This preview shows page 1. Sign up to view the full content.
Unformatted text preview: y’ s coefficient of skewness
(iii)Measure of skewness based on moments
7.10.1 Karl – Pearson’ s Coefficient of skewness:
According to Karl – Pearson, the absolute measure of
skewness = mean – mode. This measure is not suitable for making
valid comparison of the skewness in two or more distributions
because the unit of measurement may be different in different
series. To avoid this difficulty use relative measure of skewness
called Karl – Pearson’ s coefficient of skewness given by:
Mean  Mode
Karl – Pearson’ s Coefficient Skewness =
S .D.
In case of mode is ill – defined, the coefficient can be determined
by the formula:
3(Mean  Median)
Coefficient of skewness =
S .D.
Example 18:
Calculate Karl – Pearson’ s coefficient of skewness for the
following data.
25, 15, 23, 40, 27, 25, 23, 25, 20
175 Solution:
Computation of Mean and Standard deviation :
Short – cut method.
Size
Deviation from A=25
D
0
25
10
15
2
23
15
40
2
27
0
25
2
23
0
25
5
20
∑d=2 N=9 d2
0
100
4
225
4
0
4
0
25
∑d2=362 ∑d
n
−2
= 25 +
9
= 25 – 0.22 = 24.78 Mean = A + = = σ ∑d2 ∑d − n
n
362 −2 − 9 9 2 2 40.22 − 0.05
= 40.17 = 6.3
Mode = 25, as this size of item repeats 3 times
Karl – Pearson’ s coefficient of skewness
= = Mean  Mode
S .D.
176 24.78 − 25
6.3
− 0.22
=
6.3
= − 0.03
= Example 19:
Find the coefficient of skewness from the data given below
Size :
3
45
6
7
8
9
10
Frequency: 7
10 14 35
102
136
43
8
Solution:
Size Frequency
(f) 3
4
5
6
7
8
9
10 7
10
14
35
102
136
43
8
N=355 Mean = A +
= 6+ Deviation
From A=6
(d)
3
2
1
0
1
2
3
4 ∑ fd
N 480
355 = 6 + 1.35
= 7.35
Mode = 8
Coefficient of skewness = d2 fd fd2 9
4
1
0
1
4
9
16 21
20
14
0
102
272
129
32
∑fd=480 63
40
14
0
102
544
387
128
∑fd2=1278 σ=
= ∑ fd 2 ∑ fd − N
N
1278 480 − 355 355 3.6 − 1.82
= 1.78 = 1.33
= Mean  Mode
S .D.
177 2 2 = 7.35  8
1.33 = 0.65
1.33 = −0.5 Example 20:
Find Karl – Pearson’ s coefficient of skewness for the given
distribution:
X : 05
510 1015 1520 2025 2530 3035 3540
F:
2
5
7
13
21
16
8
3
Solution:
Mode lies in 2025 group which contains the maximum frequency
f...
View
Full
Document
This note was uploaded on 01/18/2014 for the course BUS 100 taught by Professor Moshiri during the Winter '08 term at UC Riverside.
 Winter '08
 Moshiri
 Business

Click to edit the document details