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Prepared by: Nurazrin Jupri
differences
will be
large
differences
will be
small
MATH0102|Nurazrin Jupri
 Two
groups of three students
Group 1
4
7
10
Group 2
7
7
7
 Mean
mark
Group 1
Group 2
4 + 7 + 10 = 21/3 = 7
7 + 7 + 7 = 21/3 = 7
 Same
mean mark, but Group 1’s marks are
widely spread, Group 2’s are all the same
 The following diagram reinforces this point
MATH0102|Nurazrin Jupri
MATH0102|Nurazrin Jupri
A
measure of the average amount by
which the values in a distribution (x)
differ from the arithmetic mean
 Average of the absolute deviations from
the arithmetic mean (ignoring the sign)
MATH0102|Nurazrin Jupri
grouped
Vertical
bars = all
differences
are taken as
positive
MATH0102|Nurazrin Jupri
ungroupe
d
 X1
= 2, X2 = 4, X3 = 3
 MD
= X1 X  X2  X  X3  X
n
 MD
=
=
2  3 4  3 3 3
=⅔
1  1  0
3
3
MATH0102|Nurazrin Jupri
Mean Deviation of grouped data
MATH0102|Nurazrin Jupri
X
MATH0102|Nurazrin Jupri
MATH0102|Nurazrin Jupri
 If
we square all the deviations from the
arithmetic mean, then we no longer need
to bother with dropping the signs since
all the values will be positive.
 Variance
is the average of the squared
deviations from the arithmetic mean
MATH0102|Nurazrin Jupri

Variance =
 X
n
i 1
i
X

2
n

1.
2.
3.
To calculate the variance
Calculate the mean value X
Subtract the mean from each value in turn,
that is, find X i  X
Square each answer to get X  X 2

MATH0102|Nurazrin Jupri
i

4. Add up all these squared values to get
 X
n
i 1
i
X

2
5. Divide the result by n to get
 X
n
i 1
1
X

2
n
6. You now have the average of the squared deviations
from the mean (in square units)
MATH0102|Nurazrin Jupri
 This
is simply the square root of the
variance
 An advantage is that we avoid the square
units of the variance
 Larger SD, larger the average dispersion
of data from the mean
 Smaller SD, smaller the average
dispersion of data from the mean
MATH0102|Nurazrin Jupri
xi
x1 - x
(x1 – x)2
4
7
10
Total
4–7=-3
7–7= 0
10 – 7 = 3
(-32) = 9
02 = 0
32 = 9
18
MATH0102|Nurazrin Jupri
X

Variance =
n
i 1
i
X
n

2
18 square units
 6
3
Standard deviation is square root of 6
= 2.449 units
MATH0102|Nurazrin Jupri
xi
xi - x
7
7
7
Total
7–7=0
7–7=0
7–7=0
MATH0102|Nurazrin Jupri
(xi – x)2
02 = 0
02 = 0
02 = 0
0
 X
n
Variance =
i 1
i
X
n

2
0
  0 square units
3
Standard deviation is square root of 0 = 0
i.e. there is no spread of values
MATH0102|Nurazrin Jupri
j
S 
2
F X
i
i 1
j
F
i 1
i

  Fi X i
  i 1 j

  Fi
 i 1
j
2
i






2
where Fi = Frequency of ith class interval
Xi = mid point of ith class interval
j = number of class intervals
MATH0102|Nurazrin Jupri
LCB
UCB
F
X
FX
FX^2
5.5
10.5
8
8
64
512
10.5
15.5
4
13
52
676
15.5
20.5
6
18
108
1944
224
3132
18
MATH0102|Nurazrin Jupri
3132  224 
S 


18  18 
2
2
S2 = 174 – 12.442
S2 = 174 – 154.86
S2 = 19.14
S = √ 19.14 = 4.375
MATH0102|Nurazrin Jupri