# Kelly’s Coefficient of Skewness for Ungrouped data | Formula | Examples

## Kelly's Coefficient of Skewness for Ungrouped data

Kelly's coefficient of skewness is based on deciles or percentiles of the data. The Bowley's coefficient of skewness is based on the middle 50 percent of the observations of data set. It means the Bowley's coefficient of skewness leaves the 25 percent observations in each tail of the data set.

Kelly suggested a measure of skewness which is based on middle 80 percent of the observations of data set.

For a symmetric distribution, the first decile namely $D_1$ and ninth decile $D_9$ are equidistant from the median i.e. $D_5$. Thus, $D_9 - D_5 = D_5 -D_1$.

The Kelley's coefficient of skewness based is defined as

 \begin{aligned} S_k &= \frac{D_9+D_1 - 2D_5}{D_9 -D_1}\\ & OR \\ S_k &=\frac{P_{90}+P_{10} - 2P_{50}}{P_{90} -P_{10}} \end{aligned}

where,

• $D1=P{10}$ is the first decile or tenth percentile,
• $D5=P{50}$ is the fifth decile or fiftieth percentile,
• $D9=P{90}$ is the ninth decile or nineteenth percentile.

#### 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 Calculator for ungrouped data

Use this calculator to find the Kelly's coefficient of skewness for ungrouped (raw) data.

Kelly's Coeff. of Skewness Calculator
Enter the X Values (Separated by comma,)
Results
Number of Obs. (n):
Ascending order of X values :
First Decile : ($D_1$)
Sample Median : ($D_5$)
Ninth Decile : ($D_9$)
Kelly's Coeff. of Skewness :

## How to calculate Kelly's Coefficient of Skewness for ungrouped data?

Step 1 - Enter the $x$ values separated by commas

Step 2 - Click on "Calculate" button to get Decile for ungrouped data

Step 3 - Gives the output as number of observations $n$

Step 4 - Gives the output as ascending order data

Step 5 - Gives the Deciles $D_1$,$D_5$ and $D_9$.

Step 6 - Gives output as Kelly's Coefficient of Skewness

## Range for Kelly's coefficient of Skewness

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 \eqref{sb} 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.

## Kelly's coefficient of skewness 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}

Ninth Decile $D_9$

The ninth 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 $\text{positively skewed}$.

## Kelly's coefficient of skewness 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}

Ninth Decile $D_9$

The ninth 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 $\text{positively skewed}$.

## Kelly's coefficient of skewness Example 3

Blood sugar level (in mg/dl) of a sample of 20 patients admitted to the hospitals are as follows:

75,89,72,78,87, 85, 73, 75, 97, 87, 84, 76,73,79,99,86,83,76,78,73.

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 sample size is $n = 20$.

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

72, 73, 73, 73, 75, 75, 76, 76, 78, 78

79, 83, 84, 85, 86, 87, 87, 89, 97, 99

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)\\ &=73+0.1\big(73 -73\big)\\ &=73\text{ mg/dl} \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.}\\ &=78.5\text{ mg/dl} \end{aligned}

Ninth Decile $D_9$

The ninth 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)\\ &=89+0.9\big(97 -89\big)\\ &=96.2\text{ mg/dl} \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{96.2+73 - 2* 78.5}{96.2 - 73}\\ &=0.5259 \end{aligned}

As the coefficient of skewness $S_k$ is $\text{greater than zero}$ (i.e., $S_k > 0$), the distribution is $\text{positively skewed}$.

## Kelly's coefficient of skewness Example 4

Diastolic blood pressure (in mmHg) of a sample of 18 patients admitted to the hospitals are as follows:

65,76,64,73,74,80, 71, 68,66, 81, 79, 75, 70, 62, 83,63, 77, 78.

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 sample size is $n = 18$.

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

62, 63, 64, 65, 66, 68, 70, 71, 73

74, 75, 76, 77, 78, 79, 80, 81, 83

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(18+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(1.9\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(1\big)^{th} \text{ obs.}\\ & +0.9 \big(\text{Value of } \big(2\big)^{th}\text{ obs.}-\text{Value of }\big(1\big)^{th} \text{ obs.}\big)\\ &=62+0.9\big(63 -62\big)\\ &=62.9\text{ mmHg} \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(18+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(9.5\big)^{th} \text{ obs.}\\ &=73.5\text{ mmHg} \end{aligned}

Ninth Decile $D_9$

The ninth 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(18+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(17.1\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(17\big)^{th} \text{ obs.}\\ & +0.1 \big(\text{Value of } \big(18\big)^{th}\text{ obs.}-\text{Value of }\big(17\big)^{th} \text{ obs.}\big)\\ &=81+0.1\big(83 -81\big)\\ &=81.2\text{ mmHg} \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{81.2+62.9 - 2* 73.5}{81.2 - 62.9}\\ &=-0.1585 \end{aligned}

As the coefficient of skewness $S_k$ is $\text{less than zero}$ (i.e., $S_k < 0$), the distribution is $\text{negatively skewed}$.

## Kelly's coefficient of skewness Example 5

The following data are the heights, correct to the nearest centimeters, for a group of children:

126, 129, 129, 132, 132, 133, 133, 135, 136, 137,
137, 138, 141, 143, 144, 146, 147, 152, 154, 161 

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 sample size is $n = 20$.

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

126, 129, 129, 132, 132, 133, 133, 135, 136, 137

137, 138, 141, 143, 144, 146, 147, 152, 154, 161

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)\\ &=129+0.1\big(129 -129\big)\\ &=129\text{ cm} \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.}\\ &=137\text{ cm} \end{aligned}

Ninth Decile $D_9$

The ninth 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)\\ &=152+0.9\big(154 -152\big)\\ &=153.8\text{ cm} \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{153.8+129 - 2* 137}{153.8 - 129}\\ &=0.3548 \end{aligned}

As the coefficient of skewness $S_k$ is $\text{greater than zero}$ (i.e., $S_k > 0$), the distribution is $\text{positively skewed}$.

## Conclusion

In this tutorial, you learned about formula for Kelly's coefficient of skewness for ungrouped data and how to calculate Kelly's coefficient of skewness for ungrouped data. You also learned about how to solve numerical problems based on Kelly's coefficient of skewness for ungrouped data.