# Quartile Deviation and its Coefficient

Quartile Deviation

Quartile deviation is based on the lower quartile ${Q_1}$ and the upper quartile ${Q_3}$. The difference ${Q_3} – {Q_1}$ is called the inter quartile range. The difference  ${Q_3} – {Q_1}$ divided by $2$ is called semi-inter-quartile range or the quartile deviation. Thus

$Q.D = \frac{{{Q_3} – {Q_1}}}{2}$

The quartile deviation is a slightly better measure of absolute dispersion than the range, but it ignores the observations on the tails. If we take difference samples from a population and calculate their quartile deviations, their values are quite likely to be sufficiently different. This is called sampling fluctuation, and it is not a popular measure of dispersion. The quartile deviation calculated from the sample data does not help us to draw any conclusion (inference) about the quartile deviation in the population.

Coefficient of Quartile Deviation

A relative measure of dispersion based on the quartile deviation is called the coefficient of quartile deviation. It is defined as:

Coefficient of Quartile Deviation $Coefficient\,of\,Quartile\,Deviation = \frac{{\frac{{{Q_3} – {Q_1}}}{2}}}{{\frac{{{Q_3} + {Q_1}}}{2}}} = \frac{{{Q_3} – {Q_1}}}{{{Q_3} + {Q_1}}}$

It is a pure number free of any units of measurement. It can be used for comparing the dispersion of two or more sets of data.

Example

The wheat production (in Kg) of 20 acres is given as: 1120, 1240, 1320, 1040, 1080, 1200, 1440, 1360, 1680, 1730, 1785, 1342, 1960, 1880, 1755, 1720, 1600, 1470, 1750, and 1885. Find the quartile deviation and coefficient of quartile deviation.

Solution:

After arranging the observations in ascending order, we get

1040, 1080, 1120, 1200, 1240, 1320, 1342, 1360, 1440, 1470, 1600, 1680, 1720, 1730, 1750, 1755, 1785, 1880, 1885, 1960.

${Q_1} = {\text{Value of }}\left( {\frac{{n + 1}}{4}} \right)th{\text{ item}}$
$= {\text{Value of }}\left( {\frac{{20 + 1}}{4}} \right)th{\text{ item}}$
$= {\text{Value of (5}}{\text{.25)}}th{\text{ item}}$
$= 5th{\text{ item}} + 0.25(6th{\text{ item}} – 5th{\text{ item}}) = 1240 + 0.25(1320 – 1240)$
${Q_1} = 1240 + 20 = 1260$

${Q_3} = {\text{Value of }}\frac{{{\text{3(n + 1)}}}}{{\text{4}}}th{\text{ item}}$
$= {\text{Value of }}\frac{{{\text{3(20 + 1)}}}}{{\text{4}}}th{\text{ item}}$
$= {\text{Value of (15}}{\text{.75)}}th{\text{ item}}$
$= 15th{\text{ item}} + 0.75(16th{\text{ item}} – 15th{\text{ item}}) = 1750 + 0.75(1755 – 1750)$
${Q_3} = 1750 + 3.75 = 1753.75$

$Q.D = \frac{{{Q_3} – {Q_1}}}{2} = \frac{{1753.75 – 1260}}{2} = \frac{{492.75}}{2} = 246.875$

$Coefficient\,of\,Quartile\,Deviation = \frac{{{Q_3} – {Q_1}}}{{{Q_3} + {Q_1}}} = \frac{{1753.75 – 1260}}{{1753.75 + 1260}} = 0.164$

Example:

Calculate the quartile deviation and coefficient of quartile deviation from the data given below:

 Maximum Load (short-tons) Number of Cables $9.3 – 9.7$ $2$ $9.8 – 10.2$ $5$ $10.3 – 10.7$ $12$ $10.8 – 11.2$ $17$ $11.3 – 11.7$ $14$ $11.8 – 12.2$ $6$ $12.3 – 12.7$ $3$ $12.8 – 13.2$ $1$

Solution:

The necessary calculations are given below:

 Maximum Load (short-tons) Number of Cables ($f$) Class Boundaries Cumulative Frequencies $9.3 – 9.7$ $2$ $9.25 – 9.75$ $2$ $9.8 – 10.2$ $5$ $9.75 – 10.25$ $2 + 5 = 7$ $10.3 – 10.7$ $12$ $10.25 – 10.75$ $7 + 12 = 19$ $10.8 – 11.2$ $17$ $10.75 – 11.25$ $19 + 17 = 36$ $11.3 – 11.7$ $14$ $11.25 – 11.75$ $36 + 14 = 50$ $11.8 – 12.2$ $6$ $11.75 – 12.25$ $50 + 6 = 56$ $12.3 – 12.7$ $3$ $12.25 – 12.75$ $56 + 3 = 59$ $12.8 – 13.2$ $1$ $12.75 – 13.25$ $59 + 1 = 60$

${Q_1} = {\text{Value of }}\left( {\frac{n}{4}} \right)th{\text{ item }} = {\text{Value of }}\left( {\frac{{60}}{4}} \right)th{\text{ item = }}15th{\text{ item}}$

${Q_1}$ lies in the class $10.25 – 10.75$

$\therefore$     ${Q_1} = l + \frac{h}{f}\left( {\frac{n}{4} – c} \right)$

Where $l = 10.25$, $h = 0.5$, $f = 12$, $n/4 = 15$ and $c = 7$

$\therefore$     ${Q_1} = 10.25 + \frac{{0.5}}{{12}}\left( {15 – 7} \right) = 10.25 + 0.33 = 10.58$

${Q_3} = {\text{Value of }}\left( {\frac{{3n}}{4}} \right)th{\text{ item }} = {\text{Value of }}\left( {\frac{{3 \times 60}}{4}} \right)th{\text{ item = 4}}5th{\text{ item}}$

${Q_3}$ lies in the class $11.25 – 11.75$

$\therefore$     ${Q_3} = l + \frac{h}{f}\left( {\frac{{3n}}{4} – c} \right)$

Where $l = 11.25$, $h = 0.5$, $f = 14$, $3n/4 = 45$ and $c = 36$

$\therefore$     ${Q_1} = 11.25 + \frac{{0.5}}{{14}}\left( {45 – 36} \right) = 11.25 + 0.32 = 11.57$

$Q.D = \frac{{{Q_3} – {Q_1}}}{2} = \frac{{11.57 – 10.58}}{2} = \frac{{0.99}}{2} = 0.495$

$Coefficient\,of\,Quartile\,Deviation = \frac{{{Q_3} – {Q_1}}}{{{Q_3} + {Q_1}}} = \frac{{11.57 – 10.58}}{{11.57 + 10.58}}$

$Coefficient\,of\,Quartile\,Deviation = \frac{{0.99}}{{22.15}} = 0.045$