3. Dispersion is the state of getting dispersed or Spread
Statistical Dispersion means the extent to which numerical data is likely to
vary about an average value.
In other words dispersion help to understand the distribution of data.
4. According to Connor ”Dispersion is a measure of the extent to which the
individual vary”
According to Kafka “ The measurement of Scatteredness of the mass of
Figures in a series about an average is called Measure of Dispersion OR
Measure of Variation.
5.
6.
7.
8. T y p e s o f M e a u r e o f D i s p e r s i o n
A b s o l u t e M e a s u r e o f D i s p e r s i o n R e l a t i v e M e a s u r e o f D i s p e r s i o n
1 .R a n g e
2 . I n t e r Q u a r t i l e R a n g e ( Q u a r t i l e D e v i a t i o n )
3 . M e a n D e v i a t i o n
4 . V a r i a n c e
5 . S t a n d a r d D e v i a t i o n
1 . C o e f f i c i e n t o f R n g e
2 . C o e f f i c i e n t o f Q u a r t i l e d e v i a t i o n
3 . C o e f f i c i e n t o f M e a n D e v i a t i o n
4 . C o e f f i c i e n t o f V a r i a n c e
5 . C o e f f i c i e n t o f S t a n d a r d D e v i a t i o n
9. Range
Range is defined as the difference between the Highest and lowest value
in a sample. Also the relative measure of Range is Known as Coefficient of
Range.
Mathematically, If H is the highest value and L is the lowest value.
Range=H-L
C
o
e
f
f
i
c
i
e
n
t
o
f
R
a
n
g
e
=
H
-
L
H
+
L
10. Advantage of Range
The Range is very Simple to Understand.
It is easy to calculate.
It is very helpful in statistical quality Control and
weather forecasting.
Take minimum time to calculate the value of
range.
11. Disadvantage of Range
It is not based on each and every item of Distribution.
Subjected to Fluctuation of Considerable Magnitude
from sample to sample.
Range can not tell anything about the character of
distribution
Range can not be calculated in the case of open end
distribution
12. Calculation of Range
In Individual Series
Question: Calculate the Range and its Coefficient from the following data
54, 45, 16,15, 75,85,28
Ans: Lowest value=15
Highest value= 85
Range = H-L
= 85-15
= 70
14. Calculation of Range in Discrete Series
(Ungrouped series)
It is obtained by Simply taking the difference between the highest and
lowest value . In this frequency of series is not taken in to consideration for
the calculation.
Example: Find the Range and Coefficient of Range from the Following
data
X 5 15 25 35 45 55
f 7 14 18 10 4 1
15. Solution:
Highest value (H)= 55
Lowest value (L)=5
Rang= H-L
= 55-5
= 50
C
o
e
f
f
i
c
i
e
n
t
o
f
R
a
n
g
e
=
H
-
L
H
+
L
=
5
5
-
5
5
5
+
5
=
5
0
6
0
=
0
.
8
3
16. Calculation of Range: Continuous Data or
Grouped Data
It is calculated by taking the difference of mid value of Highest and lowest
class interval
Example
Find the Range and Coefficient of Range from the Following Data.
Data 0-10 10-20 20-30 30-40 40-50
Frequency 1 5 10 13 9
17. Data Mid value Frequency
0-10 5 1
10-20 15 5
20-30 25 10
30-40 35 13
40-50 45 9
18. Solution
Hence H= 45 L=5
Range = H-L
=45-5=40
C
o
e
f
f
i
c
i
e
n
t
o
f
R
a
n
g
e
=
H
-
L
H
+
L
=
4
5
-
5
4
5
+
5
=
4
0
5
0
=
0
.
8
19. Standard deviation (S.D.)
It is widely used method of Studying Absolute Dispersion
Greater SD means greater Dispersion
20.
21.
22.
23.
24. Method to calculate Standard Deviation A
Step1: Calculate mean
Step 2: Find the deviation of observation from the mean i.e. dx
Step 3: take the Square of these deviation i.e. dx2
Step 4: Take Summation of Their Squared deviation i.e ∑dx2
Step 5: Apply the formula
S . D . =
d x 2
N
25. Question
Find out the Standard Deviation from the following data.
3,7,8,9,10
Solution:
N=5
X dx=X-x¯ dx2= ( X-x¯)2
3 -4.4 19.36
7 -0.4 0.16
8 0.6 0.36
9 1.6 2.56
10 2.6 6.76
∑X= 37 ∑dX2= ( X-x¯)2 =
29.2
X
N
A
M
(
x
¯
)
=
=
3
7
5
=7
.4
27. Calculation of Standard Deviation-
Discrete Series ( Ungrouped Data)
Step1: Calculate the mean of Series
Step2: Find deviations for various item from the mean dx=X-x¯
Step3: Square the deviation dx2
Step 4: Multiply frequency with dx2 and get value of fdx2
Step 5: Apply the Formula
S . D . =
f d x 2
f
28. Question
Calculate the Standard Deviation from the Following Series
Age (in years) 15 25 35 45 55 65
No. of person 7 25 20 16 11 6
29. Solution
Ag
e
(X)
No. of
person
(f)
fx dx=X-x¯ dx2 fdx2
15 7 105 -22 484 3388
25 25 625 -12 144 3600
35 20 700 -2 4 80
45 16 720 8 64 1024
55 11 605 18 324 3564
65 6 390 28 784 4704
∑ f = 85 ∑ fx=
3145
∑ fdx2 =16360
AM (x¯)=
fx
f
=
3145
85
= 37
31. Calculation of Standard Deviation-
Continuous Series( Grouped Data)
Step 1: Find out the Mid value of Each Class interval
Step 2: Assume one of the mid value as an average denote by A.
Step 3: Find out Deviation by formula
Where H=Class interval Xi= Mid value
Step 4: Find out di2, fi di and fi di2
Step 5: Calculate Standard deviation by using the Formula
di=
Xi-A
H
S . D . =
f d i 2
f i
H X
f d i
f i
2
32. Examples
In a survey of 150 families in a village, The following distribution of ages of
children was found
Age of
children
0-2 2-4 4-6 6-8 8-10
No. of Families 40 32 25 23 30
33. Solution:
Let the Assumed mean A=5 an Class Interval H=2
Class
interval
Mid
Value
(x)
Frequency
(fi)
di2 fidi fidi2
0-2 1 40 -2 4 -80 160
2-4 3 32 -1 1 -32 32
4-6 5 (A) 25 0 0 0 0
6-8 7 23 1 1 23 23
8-10 9 30 2 4 60 120
Total ∑ fi=150 ∑ fidi= -
29
∑ fidi2=335
di=
Xi-A
H
35. S.D.=
f di2
fi
H X
f di
fi
2
S.D.=
335
2 X
150
2
150
-29
S.D.= 2 X 2.233 0.03737
S.D.= 2 X 2.195
S.D.= 2 X 1.481 = 2.962
36. Variance And Coefficient of Variance
The square of Standard deviation is called Variance. It has a Significant
role in Inferential Statistics. It is denoted by σ2 or (S.D.)2 and defined as
=
V a r i a n c e
X - x ¯ 2
N
=
V a r i a n c e
d x 2
N
=
V a r i a n c e
X - x ¯ 2
N
. f
F o r R o w d a t a
F o r F r e q u e n c y D i s t r i b u t i o n
37. Coefficient of Variance
It is used to compare the variability of one character in two different group
having different magnitude of the value or two characters in the same
group by expressing in percentage
Coefficient of Variance (CV) =
S.D.
Mean
X 100
CV=
X
X 100
or
38. Question
The mean and Standard deviations of the no. of Students of two School A
and B are given below
School Mean S.D.
A 450 52
B 470 55
39. Solutions
Coefficient of Variance (CV) =
S.D.
Mean
X 100
CV of School A=
52
450
X100 = 11.55
CV of School B =
55
470
X100 = 11.70
Hence Variability of both school are nearly equal