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(MeanMedian)sd=3(¯xM)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=1nni=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=1n1(ni=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=1nni=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=1n1(ni=1x2i(ni=1xi)2n)=15(3190(138)26)=15(3190190446)=15(31903174)=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(MeanMedian)sd=3×(2323.5)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=1nni=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=1n1(ni=1x2i(ni=1xi)2n)=110(1405(121)211)=110(14051464111)=110(14051331)=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(MeanMedian)sd=3×(1110)2.7203=1.1028

As the value of sk>0, the data is positively skewed.

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