Kelly's coefficient of skewness for grouped 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, $D_9 - D_5 = D_5 -D_1$.

Kelly's coefficient of skewness is based on deciles $D_{1}$, $1^{st}$ decile, $D_{5}$, $5^{th}$ decile, and $D_{9}$, $9^{th}$ decile). Only 20% of the observations are excluded from the measure.

Formula

$S_k =\dfrac{D_9+D_1 -2*Median}{D_9-D_1}$

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.

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 $\text{negatively skewed}$.

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 $\text{negatively skewed}$.

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