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Mean square deviation Root mean square deviation Variance Standard deviation
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Mean square deviation
Root mean square deviation
Variance
Standard deviation
Choose one of the following:
• Example 1 : Method A Raw data using defintion
• Example 1 : Method B Raw data using alternative
• Example 2 : Method A Frequency distribution using defintion
• Example 2 : Method B Frequency distribution using alternative
• Summary of formulae
• Notes on this presentation
Example 1 : Method A
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Tabulate the values (x) and find the mean:
x1 6.52 5.93 5.44 6.05 6.16 5.97 5.88 5.69 5.9
Total 53.1
The sample mean
=
=
= 5.9
x
x
n
53.1
9
x
Example 1 : Method A
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Find the deviation from the mean, for each x value, x – :
x x – 1 6.5 0.62 5.9 0.03 5.4 -0.54 6.0 0.15 6.1 0.26 5.9 0.07 5.8 -0.18 5.6 -0.39 5.9 0.0
Total 53.1
x
x = 5.9
x
Example 1 : Method A
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Square the deviations from the mean and find the ‘sum of squares’ Sxx:
x x – (x – )2
1 6.5 0.6 0.362 5.9 0.0 0.003 5.4 -0.5 0.254 6.0 0.1 0.015 6.1 0.2 0.046 5.9 0.0 0.007 5.8 -0.1 0.018 5.6 -0.3 0.099 5.9 0.0 0.00
Total 53.1 0.76
x x
Sxx =2( )x x
x = 5.9 msd & rmsd
Variance & standard deviation
Example 1 : Method A Mean square deviation
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Divide the ‘sum of squares’ Sxx by n :
x x – (x – )2
1 6.5 0.6 0.362 5.9 0.0 0.003 5.4 -0.5 0.254 6.0 0.1 0.015 6.1 0.2 0.046 5.9 0.0 0.007 5.8 -0.1 0.018 5.6 -0.3 0.099 5.9 0.0 0.00
Total 53.1 0.76
x x Mean square deviation
= =
=
= 0.0844 (to 3 s.f.)
2( )x x
n
0.76
9
x = 5.9
xxS
n
2( )x x
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Divide the ‘sum of squares’ Sxx by n and take the square root :
x x – (x – )2
1 6.5 0.6 0.362 5.9 0.0 0.003 5.4 -0.5 0.254 6.0 0.1 0.015 6.1 0.2 0.046 5.9 0.0 0.007 5.8 -0.1 0.018 5.6 -0.3 0.099 5.9 0.0 0.00
Total 53.1 0.76
x x Root mean square deviation (rmsd)
= =
=
= 0.291 (to 3 s.f.)
2( )x x
n
0.76
9
RETURN
x = 5.9
xxS
n
2( )x x
Example 1 : Method A Root mean square deviation
Example 1 : Method A Variance
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Divide the ‘sum of squares’ Sxx by n – 1 :
x x – (x – )2
1 6.5 0.6 0.362 5.9 0.0 0.003 5.4 -0.5 0.254 6.0 0.1 0.015 6.1 0.2 0.046 5.9 0.0 0.007 5.8 -0.1 0.018 5.6 -0.3 0.099 5.9 0.0 0.00
Total 53.1 0.76
x x Variance
= =
=
= 0.095
2( )
1
x x
n
0.76
8
x = 5.9
1xxS
n
2( )x x
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Divide the ‘sum of squares’ Sxx by n – 1 and take the square root :
x x – (x – )2
1 6.5 0.6 0.362 5.9 0.0 0.003 5.4 -0.5 0.254 6.0 0.1 0.015 6.1 0.2 0.046 5.9 0.0 0.007 5.8 -0.1 0.018 5.6 -0.3 0.099 5.9 0.0 0.00
Total 53.1 0.76
x x Standard deviation s
= =
=
= 0.308 (to 3 s.f.)
2( )
1
x x
n
0.76
8
RETURN
x = 5.9
1xxS
n
2( )x x
Example 1 : Method A Standard deviation
Example 1 : Method B
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Tabulate the values (x) and find the mean:
x1 6.52 5.93 5.44 6.05 6.16 5.97 5.88 5.69 5.9
Total 53.1
The sample mean
=
=
= 5.9
x
x
n
53.1
9
x
Example 1 : Method B
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Tabulate the squares of the values (x2) and find the total:
x x 2
1 6.5 42.252 5.9 34.813 5.4 29.164 6.0 36.005 6.1 37.216 5.9 34.817 5.8 33.648 5.6 31.369 5.9 34.81
Total 53.1 314.05 x2
Sxx =
= 314.05 – 9 5.92
= 0.76
Example 1 : Method B
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Find the ‘sum of squares’ Sxx by subtracting from x2:
x = 5.9
x x 2
1 6.5 42.252 5.9 34.813 5.4 29.164 6.0 36.005 6.1 37.216 5.9 34.817 5.8 33.648 5.6 31.369 5.9 34.81
Total 53.1 314.05 x2
2 2x nx
2nx
msd & rmsd
Variance & standard deviation
Example 1 : Method B Mean square deviation
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Divide the ‘sum of squares’ Sxx by n :
Mean square deviation
= =
=
= 0.0844 (to 3 s.f.)
0.76
9
x x 2
1 6.5 42.252 5.9 34.813 5.4 29.164 6.0 36.005 6.1 37.216 5.9 34.817 5.8 33.648 5.6 31.369 5.9 34.81
Total 53.1 314.05
2 2x nx
n
x = 5.9
x2
xxS
n
Example 1 : Method B Root mean square deviation
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Divide the ‘sum of squares’ Sxx by n and take the square root :
Root mean square deviation (rmsd)
= =
=
= 0.291 (to 3 s.f.)
0.76
9
x x 2
1 6.5 42.252 5.9 34.813 5.4 29.164 6.0 36.005 6.1 37.216 5.9 34.817 5.8 33.648 5.6 31.369 5.9 34.81
Total 53.1 314.05
2 2x nx
n
x = 5.9
x2
RETURN
xxS
n
Example 1 : Method B Variance
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Divide the ‘sum of squares’ Sxx by n – 1 :
Variance
= =
=
= 0.095
0.76
8
x x 2
1 6.5 42.252 5.9 34.813 5.4 29.164 6.0 36.005 6.1 37.216 5.9 34.817 5.8 33.648 5.6 31.369 5.9 34.81
Total 53.1 314.05
2 2
1
x nx
n
x = 5.9
x2
1xxS
n
Example 1 : Method B Standard deviation
Extracts were taken from nine leaf cells and the pH of each was measured. The results were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
Divide the ‘sum of squares’ Sxx by n – 1 and take the square root :
Standard deviation s
= =
=
= 0.308 (to 3 s.f.)
0.76
8
x x 2
1 6.5 42.252 5.9 34.813 5.4 29.164 6.0 36.005 6.1 37.216 5.9 34.817 5.8 33.648 5.6 31.369 5.9 34.81
Total 53.1 314.05
2 2
1
x nx
n
x = 5.9
x2
RETURN
1xxS
n
Example 2 : Method A
The number of children per family, x, for a random selection of 100 families, is given by the following table:
The sample mean
=
=
= 2.1
x
xf
n
210
100
xf
First tabulate the xf values and find their mean:
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf
1 0 9 0
2 1 24 24
3 2 35 70
4 3 19 57
5 4 8 32
6 5 3 15
7 6 2 12
Total 21 100 210
n =f
Example 2 : Method A
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Find the deviation from the mean, for each x value, :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf
1 0 9 0 -2.1
2 1 24 24 -1.1
3 2 35 70 -0.1
4 3 19 57 0.9
5 4 8 32 1.9
6 5 3 15 2.9
7 6 2 12 3.9
Total 21 100 210
x = 2.1
x x
x x
Example 2 : Method A
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Square the deviations from the mean :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf
1 0 9 0 -2.1 4.41
2 1 24 24 -1.1 1.21
3 2 35 70 -0.1 0.01
4 3 19 57 0.9 0.81
5 4 8 32 1.9 3.61
6 5 3 15 2.9 8.41
7 6 2 12 3.9 15.21
Total 21 100 210
2( )x x
2( )x x
x x
x = 2.1
Example 2 : Method A
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Find the ‘sum of squares’ Sxx, the sum of :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf
1 0 9 0 -2.1 4.41 39.69
2 1 24 24 -1.1 1.21 29.04
3 2 35 70 -0.1 0.01 0.35
4 3 19 57 0.9 0.81 15.39
5 4 8 32 1.9 3.61 28.88
6 5 3 15 2.9 8.41 25.23
7 6 2 12 3.9 15.21 30.42
Total 21 100 210 169
2( )x x f
2( )x x f
2( )x x fx x 2( )x x
x = 2.1 msd & rmsd
Variance & standard deviation
Example 2 : Method A Mean square deviation
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Divide the ‘sum of squares’ Sxx by n :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf
1 0 9 0 -2.1 4.41 39.69
2 1 24 24 -1.1 1.21 29.04
3 2 35 70 -0.1 0.01 0.35
4 3 19 57 0.9 0.81 15.39
5 4 8 32 1.9 3.61 28.88
6 5 3 15 2.9 8.41 25.23
7 6 2 12 3.9 15.21 30.42
Total 21 100 210 169
Mean square deviation
= =
=
= 1.69
2( )x x f
n
169
100
2( )x x fx x 2( )x x
xxS
n
Example 2 : Method A Root mean square deviation
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Divide the ‘sum of squares’ Sxx by n and take the square root :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf
1 0 9 0 -2.1 4.41 39.69
2 1 24 24 -1.1 1.21 29.04
3 2 35 70 -0.1 0.01 0.35
4 3 19 57 0.9 0.81 15.39
5 4 8 32 1.9 3.61 28.88
6 5 3 15 2.9 8.41 25.23
7 6 2 12 3.9 15.21 30.42
Total 21 100 210 169
Root mean square deviation
= =
=
= 1.3
2( )x x f
n
169
100
RETURN
2( )x x fx x 2( )x x
xxS
n
Example 2 : Method A Variance
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Divide the ‘sum of squares’ Sxx by n – 1 :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf
1 0 9 0 -2.1 4.41 39.69
2 1 24 24 -1.1 1.21 29.04
3 2 35 70 -0.1 0.01 0.35
4 3 19 57 0.9 0.81 15.39
5 4 8 32 1.9 3.61 28.88
6 5 3 15 2.9 8.41 25.23
7 6 2 12 3.9 15.21 30.42
Total 21 100 210 169
Variance
= =
=
= 1.71 (to 3 s.f.)
2( )
1
x x f
n
169
99
2( )x x fx x 2( )x x
1xxS
n
Example 2 : Method A Standard deviation
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Divide the ‘sum of squares’ Sxx by n – 1 and take the square root :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf
1 0 9 0 -2.1 4.41 39.69
2 1 24 24 -1.1 1.21 29.04
3 2 35 70 -0.1 0.01 0.35
4 3 19 57 0.9 0.81 15.39
5 4 8 32 1.9 3.61 28.88
6 5 3 15 2.9 8.41 25.23
7 6 2 12 3.9 15.21 30.42
Total 21 100 210 169
Standard deviation s
= =
=
= 1.31 (to 3 s.f.)
2( )
1
x x f
n
169
99
RETURN
2( )x x fx x 2( )x x
1xxS
n
Example 2 : Method B
The number of children per family, x, for a random selection of 100 families, is given by the following table:
The sample mean
=
=
= 2.1
x
xf
n
210
100
xf
Tabulate the xf values and find their mean:
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf
1 0 9 0
2 1 24 24
3 2 35 70
4 3 19 57
5 4 8 32
6 5 3 15
7 6 2 12
Total 21 100 210
n =f
Example 2 : Method B
The number of children per family, x, for a random selection of 100 families, is given by the following table:
x2f
Tabulate the squares of the values (x2f) and find the total:
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
n =f
x f xf x2f
1 0 9 0 0
2 1 24 24 24
3 2 35 70 140
4 3 19 57 171
5 4 8 32 128
6 5 3 15 75
7 6 2 12 72
Total 21 100 210 610
Example 2 : Method B
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Find the ‘sum of squares’ by subtracting from x2f :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf x2f
1 0 9 0 0
2 1 24 24 24
3 2 35 70 140
4 3 19 57 171
5 4 8 32 128
6 5 3 15 75
7 6 2 12 72
Total 21 100 210 610
2nx
2 2x f nx Sxx =
= 610 – 100 2.12
= 169
x = 2.1
n =f
x2f
msd & rmsd
Variance & standard deviation
Example 2 : Method B Mean square deviation
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Divide the ‘sum of squares’ Sxx by n :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf x2f
1 0 9 0 0
2 1 24 24 24
3 2 35 70 140
4 3 19 57 171
5 4 8 32 128
6 5 3 15 75
7 6 2 12 72
Total 21 100 210 610
x = 2.1
n =f
Mean square deviation
= =
=
= 1.69
169
100
2 2x f nx
n
x2f
xxS
n
Example 2 : Method B Root mean square deviation
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Divide the ‘sum of squares’ Sxx by n and take the square root :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf x2f
1 0 9 0 0
2 1 24 24 24
3 2 35 70 140
4 3 19 57 171
5 4 8 32 128
6 5 3 15 75
7 6 2 12 72
Total 21 100 210 610
x = 2.1
n =f
Root mean square deviation
= =
=
= 1.3
169
100
2 2x f nx
n
x2fRETURN
xxS
n
Example 2 : Method B Variance
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Divide the ‘sum of squares’ Sxx by n – 1 :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf x2f
1 0 9 0 0
2 1 24 24 24
3 2 35 70 140
4 3 19 57 171
5 4 8 32 128
6 5 3 15 75
7 6 2 12 72
Total 21 100 210 610
x = 2.1
n =f
Variance
= =
=
= 1.71 (to 3 s.f.)
169
99
2 2
1
x f nx
n
x2f
1xxS
n
Example 2 : Method B Standard deviation
The number of children per family, x, for a random selection of 100 families, is given by the following table:
Divide the ‘sum of squares’ Sxx by n – 1 and take the square root :
No. of children, x 0 1 2 3 4 5 6 >6
Frequency, f 9 24 35 19 8 3 2 0
x f xf x2f
1 0 9 0 0
2 1 24 24 24
3 2 35 70 140
4 3 19 57 171
5 4 8 32 128
6 5 3 15 75
7 6 2 12 72
Total 21 100 210 610
x = 2.1
n =f
Standard deviation s
= =
=
= 1.31 (to 3 s.f.)
169
99
2 2
1
x f nx
n
x2fRETURN
1xxS
n
Sxx DefinitionAlternative
version
Raw data
Frequency distribution
2( )x x 2 2x nx
Various forms of ‘sum of squares’ Sxx
2( )x x f 2 2x f nx
Variance Standard deviation s 1
xxS
n
1xxS
n
Mean square Root mean squaredeviation deviation (rmsd)
xxS
nxxS
n
RETURN
Notes on using the presentation
The presentation covers calculations using
Raw data (Example 1) or a
Frequency distribution (Example 2).
The ‘sum of squares’ Sxx is evaluated using the
Definition formula (Method A) or the
Alternative formula (Method B).
For each example and each method the ‘sum of squares’ Sxx is used to calculate the
mean square deviation and root mean square deviation or the
variance and standard deviation
Use the links in the presentation to choose the appropriate example and method, together with the desired calculations. RETURN
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