Kelly’s Coefficient of Skewness for ungrouped data
For a symmetric distribution, the first decile namely $D_1$ and nineth decile $D_9$ are equidistance from the median i.e. $D_5$. Thus, $D_9 - D_5 = D_5 -D_1$.
The coefficient of skewness based on deciles is defined as
$S_k = \dfrac{D_9+D_1 - 2D_5}{D_9 -D_1}$
OR
$S_k =\dfrac{P_{90}+P_{10} -2*Median}{P_{90}-P_{10}}$
where,
$D_1 (=P_{10})$is the first decile (or tenth percentile) of the data$Median=D_5 (=P_{50})$is the median of the data$D_9 (=P_{90})$is the ninth decile (or nineteenth percentile) of the data.
Interpretation
- If $S_k<0$, the data is negatively skewed.
- If $S_k=0$, the data is symmetric(i.e., not skewed).
- If $S_k>0$, the data is positively skewed.
Kelly’s coefficient of skewness ranges from -1 to +1.
Proof
We know that, if $a>0$ and $b>0$, then $|a-b|\leq |a+b|$, $$ \begin{aligned} & \text{i.e., } \bigg|\dfrac{a-b}{a+b} \bigg| \leq 1 \end{aligned} $$
Now, taking $a= D_9 - D_5$ and $b= D_5-D_1$ in above inequality we get
$$ \begin{aligned} & \bigg|\dfrac{(D_9 - D_5)-(D_5-D_1)}{(D_9 - D_5)+(D_5-D_1)}\bigg| \leq 1\\ &\Rightarrow \bigg|\dfrac{D_9 + D_1-2D_5}{D_9 -D_1}\bigg| \leq 1\\ & \Rightarrow |S_k|\leq 1\\ & \Rightarrow -1\leq S_k\leq 1. \end{aligned} $$
Thus, Kelly’s coefficient of skewness ranges from -1 and +1.
Example 1
The marks obtained by a sample of 20 students in a class test are as follows:
20, 30, 21, 29, 10, 17, 18, 15, 27, 25,
16, 15, 19, 22, 13, 17, 14, 18, 12, 9.
Find Kelly’s Coefficient of Skewness.
Solution
Kelly’s coefficient of skewness is
$$ \begin{aligned} S_k &= \frac{D_9+D_1 - 2D_5}{D_9 -D_1}\\ \end{aligned} $$
The formula for $i^{th}$ decile is
$D_i =$ Value of $\bigg(\dfrac{i(n+1)}{10}\bigg)^{th}$ observation, $i=1,2,3,\cdots, 9$
where $n$ is the total number of observations.
Arrange the data in ascending order
9, 10, 12, 13, 14, 15, 15, 16, 17, 17
18, 18, 19, 20, 21, 22, 25, 27, 29, 30
First Decile $D_1$
The first decile $D_1$ can be computed as follows:
$$ \begin{aligned} D_1 &=\text{Value of }\bigg(\dfrac{1(n+1)}{10}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{1(20+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(2.1\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(2\big)^{th} \text{ obs.}+0.1 \big(\text{Value of } \big(3\big)^{th}\text{ obs.}-\text{Value of }\big(2\big)^{th} \text{ obs.}\big)\\ &=10+0.1\big(12 -10\big)\\ &=10.2 \end{aligned} $$
Fifth Decile $D_5$
The fifth decile $D_5$ can be computed as follows:
$$ \begin{aligned} D_5 &=\text{Value of }\bigg(\dfrac{5(n+1)}{10}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{5(20+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(10.5\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(10\big)^{th} \text{ obs.}+0.5 \big(\text{Value of } \big(11\big)^{th}\text{ obs.}-\text{Value of }\big(10\big)^{th} \text{ obs.}\big)\\ &=17+0.5\big(18 -17\big)\\ &=17.5 \end{aligned} $$
Nineth Decile $D_9$
The nineth decile $D_9$ can be computed as follows:
$$ \begin{aligned} D_9 &=\text{Value of }\bigg(\dfrac{9(n+1)}{10}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{9(20+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(18.9\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(18\big)^{th} \text{ obs.}+0.9 \big(\text{Value of } \big(19\big)^{th}\text{ obs.}-\text{Value of }\big(18\big)^{th} \text{ obs.}\big)\\ &=27+0.9\big(29 -27\big)\\ &=28.8 \end{aligned} $$
Kelly’s coefficient of skewness
Kelly’s coefficient of skewness is
$$ \begin{aligned} S_k &= \frac{D_9+D_1 - 2D_5}{D_9 -D_1}\\ &=\frac{28.8+10.2 - 2* 17.5}{28.8 - 10.2}\\ &=0.2151 \end{aligned} $$
As the coefficient of skewness $S_k$ is $\text{greater than zero}$ (i.e., $S_k > 0$), the distribution is $\text{positively skewed}$.
Example 2
The following data gives the hourly wage rates (in dollars) of 25 employees of a company.
20, 28, 30, 18, 27, 19, 22, 21, 24, 25,
18, 25, 20, 27, 24, 20, 23, 32, 20, 35,
22, 26, 25, 28, 31.
Find the Kelly’s coefficient of skewness.
Solution
Kelly’s coefficient of skewness is
$$ \begin{aligned} S_k &= \frac{D_9+D_1 - 2D_5}{D_9 -D_1}\\ \end{aligned} $$
The sample size is $n = 25$.
The formula for $i^{th}$ decile is
$D_i =$ Value of $\bigg(\dfrac{i(n+1)}{10}\bigg)^{th}$ observation, $i=1,2,3,\cdots, 9$
where $n$ is the total number of observations.
Arrange the data in ascending order
18, 18, 19, 20, 20, 20, 20, 21, 22, 22
23, 24, 24, 25, 25, 25, 26, 27, 27, 28
28, 30, 31, 32, 35
First Decile $D_1$
The first decile $D_1$ can be computed as follows:
$$ \begin{aligned} D_1 &=\text{Value of }\bigg(\dfrac{1(n+1)}{10}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{1(25+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(2.6\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(2\big)^{th} \text{ obs.}+0.6 \big(\text{Value of } \big(3\big)^{th}\text{ obs.}-\text{Value of }\big(2\big)^{th} \text{ obs.}\big)\\ &=18+0.6\big(19 -18\big)\\ &=18.6\text{ dollars} \end{aligned} $$
Fifth Decile $D_5$
The fifth decile $D_5$ can be computed as follows:
$$ \begin{aligned} D_5 &=\text{Value of }\bigg(\dfrac{5(n+1)}{10}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{5(25+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(13\big)^{th} \text{ obs.}\\ &=24\text{ dollars} \end{aligned} $$
Nineth Decile $D_9$
The nineth decile $D_9$ can be computed as follows:
$$ \begin{aligned} D_9 &=\text{Value of }\bigg(\dfrac{9(n+1)}{10}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{9(25+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(23.4\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(23\big)^{th} \text{ obs.}+0.4 \big(\text{Value of } \big(24\big)^{th}\text{ obs.}-\text{Value of }\big(23\big)^{th} \text{ obs.}\big)\\ &=31+0.4\big(32 -31\big)\\ &=31.4\text{ dollars} \end{aligned} $$
Kelly’s coefficient of skewness
Kelly’s coefficient of skewness is
$$ \begin{aligned} S_k &= \frac{D_9+D_1 - 2D_5}{D_9 -D_1}\\ &=\frac{31.4+18.6 - 2* 24}{31.4 - 18.6}\\ &=0.1562 \end{aligned} $$
As the coefficient of skewness $S_k$ is $\text{greater than zero}$ (i.e., $S_k > 0$), the distribution is $\text{positively skewed}$.
