Karl Pearson coefficient of skewness for ungrouped data
Let xi,i=1,2,⋯,n be n observations.
Formula
The Karl Pearson’s coefficient skewness is given by
SK=3(Mean−Median)sd=3(¯x−M)sx
where,
- ¯x is the sample mean,
- M is the median,
- sx is the sample standard deviation.
Sample mean
The sample mean ¯x is given by
¯x=1nn∑i=1xi
Sample Median
Arrange the data in ascending order of magnitude.
Median of X is given by
Md={value of (n+12)th obs.,if n is odd;average of (n2)th and (n2+1)th obs.,if n is even.
Sample Standard deviation
sample standard deviation is given by
sx=√s2x=√1n−1(n∑i=1x2i−(∑ni=1xi)2n)
Example 1
The age (in years) of 6 randomly selected students from a class are
22,25,24,23,24,20.
Find the Karl Pearson’s coefficient of skewness.
Solution
xi | x2i | |
---|---|---|
22 | 484 | |
25 | 625 | |
24 | 576 | |
23 | 529 | |
24 | 576 | |
20 | 400 | |
Total | 138 | 3190 |
Sample mean
The sample mean of X is
¯x=1nn∑i=1xi=1386=23 years
The average of age of students is 23 years.
Sample Median The data in ascending order of magnitude is 20,22,23,24,24,25.
Here n=6 which is even.
Sample median = average of (n2)th and (n2+1)th observations.
Thus the median age of students is
M=(62)thObs.+(62+1)thObs.2=(3)thObs.+(4)thObs.2=23+242=23.5 years.
The median age of students is M=23.5 years.
Sample variance
Sample variance of X is
s2x=1n−1(n∑i=1x2i−(∑ni=1xi)2n)=15(3190−(138)26)=15(3190−190446)=15(3190−3174)=165=3.2
Sample standard deviation
The standard deviation is the positive square root of the variance.
The sample standard deviation is
sx=√s2x=√3.2=1.7889 years
Thus the standard deviation of age of students is 1.7889 years.
Karl Pearson’s coefficient of skewness
The Karl Pearson’s coefficient skewness is
sk=3(Mean−Median)sd=3×(23−23.5)1.7889=−0.8385
As the value of sk<0, the data is negatively skewed.
Example 2
A random sample of 11 patients yielded the following data on the length of stay (in days) in the hospital.
12,9,10,15,10,14,7,10,8,11,15
Find the Karl Pearson’s coefficient of skewness.
Solution
xi | x2i | |
---|---|---|
12 | 144 | |
9 | 81 | |
10 | 100 | |
15 | 225 | |
10 | 100 | |
14 | 196 | |
7 | 49 | |
10 | 100 | |
8 | 64 | |
11 | 121 | |
15 | 225 | |
Total | 121 | 1405 |
Sample mean
The sample mean of X is
¯x=1nn∑i=1xi=12111=11 days
The average of length of stay in the hospital is 11 days.
Sample Median
n=11 which is odd.
The data in ascending order of magnitude is 7,8,9,10,10,10,11,12,14,15,15.
Sample median = average of (n2)th and (n2+1)th observations
That is
M=value of (n+12)th obs.=value of (11+12)th obs.=value of (6)thObs.=10 days
The median length of stay in the hospital is M=10 days.
Sample variance
Sample variance of X is
s2x=1n−1(n∑i=1x2i−(∑ni=1xi)2n)=110(1405−(121)211)=110(1405−1464111)=110(1405−1331)=7410=7.4
Sample standard deviation
The standard deviation is the positive square root of the variance.
The sample standard deviation is
sx=√s2x=√7.4=2.7203 days
Thus the standard deviation of length of stay in the hospital is 2.7203 days.
Karl Pearson’s coefficient of skewness
The Karl Pearson’s coefficient skewness is
sk=3(Mean−Median)sd=3×(11−10)2.7203=1.1028
As the value of sk>0, the data is positively skewed.