Measures of Skewness.pptx

UNIT : IV
MEASURES
OF
SKEWNESS
Mrs. D. Melba Sahaya Sweety RN,RM
PhD Nursing , MSc Nursing (Pediatric Nursing), BSc
Nursing
Associate Professor
Department of Pediatric Nursing
Enam Nursing College, Savar,
1

INTRODUCTION
• The skewness of a distribution is defined as the lack of
symmetry. A distribution is said to be 'skewed' when the
mean and the median fall at different points in the
distribution, and the balance (or centre of gravity) is
shifted to one side or the other-to left or right.
• Measure of Dispersion tells us about the variation of the
data set. Skewness tells us about the direction of
variation of the data set.
2

CONCEPT OF SKEWNESS
The concept of skewness helps us to understand the
relationship between three measures; mean, median
and mode.
• If, in a distribution, Mean = Median = Mode, then
that distribution is known as Symmetrical
Distribution.
• If, in a distribution, Mean ≠ Median ≠ Mode, then it
is not a symmetrical distribution and it is called a
Skewed Distribution
• When mean > median > mode, skewness will be
positively Skewed Distribution.
• When mean < median < mode, skewness will be
negatively skewed Distribution.
3

TYPESOF DISTRIBUTION
Types of Distribution
Symmetrical
Distribution
Skewed
Distribution
Positively
Skewed
Negatively
Skewed
J shape
Skewed
4

A frequency distribution is said to be
symmetrical if the frequencies are
equally distributed on both the sides
of central value. A symmetrical
distribution may be either bell –
shaped or U shaped.
A symmetrical distribution the
frequencies are first steadily rise
and then steadily fall. There is only
one mode and the values of mean,
median and mode are equal.
SYMMETRICAL
DISTRIBUTION
Mean = Median = Mode
5

A frequency distribution is
said to be positively skewed
distribution if the
frequencies are distributed
on right sides of central
value. In positive skewed
distribution the right tail is
longer.
POSITIVELYSKEWED
DISTRIBUTION
Mean > Median >Mode
6

A frequency distribution is
said to be positively skewed
distribution if the
frequencies are distributed
on right sides of central
value. In positive skewed
distribution the right tail is
longer.
NEGATIVELY SKEWED
DISTRIBUTION
Mean < Median < Mode
7

The case of extreme positive
skewness would arise when
frequencies are highest in the lowest
values and then they steadily fall as
the values increase. Similarly, the
extreme negative skewness would
arise when frequencies are lowest in
the lower values and they steadily
increase as values increase the
highest frequency representing the
highest values: Such distribution is
called 'J' shaped Skewed
distribution.
J SHAPED SKEWED
DISTRIBUTION
Mean ˃ Mode
Mean ˃ Median
Mean < Mode
Mean < Median
8

Measures
of
Skewness
Absolute Measures
of Skewness
Relative Measures
of Skewness
MEASURES OF SKEWNESS
9

• The absolute measures of skewness is based on the
difference between mean and mode or mean and median
• Following are the absolute measures of skewness:
• 1. Skewness (Sk) = Mean – Median
• 2. Skewness (Sk) = Mean – Mode
• 3. Skewness (Sk) = (Q3 - Q2) - (Q2 - Q1) .
Absolute Measures of
Skewness
10

In order to make valid comparison between the skewness of two or more
distributions we have to eliminate the distributing influence of variation. Such
elimination can be done by dividing the absolute skewness by standard
deviation. The following are the important methods of measuring relative
skewness:
[1] Karl – Pearson’ s coefficient of skewness
[2] Bowley’ s coefficient of skewness.
[3] Kelly's Measure of Skewness
[4] Moment Coefficient of skewness
Relative Measures of
Skewness
11

• This method is most frequently used for measuring skewness. The formula
for measuring coefficient of skewness is
• The value of this coefficient would be zero in a symmetrical distribution.
If mean is greater than mode, coefficient of skewness would be positive
otherwise negative. The value of the Karl Pearson’s coefficient of
skewness usually lies between ±1 for moderately skewed distubution.
Karl – Pearson’s coefficient of skewness
12

• If mode is not well defined, we use the formula
• Coefficient usually lies between -3 and +3 In practice it is rarely
obtained.
13

• Example 1, Compute the Karl Pearson's coefficient of skewness from the
following data 10 15 15 15 15 20 20 25 35
X = 10 + 15+ 15 + 15+ 20+ 20 + 25 + 35
9
X = 155/9 = 17.2
X X – X (x-x )2
10 10 – 17.2 = - 7.2 51.84
15 15 – 17.2 = - 2.2 4.84
15 15 – 17.2 = - 2.2 4.84
15 15 – 17.2 = - 2.2 4.84
20 20 – 17.2 = 2.8 7.84
20 20 – 17.2 = 2.8 7.84
25 25 – 17.2 = 7.8 60.84
35 35 – 17.2 = 17.8 316.8
459.72
σ =√459.72
9
σ = √51.08
σ = 7.1
Skp = 17.2 – 15
7.1
= 0.30
14

Example 2, Compute the Karl Pearson's coefficient
of skewness from the following data:
Marks scored by the students Number of students
58 10
59 18
60 30
61 42
62 35
63 28
64 16
65 8 15

x f d = X- A d2 fd fd2
58 10 58 – 61 = -3 9 -30 90
59 18 59 – 61 = -2 4 -36 72
60 30 60 – 61 = -1 1 -30 30
61 42 61 – 61 = 0 0 0 0
62 35 62 – 61 = 1 1 35 35
63 28 63 – 61 = 2 4 56 112
64 16 64 – 61 = 3 9 48 144
65 8 65 – 61 = 4 16 32 128
N =
187
Σfd
=75
Σfd2 =
611
16

Mode = 61
Mean
X = 61 + 75 /187
X = 61 + 0.40
X = 61.4
Standard deviation
σ =√611 - 75 2
187 187
σ = √3.26 – 0.17
σ = √3.09
σ = 1.76
Skp = 61.4 – 61 = 0.22
1.76
Thus the
distribution is
positively
Skewed
17

• Example 3 ,From the marks secured by 120 students
in Sections A and B of a class of 120 students, the
following measures are obtained:
• SectionA : X= 46.83, σ = 14.8, Mode = 51.67
SectionB : X = 47.83, σ = 14.8, Mode = 47.07
Determine which distribution of marks is more
skewed
18

Section A
Skp = 46.83 – 51.67
14.8
Skp = - 4.84
14.8
Skp = - 0.37
Section B
Skp = 47.83 – 47.07
14.8
Skp = 0.76
14.8
Skp = 0.05
Hence the
distribution of marks
in Section A is more
skewed. The
skewness for Section
Ais negative, while
that of B is positive.
19

Class 0 – 10 10 - 20 20 - 30 30 - 40 40 - 50 50 – 60 60 - 70 70 - 80
f 15 15 23 22 25 10 5 10
20

Mid
x
f d = X - A
h
fd d2 fd2
5 15 -3 - 45 9 135
15 15 - 2 - 30 4 60
25 23 -1 - 23 1 23
35 22 0 0 0 0
45 25 1 25 1 25
55 10 2 20 4 40
65 5 3 15 9 45
75 10 4 40 16 160
N =
125
Σfd = 2 Σfd2 = 488
21

• Mode
l = 40, f1 = 25, f2 = 10, f0 = 22, h = 10
Z = 40 + 25 – 22 x 10
(2 x 25) – (22 – 10)
Z = 40 + 3/ 38 x 10
Z = 40 + 0.79
Z = 40.79
Mean
X = 35 + 2/125 x 10
X = 35+ 0.016 x 10
X = 35 + 0.16
X = 35.16
Standard deviation
σ =√488 - 2 2
125 125 x 10
σ = √3.9 – 0.000256 x 10
σ = √3.9 x 10
σ = 1.97 x10 = 19.7
Karl pearson
coefficient of skewnes
Skp = 35.16 – 40.79
19.7
= - 0.28
22

CI f
0 - 10 10
10 - 20 40
20 - 30 20
30 - 40 0
40 - 50 10
50 - 60 40
60 - 70 16
70 - 80 14 23

CI f LCF Mid
x
d = X – A
h
(A= 35)(h = 10)
d2 fd fd2
0 - 10 10 10 5 - 3 9 - 30 90
10 - 20 40 50 15 - 2 4 - 80 160
20 - 30 20 70 25 - 1 1 - 20 20
30 - 40 0 70 35 0 0 0 0
40 - 50 10 80 45 1 1 10 10
50 - 60 40 120 55 2 4 80 160
60 - 70 16 136 65 3 9 48 144
70 - 80 14 150 75 4 16 56 244
N= 150 Σfd = 64 Σfd2 = 828 24

• Median
N/2 = 150/2 = 75 l = 40, m = 70, h = 10
M = 40 + (75 – 70 ) x 10
10
M = 40 + 5/ 10 x 10
M = 40 + 0.5 x 10
M = 40 + 5 = 45
Mean
X = 35 + 64/150 x 10
X = 35+ 0.42 x 10
X = 35 + 4.2
X = 39.2
Standard deviation
σ =√828 - 64 2
150 150 x 10
σ = √5.52 – 0.1849 x 10
σ = √5.3351 x 10
σ = 2.3 x10 = 23
Karl pearson coefficient of
skewnes
Skp = 3x (39.2 – 45)
23
Skp = 3 x (- 0.25)
Skp = - 0.75
The distribution is negatively Skewed
25

• In Karl- Pearson’s method of measuring skewness the whole of the series is
needed. Prof. Bowley has suggested a formula based on relative position of
quartiles. In a symmetrical distribution, the quartiles are equidistant from the
value of the median; ie., Median – Q1 = Q3 – Median.
• But in a skewed distribution, the quartiles will not be equidistant from the
median. Hence Bowley has suggested the following formula:
Bowley’s coefficient of skewness
If Q3 +Q1 > 2M then coefficient of skewness is Positive
If Q3 +Q1 < 2M then coefficient of skewness is Negative
26

• Example : 1, Find Bowley’ s coefficient of skewness of the following series 2, 4, 6, 8, 10,
12, 14, 16, 18, 20, 22
Q1 = 11 + 1 th
4 term
= (12/4)th term
= (3)th term
Q1 = 6
Q3 = 3 11+ 1 th
4 term
= (3 x 12/4)th term
= (36 / 4 )th term
= (9)th term
Q3 = 18
Median = 11 + 1 th
2 term
= (12/2)th term
= (6)th term
M = 12
SkB = (18+6) – 2x12
18 - 6
= 24 -24
12
SkB = 0
The given data is
symmetrical 27

Example 2, The following table shows the distribution of 128 families
according to the number of children. Find the Bowley’s coefficient of
Skewness
No. of Children No. of families
0 20
1 15
2 25
3 30
4 18
5 10
6 6
7 3
8 1 28

x f CF
0 20 20
1 15 35
2 25 60
3 30 90
4 18 108
5 10 118
6 6 124
7 3 127
8 1 128
N = 128
N = 128
Median = 128 + 1 th
2 term
= (129/2)th term
= (64.5)th term
M = 3
Median
29

Q1 = 128 + 1 th
4 term
= (129/4) th term
= (32.25) th term
Q1 = 1
Q3 = 3 128 + 1 th
4 term
= (3 x 129/4) th term
= (387 / 4 ) th term
= (96.75)th term
Q3 = 4
SkB = (4+1) – 2 x 3
4 - 1
= 5 - 6
3
SkB = - 0.333
Since SkB < 0
distribution is skewed
left.
30

Example 3, Find Bowley’s Coefficient of skewness from the following
data.
(x) (f)
0 - 100 25
100 - 200 6
200 - 300 17
300 - 400 37
400 - 500 40
500 - 600 20
600 - 700 15
31

(x) (f) CF
0 - 100 25 25
100 - 200 6 31
200 - 300 17 48 Q1
300 - 400 37 85
400 - 500 40 125 Q3
500 - 600 20 145
600 - 700 15 160
N= 160
Median
N/2 = 160/2 = 80 , l = 300 ,
m = 48, C = 100 , f = 37
M = 300 + (80 – 48) / 37 x 100
M = 300 + (32/37) x 100
M = 300 + 0.86 x 100
M = 300 + 86
M = 386
32

N/4 = 160/4 = 40, l1= 200,
m1 = 31, f1 = 17, C = 100
Q1 = 200 + (40 – 31) / 17 x 100
= 200 + (9/17) x 100
= 200 + 52.9
Q1 = 252.9
SkB = (487.5+252.9) – 2 x286
487.5 – 252.9
= 740.4 - 572
234.6
SkB = 0. 717
Since SkB > 0 distribution is
skewed right. The distribution
is Positively Skewed
3 x N/4 = 3x 40 = 120, l1= 400,
m1 = 85, f1 = 40
Q3 = 400 + 120 - 85 x 100
40
= 400 + 35 x 100
40
= 400 + 0.875 x 100
= 400 + 87.5
Q3 = 487.5
33

• Bowley’s measure of skewness is based on the middle 50% of the observation
because it leaves 25% of the observations on each extreme of the distribution. As
an improvement over Bowley’s measure, Kelly has suggested a measure based
on P10 and P90 so that only 10% of the observations on each extreme are
ignored. Kelly’s coefficient of skewness, denoted by Skk, is given by
Kelly’s coefficient of skewness
If SKk > 0 Positively Skewed
If SKk < 0 Negatively Skewed
OR
34

• Example : 1 find Kelly’s coefficient of Skewness by the following
data 282, 754, 125, 765,875,645,985, 235,175,895,905,112,155
Solution : Arrange the data in a ascending order
112,125,155,175,235,282,645,754,765, 875,895,905,935
Percentile = n+1 = 13+1 = 14/100 = 0.14
100 100
P90 = 90 x n + 1 = 90 x 0.14 = 12.6
100
P90 = 12th item + 0.6 (13th item – 12th item)
= 905 +0.6 (985 – 905)
= 905 + 0.6 (80) = 905 + 48
P90 = 953
35

P10 = 10 x n + 1 = 10 x 0.14 = 1.4
100
P10 = 1th item + 0.4 (2th item – 1th item)
= 112 +0.4 (125 – 112)
= 112 + 0.4 (13) = 112 + 5.2
P10 = 117.2
Median = n + 1 th
2 term
M = 13 +1 th
2 term
= (7 )th term
Median = 645
SKK = (953 + 117.2) - (2 x 645) = (953 + 117.2) - 1290
953 – 117.2 835.8
SKK = 1070.2 – 1290 = - 219.8
835.8 835.8
SKK = - 0.26 it is negatively Skewed 36

• Example – 2, find the Kelly’s coefficient of skewness by the following data
x f
30 8
32 12
35 20
38 10
40 5
Solution:
x f c.f
30 8 8
32 12 20
35 20 40
38 10 50
40 5 55
N = 55
Percentile = N+1 = 55+1
100 100
= 56/100 = 0.56
P90 = 90 x N + 1
100
= 90 x 0.56 = 50.4
P90 = 40
P10 = 10 x N + 1
100
= 10 x 0.56 = 5.6
P10 = 30
Median = N + 1 th
2 term
M = 55 +1 th
2 term
= (28 )th term
Median = 35
SKK = (40 + 30) - (2 x 35) = 70 - 70
40 – 30 10
SKK = 0/10
SKK = 0 there is no skewness 37

CI f
10-20 6
20-30 8
30-40 12
40-50 10
50-60 5
60-70 4
Solution:
x f c.f
10-20 6 6
20-30 8 14
30-40 12 26
40-50 10 36
50-60 5 41
60-70 4 45
N = 45
Percentile = N = 45
100 100
= 0.45
90 x N = 90 x 0.45 = 40.5 , l = 50 , m = 36, f = 5, c = 10
100
P90 = 50 + (40.5 – 36) x 10 = 50 + (4.5) x 10
5 5
P90 = 50 + 0.9 x 10
P90 = 50 + 9
P90 = 59
38

10 x N = 10 x 0.45 = 4.5
100
L = 10, m = 0, c = 10 , f = 6
P10 = 10 + (4.5 – 0) x10
6
P10 = 10 + 4.5 x 10 = 10 + 0.75 x 10
6
P10 = 10 + 7.5
P10 = 17.5
N/2 = 45/2 = 22.5 , l = 30 ,
m = 14, C = 10 , f = 12
M = 30 + (22.5 – 14) x 10
12
M = 30 + (8.5 /12) x 10
M = 30 + 0.708x 10
M = 30 + 7.08
M = 37.08
SKK = (59 + 17.5) - (2 x 37.08)
59 – 17.5
SKK = 76.5 – 74.166
41.5
SKK = 2.334
41.5
SKK = 0.056
Positively skewed
39

• Example : 4 find Kelly’s coefficient of Skewness by the following
data by deciles formula 282, 754, 125, 765,875,645,985,
235,175,895,905,112,155
Solution : Arrange the data in a ascending order
112,125,155,175,235,282,645,754,765, 875,895,905,935
D1 = n + 1 th = 13 +1 th = 14 th = (1.4 ) th term
10 term 10 term 10 term
D1 = 1th item + 0.4 (2th item – 1th item)
= 112 + 0.4 (125 – 112)
= 112 + 0.4 (13)
= 112 + 5.2
D1 = 117.2
40

D9 = 9 x n + 1 th = 9 13 + 1 th
10 term 10 term
= 9 x (14/10)th term = 9 x (1.4)th term
= ( 12.6 )th term
D9 = 12th item + 0.6 (13th item – 12th item)
= 905 +0.6 (985 – 905)
= 905 + 0.6 (80) = 905 + 48
D9 = 953
Median = n + 1 th
2 term
M = 13 +1 th
2 term
= (7 )th term
Median = 645
SKK = (953 + 117.2) - (2 x 645) = 1070.2 - 1290 = 219.8
953 – 117.2 835.8 835.
SKK = - 0.26 it is negatively Skewed 41

CI f
10-20 6
20-30 8
30-40 12
40-50 10
50-60 5
60-70 4
Solution:
x f c.f
10-20 6 6
20-30 8 14
30-40 12 26
40-50 10 36
50-60 5 41
60-70 4 45
N = 45
9 x N = 9 x 45 = 405 = 40.5, l = 50 , m = 36, f = 5, c = 10
10 10 10
D9 = 50 + (40.5 – 36) x 10 = 50 + (4.5) x 10
5 5
D9 = 50 + 0.9 x 10
D9 = 50 + 9
D9 = 59
42

N = 45 = 4.5
10 10
L = 10, m = 0, c = 10 , f = 6
D1= 10 + (4.5 – 0) x10
6
D1 = 10 + 4.5 x 10 = 10 + 0.75 x 10
6
D1 = 10 + 7.5
D1 = 17.5
N/2 = 45/2 = 22.5 , l = 30 ,
m = 14, C = 10 , f = 12
M = 30 + (22.5 – 14) x 10
12
M = 30 + (8.5 /12) x 10
M = 30 + 0.708x 10
M = 30 + 7.08
M = 37.08
SKK = (59 + 17.5) - (2 x 37.08)
59 – 17.5
SKK = 76.5 – 74.166
41.5
SKK = 2.334
41.5
SKK = 0.056
Positively skewed
43

Moments are a set of statistical parameters to measure a distribution. or
Moments are the mean of various powers of deviation of items. If the
deviations are about the arithmetic mean, the moments are called central
moments.
Whenever the deviation are taken from values other than the mean, the
moments are called raw moments or arbitrary moments or non – central
moments.
Karl Pearson defined the β and γ coefficients of skewness, based upon the
second and third central moments: It is used as measure of skewness.
Pearson’s Moment coefficient of skewness
44

For a symmetrical distribution, β1 shall be zero. β1as a measure of skewness does not
tell about the direction of skewness, i.e. positive or negative. Because µ3being the
sum of cubes of the deviations from mean may be positive or negative but µ3
2 is
always positive. Also, µ2 being the variance always positive. Hence, β1 would be
always positive. This drawback is removed if we calculate Karl Pearson’s Gamma
coefficient γ1 which is the square root of β1 i. e.
45

These coefficients are pure numbers independent of units of
measurement and as such can be conveniently used for comparative
studies
Interpretation:
[1] If 1 γ1 < 0 , the distribution is negatively skewed.
[2] If 1 γ1 = 0 , the distribution is symmetric.
[3] If 1 γ1 > 0 , the distribution is positively skewed.
46

• Example : 1 calculate Pearson's moment coefficient of skewness
from the following data 2, 3, 7, 8, 10, 12, 14
•
Mean
X = 2+3+7+8+10+12+14 = 56
7 7
X = 8
X X - X (X – X)2 (X – X)3
2 -6 36 -216
3 -5 25 -125
7 -1 1 -1
8 0 0 0
10 2 4 8
12 4 8 64
14 6 36 216
110 - 54
µ3 = - 54 / 7 = - 7.7
µ2 = 110 / 7 = 15.7
γ1 = - 7.7 = -7.7
√(15.7 ) 62.2
γ1 = - 0.12 47
3

• Example – 2, find the Pearson’s moment coefficient of skewness by the following data
Solution:
x f fx (X – X) (X – X)2 f (X – X)2 (X – X)3 f(X – X)3
30 8 240 -4.6 21.16 169.28 -97.336 -778.688
32 12 384 -2.6 6.76 81.12 -17.576 -210.912
35 20 700 0.4 0.16 3.2 0.064 1.28
38 10 380 3.4 11.56 115.6 39.304 393.04
40 5 200 5.4 29.16 145.8 157.464 787.32
N = 55 1904 515 192.04
Wages per day in taka 30 32 35 38 40
No .of workers 8 12 20 10 5
48

µ2 = 515
55
µ2 = 9.36
µ2
3 = 192.04
55
µ2
3 = (3.49 )
µ2
3 = 42.50
γ1 = 3.49
√ 42.50
γ1 = 3.49
6.51
γ1 = 0.53
49
X = Σfx = 1904
Σf 55
X = 34.6
Mean
3
3

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