# Kelly’s coefficient of skewness for grouped data | Formula | Examples

## Kelly's coefficient of skewness for grouped 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,

• $D_1=P_{10}$ is the first decile or tenth percentile,
• $D_5=P_{50}$ is the fifth decile or fiftieth percentile,
• $D_9=P_{90}$ 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 grouped data

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

Kelly's Coeff. of Skewness Calculator
Type of Freq. Dist. DiscreteContinuous
Enter the Classes for X (Separated by comma,)
Enter the frequencies (f) (Separated by comma,)
Results
Number of Obs. (n):
First Decile : ($D_1$)
Sample Median : ($D_5$)
Nineth Decile : ($D_9$)
Kelly's Coeff. of Skewness :

## How to find Kelly's coefficient of skewness for grouped data?

Step 1 - Select type of frequency distribution (Discrete or continuous)

Step 2 - Enter the Range or classes (X) seperated by comma (,)

Step 3 - Enter the Frequencies (f) seperated by comma

Step 4 - Click on "Calculate" button for decile calculation

Step 5 - Gives output as number of observation (N)

Step 6 - Gives output as $D_1$, $D_5$ and $D_9$

Step 7 - Gives output as Kelly's Coefficient of Skewness

## Kelly's Coefficient of Skewness Example 1

A librarian keeps the records about the amount of time spent (in minutes) in a library by college students. Data is as follows:

Time spent 30 32 35 38 40
No. of students 8 12 20 10 5

Calculate Kelly's coefficient of skewness.

#### Solution

$x_i$ $f_i$ $cf$
30 8 8
32 12 20
35 20 40
38 10 50
40 5 55
Total 55

Deciles

The formula for $i^{th}$ deciles is

$D_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,\cdots, 9$

where $N$ is the total number of observations.

First Decile $D_1$

 \begin{aligned} D_{1} &=\bigg(\dfrac{1(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(55)}{10}\bigg)^{th}\text{ value}\\ &=\big(5.5\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $5.5$ is $8$. The corresponding value of $X$ is the $1^{st}$ decile. That is, $D_1 =30$ minutes.

Fifth Decile $D_5$

 \begin{aligned} D_{5} &=\bigg(\dfrac{5(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{5(55)}{10}\bigg)^{th}\text{ value}\\ &=\big(27.5\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $27.5$ is $40$. The corresponding value of $X$ is the $5^{th}$ decile. That is, $D_5 =35$ minutes.

Ninth Decile $D_9$

 \begin{aligned} D_{9} &=\bigg(\dfrac{9(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{9(55)}{10}\bigg)^{th}\text{ value}\\ &=\big(49.5\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $49.5$ is $50$. The corresponding value of $X$ is the $9^{th}$ decile. That is, $D_9 =38$ minutes.

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{38+30 - 2* 35}{38 - 30}\\ &=\frac{-2}{8}\\ &=-0.25 \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 2

The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students.

Time spent on Internet ($x$) 10-12 13-15 16-18 19-21 22-24
No. of students ($f$) 3 12 15 24 2

Calculate Kelly's coefficient of skewness.

#### Solution

Let $X$ denote the amount of time (in minutes) spent on the internet.

Here the classes are inclusive. To make them exclusive type subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.

Class Interval Class Boundries $f_i$ $cf$
10-12 9.5-12.5 3 3
13-15 12.5-15.5 12 15
16-18 15.5-18.5 15 30
19-21 18.5-21.5 24 54
22-24 21.5-24.5 2 56
Total 56

The formula for $i^{th}$ decile is

$D_i =\bigg(\dfrac{i(N)}{10}\bigg)^{th}$ value, $i=1,2,\cdots, 9$

where $N$ is the total number of observations.

First Decile ($D_1$)

 \begin{aligned} D_{1} &=\bigg(\dfrac{1(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(56)}{10}\bigg)^{th}\text{ value}\\ &=\big(5.6\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $5.6$ is $15$, the corresponding class $12.5-15.5$ is the $1^{st}$ decile class.

Thus

• $l = 12.5$, the lower limit of the $1^{st}$ decile class
• $N=56$, total number of observations
• $f =12$, frequency of the $1^{st}$ decile class
• $F_< = 3$, cumulative frequency of the class previous to $1^{st}$ decile class
• $h =3$, the class width

The first decile $D_1$ can be computed as follows:

 \begin{aligned} D_1 &= l + \bigg(\frac{\frac{1(N)}{10} - F_<}{f}\bigg)\times h\\ &= 12.5 + \bigg(\frac{\frac{1*56}{10} - 3}{12}\bigg)\times 3\\ &= 12.5 + \bigg(\frac{5.6 - 3}{12}\bigg)\times 3\\ &= 12.5 + \big(0.2167\big)\times 3\\ &= 12.5 + 0.65\\ &= 13.15 \text{ minutes} \end{aligned}

Fifth Decile ($D_5$)

 \begin{aligned} D_{5} &=\bigg(\dfrac{5(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{5(56)}{10}\bigg)^{th}\text{ value}\\ &=\big(28\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $28$ is $30$, the corresponding class $15.5-18.5$ is the $5^{th}$ decile class.

Thus

• $l = 15.5$, the lower limit of the $5^{th}$ decile class
• $N=56$, total number of observations
• $f =15$, frequency of the $5^{th}$ decile class
• $F_< = 15$, cumulative frequency of the class previous to $5^{th}$ decile class
• $h =3$, the class width

The fifth decile $D_5$ can be computed as follows:

 \begin{aligned} D_5 &= l + \bigg(\frac{\frac{5(N)}{10} - F_<}{f}\bigg)\times h\\ &= 15.5 + \bigg(\frac{\frac{5*56}{10} - 15}{15}\bigg)\times 3\\ &= 15.5 + \bigg(\frac{28 - 15}{15}\bigg)\times 3\\ &= 15.5 + \big(0.8667\big)\times 3\\ &= 15.5 + 2.6\\ &= 18.1 \text{ minutes} \end{aligned}

Ninth Decile ($D_9$)

 \begin{aligned} D_{9} &=\bigg(\dfrac{9(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{9(56)}{10}\bigg)^{th}\text{ value}\\ &=\big(50.4\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $50.4$ is $54$, the corresponding class $18.5-21.5$ is the $9^{th}$ decile class.

Thus

• $l = 18.5$, the lower limit of the $9^{th}$ decile class
• $N=56$, total number of observations
• $f =24$, frequency of the $9^{th}$ decile class
• $F_< = 30$, cumulative frequency of the class previous to $9^{th}$ decile class
• $h =3$, the class width

The ninth decile $D_9$ can be computed as follows:

 \begin{aligned} D_9 &= l + \bigg(\frac{\frac{9(N)}{10} - F_<}{f}\bigg)\times h\\ &= 18.5 + \bigg(\frac{\frac{9*56}{10} - 30}{24}\bigg)\times 3\\ &= 18.5 + \bigg(\frac{50.4 - 30}{24}\bigg)\times 3\\ &= 18.5 + \big(0.85\big)\times 3\\ &= 18.5 + 2.55\\ &= 21.05 \text{ minutes} \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{21.05+13.15 - 2* 18.1}{21.05 - 13.15}\\ &=\frac{-2}{7.9}\\ &=-0.25316 \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 3

The Scores of students in a Math test is given in the table below :

Class Interval 10-20 20-30 30-40 40-50 50-60 60-70
Frequency ($f$) 6 8 12 10 5 4

Find Kelly's Coefficient of Skewness.

#### Solution

Let $X$ denote the scores in Math Test.

Class Interval Class Boundries $f_i$ $cf$
10-20 10-20 6 6
20-30 20-30 8 14
30-40 30-40 12 26
40-50 40-50 10 36
50-60 50-60 5 41
60-70 60-70 4 45
Total 45

The formula for $i^{th}$ decile is

$D_i =\bigg(\dfrac{i(N)}{10}\bigg)^{th}$ value, $i=1,2,\cdots, 9$

where $N$ is the total number of observations.

First Decile ($D_1$)

 \begin{aligned} D_{1} &=\bigg(\dfrac{1(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(45)}{10}\bigg)^{th}\text{ value}\\ &=\big(4.5\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $4.5$ is $6$, the corresponding class $10-20$ is the $1^{st}$ decile class.

Thus

• $l = 10$, the lower limit of the $1^{st}$ decile class
• $N=45$, total number of observations
• $f =6$, frequency of the $1^{st}$ decile class
• $F_< = 0$, cumulative frequency of the class previous to $1^{st}$ decile class
• $h =10$, the class width

The first decile $D_1$ can be computed as follows:

 \begin{aligned} D_1 &= l + \bigg(\frac{\frac{1(N)}{10} - F_<}{f}\bigg)\times h\\ &= 10 + \bigg(\frac{\frac{1*45}{10} - 0}{6}\bigg)\times 10\\ &= 10 + \bigg(\frac{4.5 - 0}{6}\bigg)\times 10\\ &= 10 + \big(0.75\big)\times 10\\ &= 10 + 7.5\\ &= 17.5 \text{ Scores} \end{aligned}

Fifth Decile ($D_5$)

 \begin{aligned} D_{5} &=\bigg(\dfrac{5(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{5(45)}{10}\bigg)^{th}\text{ value}\\ &=\big(22.5\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $22.5$ is $26$, the corresponding class $30-40$ is the $5^{th}$ decile class.

Thus

• $l = 30$, the lower limit of the $5^{th}$ decile class
• $N=45$, total number of observations
• $f =12$, frequency of the $5^{th}$ decile class
• $F_< = 14$, cumulative frequency of the class previous to $5^{th}$ decile class
• $h =10$, the class width

The fifth decile $D_5$ can be computed as follows:

 \begin{aligned} D_5 &= l + \bigg(\frac{\frac{5(N)}{10} - F_<}{f}\bigg)\times h\\ &= 30 + \bigg(\frac{\frac{5*45}{10} - 14}{12}\bigg)\times 10\\ &= 30 + \bigg(\frac{22.5 - 14}{12}\bigg)\times 10\\ &= 30 + \big(0.7083\big)\times 10\\ &= 30 + 7.0833\\ &= 37.0833 \text{ Scores} \end{aligned}

Ninth Decile ($D_9$)

 \begin{aligned} D_{9} &=\bigg(\dfrac{9(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{9(45)}{10}\bigg)^{th}\text{ value}\\ &=\big(40.5\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $40.5$ is $41$, the corresponding class $50-60$ is the $9^{th}$ decile class.

Thus

• $l = 50$, the lower limit of the $9^{th}$ decile class
• $N=45$, total number of observations
• $f =5$, frequency of the $9^{th}$ decile class
• $F_< = 36$, cumulative frequency of the class previous to $9^{th}$ decile class
• $h =10$, the class width

The ninth decile $D_9$ can be computed as follows:

 \begin{aligned} D_9 &= l + \bigg(\frac{\frac{9(N)}{10} - F_<}{f}\bigg)\times h\\ &= 50 + \bigg(\frac{\frac{9*45}{10} - 36}{5}\bigg)\times 10\\ &= 50 + \bigg(\frac{40.5 - 36}{5}\bigg)\times 10\\ &= 50 + \big(0.9\big)\times 10\\ &= 50 + 9\\ &= 59 \text{ Scores} \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{59+17.5 - 2* 37.0833}{59 - 17.5}\\ &=\frac{2.3334}{41.5}\\ &=0.05623 \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

The following data shows the distribution of maximum loads in short tons supported by certain cables produced by a company:

9.25-9.75 2
9.75-10.25 5
10.25-10.75 12
10.75-11.25 17
11.25-11.75 14
11.75-12.25 6
12.25-12.75 3
12.75-13.25 1

Find Kelly's Coefficient of Skewness.

#### Solution

Let $X$ denote the maximum load.

Here the classes are inclusive. To make them exclusive type subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.

Class Interval Class Boundries $f_i$ $cf$
9.25-9.75 9.25-9.75 2 2
9.75-10.25 9.75-10.25 5 7
10.25-10.75 10.25-10.75 12 19
10.75-11.25 10.75-11.25 17 36
11.25-11.75 11.25-11.75 14 50
11.75-12.25 11.75-12.25 6 56
12.25-12.75 12.25-12.75 3 59
12.75-13.25 12.75-13.25 1 60
Total 60

The formula for $i^{th}$ decile is

$D_i =\bigg(\dfrac{i(N)}{10}\bigg)^{th}$ value, $i=1,2,\cdots, 9$

where $N$ is the total number of observations.

First Decile ($D_1$)

 \begin{aligned} D_{1} &=\bigg(\dfrac{1(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(60)}{10}\bigg)^{th}\text{ value}\\ &=\big(6\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $6$ is $7$, the corresponding class $9.75-10.25$ is the $1^{st}$ decile class.

Thus

• $l = 9.75$, the lower limit of the $1^{st}$ decile class
• $N=60$, total number of observations
• $f =5$, frequency of the $1^{st}$ decile class
• $F_< = 2$, cumulative frequency of the class previous to $1^{st}$ decile class
• $h =0.5$, the class width

The first decile $D_1$ can be computed as follows:

 \begin{aligned} D_1 &= l + \bigg(\frac{\frac{1(N)}{10} - F_<}{f}\bigg)\times h\\ &= 9.75 + \bigg(\frac{\frac{1*60}{10} - 2}{5}\bigg)\times 0.5\\ &= 9.75 + \bigg(\frac{6 - 2}{5}\bigg)\times 0.5\\ &= 9.75 + \big(0.8\big)\times 0.5\\ &= 9.75 + 0.4\\ &= 10.15 \text{ tons} \end{aligned}

Fifth Decile ($D_5$)

 \begin{aligned} D_{5} &=\bigg(\dfrac{5(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{5(60)}{10}\bigg)^{th}\text{ value}\\ &=\big(30\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $30$ is $36$, the corresponding class $10.75-11.25$ is the $5^{th}$ decile class.

Thus

• $l = 10.75$, the lower limit of the $5^{th}$ decile class
• $N=60$, total number of observations
• $f =17$, frequency of the $5^{th}$ decile class
• $F_< = 19$, cumulative frequency of the class previous to $5^{th}$ decile class
• $h =0.5$, the class width

The fifth decile $D_5$ can be computed as follows:

 \begin{aligned} D_5 &= l + \bigg(\frac{\frac{5(N)}{10} - F_<}{f}\bigg)\times h\\ &= 10.75 + \bigg(\frac{\frac{5*60}{10} - 19}{17}\bigg)\times 0.5\\ &= 10.75 + \bigg(\frac{30 - 19}{17}\bigg)\times 0.5\\ &= 10.75 + \big(0.6471\big)\times 0.5\\ &= 10.75 + 0.3235\\ &= 11.0735 \text{ tons} \end{aligned}

Ninth Decile ($D_9$)

 \begin{aligned} D_{9} &=\bigg(\dfrac{9(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{9(60)}{10}\bigg)^{th}\text{ value}\\ &=\big(54\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $54$ is $56$, the corresponding class $11.75-12.25$ is the $9^{th}$ decile class.

Thus

• $l = 11.75$, the lower limit of the $9^{th}$ decile class
• $N=60$, total number of observations
• $f =6$, frequency of the $9^{th}$ decile class
• $F_< = 50$, cumulative frequency of the class previous to $9^{th}$ decile class
• $h =0.5$, the class width

The ninth decile $D_9$ can be computed as follows:

 \begin{aligned} D_9 &= l + \bigg(\frac{\frac{9(N)}{10} - F_<}{f}\bigg)\times h\\ &= 11.75 + \bigg(\frac{\frac{9*60}{10} - 50}{6}\bigg)\times 0.5\\ &= 11.75 + \bigg(\frac{54 - 50}{6}\bigg)\times 0.5\\ &= 11.75 + \big(0.6667\big)\times 0.5\\ &= 11.75 + 0.3333\\ &= 12.0833 \text{ tons} \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{12.0833+10.15 - 2* 11.0735}{12.0833 - 10.15}\\ &=\frac{0.0863}{1.9333}\\ &=0.04464 \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 5

Following table shows the weight of 100 pumpkin produced from a farm :

Weight ('00 grams) Frequency
$4 \leq x < 6$ 4
$6 \leq x < 8$ 14
$8 \leq x < 10$ 34
$10 \leq x < 12$ 28
$12 \leq x < 14$ 20

Find Kelly's Coefficient of Skewness.

#### Solution

Let $X$ denote the weight of pumpkin.

Here the classes are inclusive. To make them exclusive type subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.

Class Interval Class Boundries $f_i$ $cf$
4-6 4-6 4 4
6-8 6-8 14 18
8-10 8-10 34 52
10-12 10-12 28 80
12-14 12-14 20 100
Total 100

The formula for $i^{th}$ decile is

$D_i =\bigg(\dfrac{i(N)}{10}\bigg)^{th}$ value, $i=1,2,\cdots, 9$

where $N$ is the total number of observations.

First Decile ($D_1$)

 \begin{aligned} D_{1} &=\bigg(\dfrac{1(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(100)}{10}\bigg)^{th}\text{ value}\\ &=\big(10\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $10$ is $18$, the corresponding class $6-8$ is the $1^{st}$ decile class.

Thus

• $l = 6$, the lower limit of the $1^{st}$ decile class
• $N=100$, total number of observations
• $f =14$, frequency of the $1^{st}$ decile class
• $F_< = 4$, cumulative frequency of the class previous to $1^{st}$ decile class
• $h =2$, the class width

The first decile $D_1$ can be computed as follows:

 \begin{aligned} D_1 &= l + \bigg(\frac{\frac{1(N)}{10} - F_<}{f}\bigg)\times h\\ &= 6 + \bigg(\frac{\frac{1*100}{10} - 4}{14}\bigg)\times 2\\ &= 6 + \bigg(\frac{10 - 4}{14}\bigg)\times 2\\ &= 6 + \big(0.4286\big)\times 2\\ &= 6 + 0.8571\\ &= 6.8571 \text{ ('00 grams)} \end{aligned}

Fifth Decile ($D_5$)

 \begin{aligned} D_{5} &=\bigg(\dfrac{5(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{5(100)}{10}\bigg)^{th}\text{ value}\\ &=\big(50\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $50$ is $52$, the corresponding class $8-10$ is the $5^{th}$ decile class.

Thus

• $l = 8$, the lower limit of the $5^{th}$ decile class
• $N=100$, total number of observations
• $f =34$, frequency of the $5^{th}$ decile class
• $F_< = 18$, cumulative frequency of the class previous to $5^{th}$ decile class
• $h =2$, the class width

The fifth decile $D_5$ can be computed as follows:

 \begin{aligned} D_5 &= l + \bigg(\frac{\frac{5(N)}{10} - F_<}{f}\bigg)\times h\\ &= 8 + \bigg(\frac{\frac{5*100}{10} - 18}{34}\bigg)\times 2\\ &= 8 + \bigg(\frac{50 - 18}{34}\bigg)\times 2\\ &= 8 + \big(0.9412\big)\times 2\\ &= 8 + 1.8824\\ &= 9.8824 \text{ ('00 grams)} \end{aligned}

Ninth Decile ($D_9$)

 \begin{aligned} D_{9} &=\bigg(\dfrac{9(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{9(100)}{10}\bigg)^{th}\text{ value}\\ &=\big(90\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $90$ is $100$, the corresponding class $12-14$ is the $9^{th}$ decile class.

Thus

• $l = 12$, the lower limit of the $9^{th}$ decile class
• $N=100$, total number of observations
• $f =20$, frequency of the $9^{th}$ decile class
• $F_< = 80$, cumulative frequency of the class previous to $9^{th}$ decile class
• $h =2$, the class width

The ninth decile $D_9$ can be computed as follows:

 \begin{aligned} D_9 &= l + \bigg(\frac{\frac{9(N)}{10} - F_<}{f}\bigg)\times h\\ &= 12 + \bigg(\frac{\frac{9*100}{10} - 80}{20}\bigg)\times 2\\ &= 12 + \bigg(\frac{90 - 80}{20}\bigg)\times 2\\ &= 12 + \big(0.5\big)\times 2\\ &= 12 + 1\\ &= 13 \text{ ('00 grams)} \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{13+6.8571 - 2* 9.8824}{13 - 6.8571}\\ &=\frac{0.0923}{6.1429}\\ &=0.01503 \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 grouped data and how to calculate Kelly's coefficient of skewness for grouped data. You also learned about how to solve numerical problems based on Kelly's coefficient of skewness for grouped data. 