Measure of Dispersion final.pdf

Measure of Dispersion
DR. ANSHU DUDHE
PROFESSOR,
NAGPUR COLLEGE OF PHARMACY, NAGPUR

 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.

 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.

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

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

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.

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

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

C
o
e
f
f
i
c
i
e
n
t
o
f
R
a
n
g
e
=
H
-
L
H
+
L
=
8
5
-
1
5
8
5
+
1
5
=
7
0
1
0
0
=
0
.
7

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

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

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

Data Mid value Frequency
0-10 5 1
10-20 15 5
20-30 25 10
30-40 35 13
40-50 45 9

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

Standard deviation (S.D.)
It is widely used method of Studying Absolute Dispersion
Greater SD means greater Dispersion

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

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

S
.
D
.
=
d
x
2
N
S
.
D
.
=
2
9
.
2
5
S
.
D
.
=
5
.
8
4
S
.
D
.
=
2
.
4
1
6

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

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

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

S
.
D
.
=
f
d
x
2
S
.
D
.
=
1
6
3
6
0
8
5
S
.
D
.
=
1
9
2
,
5
S
.
D
.
=
1
3
.
8
7
f

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

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

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

M
e
a
n
=
A
+
f
d
i
=
5
+
N
X
H
-
2
9
1
5
0
X
2
=
5
+
(
-
0
.
3
8
8
6
)
=
4
.
6
1
1
4

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

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

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

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

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

Measure of Dispersion final.pdf

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Measure of Dispersion final.pdf