The Standard Deviation and Other Measures

of Dispersion

Members of Group:
1. Daffa Alim Sajid (4101418112)
2. Yusuf Rhamadan (4101418132)
3. Myrat Hayytjanov (4101418228)
 A. The Range, Interquartile Range, and Semi-Interquartile Range
The range of a set of numbers is the difference between the largest and smallest numbers in the
set.
T he Range = Largest Data − S mallest Data

The interquartile range is difference between third quartile and first quartile. The formula is
I nterquartile Range = Q3 − Q1

The semi-interquartile range, or quartile deviation, of a set of data is denoted by Q and is dened
by
Q −Q
Q = 32 1

Example
Find the range of the set 7,5,10,15,12,7,8,10

Solution
5,7,7,8,10,10,12,15
The range = Largest data - Smallest data
= 15-5
= 10
Example
Find the range and the quartile deviation of heights of the students at Alpha University as
given in table above.

Height (cm) Number of Students

151-155 5
156-160 18
161-165 42
166-170 27
171-175 8
Total 100
Solution

Height (cm) Number of Students Class mark (xi )

151-155 5 153
156-160 18 158
161-165 42 163
166-170 27 168
171-175 8 173
Total 100 -

The range = Class mark of the highest class – Class mark of the lowest class
= 173-153
= 20

1(100)
T he P osition of Q1 = 4
= 25

( )
1×100 −23
Q1 = 160, 5 + 5 4
42

2
Q1 = 160, 5 + 5 ( 42 )

Q1 = 160, 5 + 0, 2

Q1 = 160, 7

3(100)
T he P osition of Q3 = 4
= 75
 ( )
3×100 −65
Q3 = 165, 5 + 5 4
27

10
Q3 = 165, 5 + 5 ( 27 )

Q3 = 165, 5 + 1, 9

Q3 = 167.4
Q3 −Q1
Q= 2

167,4−160,7
Q= 2

Q = 3, 3

B. The Mean Deviation

The mean deviation, or average deviation, of a set of n numbers X1, X2, ..., Xn is abbreviated
MD and is dened by
∑ |x−x|
MD = n

Then, for the group data, the formula is defined by

∑ f |xi −x|
MD = n

Example
Find the mean deviation of the set number above
5,7,7,8,10,10,12,15

Solution
x = 5+7+7+8+10+10+12+15
8
= 9, 25

∑ |x−x|
MD = n
 |5−9,25|+|7−9,25|+|7−9,25|+|8−9,25|+ |10−9,25|+|10−9,25|+|12−9,25|+|15−9,25|
MD = 8

20
MD = 8

M D = 2, 5

Example
Find the mean deviation of heights of the students at Alpha University as given in table
above

Height (cm) Number of Students

151-155 5
156-160 18
161-165 42
166-170 27
171-175 8
Total 100
Solution
x = 153.5+158.18+163.42+168.27+173.8
100

765+2844+6846+4536+1384
x= 100

16375
x= 100

x = 163, 75
Height (cm) ass mark ( xi ) | xi − x | Frequency (f ) f |xi − x|
151-155 153 10,75 5 53,75

156-160 158 5,75 18 103,5

161-165 163 0,75 42 31,5
166-170 168 4,25 27 114,75
171-175 173 9,25 8 74
Total 100 377,5

∑ f |xi −x|
MD = n

377,5
MD = 100

M D = 3, 775 cm

C. The Standard Deviation

The standard deviation of a set of n numbers X1, X2, ..., Xn is denoted by s and is dened by
∑(x−x)2
2
s = n

2
n ∑ x2 −(∑ x)
2
s = n.n

Then, for group data, the formula is defined by

∑ f (xi −x)2
2
s = n

2
n ∑ f x2i −(∑ f xi )
2
s = n.n
Example
You and your friends have just measured the heights of your dog (in millimetres)
The heights (at the shoulders) are: 600 mm, 470 mm, 170 mm, 430 mm, and 300 mm.
Find out the standard deviation
Solution
600+470+170+430+300
x= 5

1970
x= 5

x = 394

∑(x−x)2
2
s = n

2 2 2 2 2
(600−394) +(470−394) +(170−394) +(430−394) +(300−394)
s2 = 5

2 2 2 2 2
206 +76 +(−224) +36 +(−94)
s2 = 5

42436+5776+50176+1296+8836
s2 = 5

108520
s2 = 5

s2 = 21704

s = √21704

s = 147 mm

Using another formula

2
n ∑ x2 −(∑ x)
2
s = n
2
2 5(6002 +4702 +1702 +4302 +3002 ) −(600+470+170+430+300)
s = 5.5
 2
5(360000+220900+28900+184900+90000)−(1970)
s2 = 25
5(884700)−3880900
s2 = 25
4423500−3880900
s2 = 25
542600
s2 = 25

s2 = 21704

s = √21704

s = 147 mm

Sometimes, the standard deviation of a sample’s data is dened with ​(n-1) replacing ​n in the
denominators of the expressions in the formula because the resulting value represents a
better estimate of the standard deviation of a population from which the sample is taken.

Our example has been for a ​Population ​(the 5 dogs are the only dogs we are instead in). But if
the data is a ​Sample ​(a selection taken from a bigger population), for example if our 5
dogs are just a sample of bigger population of dogs, we divide by ​(5-1) ​not ​5.

Example
If our 5 dogs are just a sample of a bigger population of dogs, then find the standard
deviation

Solution

∑(x−x)2
s2 = n−1

2 2 2 2 2
(600−394) +(470−394) +(170−394) +(430−394) +(300−394)
s2 = 4

2 2 2 2 2
206 +76 +(−224) +36 +(−94)
s2 = 4

42436+5776+50176+1296+8836
s2 = 4

108520
s2 = 4
 s2 = 27130

s = √27130

s = 164 mm

Example
Find the standard deviation of heights of the students at Alpha University as given in
table above

Height (cm) Number of Students

151-155 5
156-160 18
161-165 42
166-170 27
171-175 8
Total 100

Solution
x = 153.5+158.18+163.42+168.27+173.8
100

765+2844+6846+4536+1384
x= 100

16375
x= 100

x = 163, 75

eight (cm) ass mark ( xi ) xi − x (xi − x)2 requency (f ) f (xi − x)2

151-155 153 -10,75 15,5625 5 577,8125
156-160 158 -5,75 33,0625 18 595,125
161-165 163 -0,75 0,5625 42 23,625
166-170 168 4,25 18,0625 27 487,6875
171-175 173 9,25 85,5625 8 684,5
Total 100 2368,75

∑ f (xi −x)2
2
s = n

2368,75
s2 = 100

s2 = 23, 6875

s = √23, 6875

s = 4, 87
 D. The Variance
The variance of a set of data is dened as the square of the standard deviation. Then the formula
is defined by

∑(x−x)2
o= n

For group data, the formula of the variance is defined by

∑ f (xi −x)2
o= n

Example
You and your friends have just measured the heights of your dog (in millimetres)
The heights (at the shoulders) are: 600 mm, 470 mm, 170 mm, 430 mm, and 300 mm.
Find out the variance.

Solution

∑(x−x)2
o= n−1

2 2 2 2 2
(600−394) +(470−394) +(170−394) +(430−394) +(300−394)
o= 4

2 2 2 2 2
206 +76 +(−224) +36 +(−94)
o= 4

42436+5776+50176+1296+8836
o= 4

108520
o= 4

o = 27130
Example
Find the variance of heights of the students at Alpha University as given in table above

Height (cm) Number of Students

151-155 5
156-160 18
161-165 42
166-170 27
171-175 8
Total 100

Solution
x = 153.5+158.18+163.42+168.27+173.8
100

765+2844+6846+4536+1384
x= 100

16375
x= 100

x = 163, 75

eight (cm) ass mark ( xi ) xi − x (xi − x)2 requency (f ) f (xi − x)2

151-155 153 -10,75 15,5625 5 577,8125
156-160 158 -5,75 33,0625 18 595,125
161-165 163 -0,75 0,5625 42 23,625
166-170 168 4,25 18,0625 27 487,6875
171-175 173 9,25 85,5625 8 684,5
Total 100 2368,75
 ∑ f (xi −x)2
o= n

2368,75
o= 100

o = 23, 6875

E. Absolute and Relative Dispersion; Coefficient of Variance

The actual variation, or dispersion, as determined from the standard deviation or other measure
of dispersion is called the absolute dispersion.

If the absolute dispersion is the standard deviation ​s​ and if the average is the mean x ,then the
relative dispersion is called the coefficient of variation, or coefficient of dispersion; it is
denoted by V and is given by

V = xs ×100%

Example
An electron lamp has the average of lifetime about 3500 hours with 1050 hours standard
deviation. Another type has average of lifetime about 10000 hours with 2000 hours
standard deviation.

Then, we can compute it. For first lamp, we get:

V = xs ×100%

1050
V = 3500
×100%

V = 30%

From the second lamp, we get:

V = xs ×100%

2000
V = 10000
×100%
V = 20%

Thus, we can conclude that those lamps have different using period or lifetime.

F. Empirical Relations Between Measures Dispersion

For moderately skewed distributions, we have the empirical formulas

4
M ean Deviation = 5
(Standard Deviation)

2
S emi − I nterquartile Range = 3
(Standard Deviation)

Example
From the first example, we got the standard deviation from heights of the students at Alpha
University table is 4,87 or 5 (only 1 digit).
2
S emi − I nterquartile Range = 3
(Standard Deviation)
2
S emi − I nterquartile Range = 3
(5)

S emi − I nterquartile Range = 3, 3