following distribution of marks by using all the methods.

Marks
>No. of Students
>1 - 3
>40
>3 - 5
>30
>5 - 7
>20
>7 - 9
>10

Solution:

Method-I: Actual Mean Method

>Marks
>f
>X
>fX
>{\left( {X - \overline X } \right)^2}
>f{\left( {X - \overline X } \right)^2}
>1 - 3
>40
>2
>80
>4
>160
>3 - 5
>30
>4
>120
>0
>0
>5 - 7
>20
>6
>120
>4
>80
>7 - 9
>10
>8
>80
>16
>160
>Total
>100
> 
>400
> 
>400

\overline X = \frac{{\sum fX}}{{\sum f}} = \frac{{400}}{{100}} = 4
S = \sqrt {\frac{{\sum f{{\left( {X - \overline X } \right)}^2}}}{{\sum f}}} = \sqrt {\frac{{400}}{{100}}} = \sqrt 4 = 2 Marks

Method-II: Taking assumed mean as 2

>Marks
>f
>X
>D = \left( {X - 2} \right)
>fD
>f{D^2}
>1 - 3
>40
>2
>0
>0
>0
>3 - 5
>30
>4
>2
>60
>120
>5 - 7
>20
>6
>4
>80
>320
>7 - 9
>10
>8
>6
>60
>160
>Total
>100
> 
 
>200
>800

S = \sqrt {\frac{{\sum f{D^2}}}{{\sum f}} - {{\left( {\frac{{\sum fD}}{{\sum f}}} \right)}^2}} = \sqrt {\frac{{800}}{{100}} - {{\left( {\frac{{200}}{{100}}} \right)}^2}}
S = \sqrt {8 - 4} = \sqrt 4 = 2 Marks

Method-III: Using assumed mean as Zero

>Marks
>f
>X
>fX
>f{X^2}
>1 - 3
>40
>2
>80
>160
>3 - 5
>30
>4
>120
>480
>5 - 7
>20
>6
>120
>720
>7 - 9
>10
>8
>80
>640
>Total
>100
> 
>400
>2000

S = \sqrt {\frac{{\sum f{X^2}}}{{\sum f}} - {{\left( {\frac{{\sum fX}}{{\sum f}}} \right)}^2}} = \sqrt {\frac{{2000}}{{100}} - {{\left( {\frac{{400}}{{100}}} \right)}^2}}
S = \sqrt {20 - 16} = \sqrt 4 = 2 Marks

Method-IV: By taking 2 as the common divisor

>Marks
>f
>X
>U = \left( {X - 2} \right)/2
>fU
>f{U^2}
>1 - 3
>40
>2
> - 2
> - 80
>160
>3 - 5
>30
>4
> - 1
> - 30
>30
>5 - 7
>20
>6
>0
>0
>0
>7 - 9
>10
>8
>1
>10
>10
>Total
>100
> 
 
> - 100
>200

S = \sqrt {\frac{{\sum f{U^2}}}{{\sum f}} - {{\left( {\frac{{\sum fU}}{{\sum f}}} \right)}^2}} \times h = \sqrt {\frac{{200}}{{100}} - {{\left( {\frac{{ - 100}}{{100}}} \right)}^2}} \times 2
S = \sqrt {2 - 1} \times 2 = \sqrt 1 \times 2 = 1 \times 2 = 2 Marks