# Quartiles

There are three quartiles called the first quartile, second quartile and third quartile. The quartiles divide the set of observations into four equal parts. The second quartile is equal to the median. The first quartile is also called the lower quartile and is denoted by ${Q_1}$. The third quartile is also called the upper quartile and is denoted by ${Q_3}$. The lower quartile ${Q_1}$ is a point which has 25% of the observations below it and 75% of the observations above it. The upper quartile ${Q_3}$ is a point with 75% of the observations below it and 25% of the observations above it.

Quartile for Individual Observations (Ungrouped Data)

Quartile for a Frequency Distribution (Discrete Data)

Quartile for Grouped Frequency Distribution

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 the 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.

${Q_1} = {\text{Value of }}\left( {\frac{{n + 1}}{4}} \right)th{\text{ item}}$
$= {\text{Value of }}\left( {\frac{{20 + 1}}{4}} \right)th{\text{ item}}$
$= {\text{Value of (5}}{\text{.25)}}th{\text{ item}}$
$= 5th{\text{ item}} + 0.25(6th{\text{ item}} - 5th{\text{ item}}) = 1240 + 0.25(1320 - 1240)$
${Q_1} = 1240 + 20 = 1260$
${Q_3} = {\text{Value of }}\frac{{{\text{3(n + 1)}}}}{{\text{4}}}th{\text{ item}}$
$= {\text{Value of }}\frac{{{\text{3(20 + 1)}}}}{{\text{4}}}th{\text{ item}}$
$= {\text{Value of (15}}{\text{.75)}}th{\text{ item}}$
$= 15th{\text{ item}} + 0.75(16th{\text{ item}} - 15th{\text{ item}}) = 1750 + 0.75(1755 - 1750)$
${Q_3} = 1750 + 3.75 = 1753.75$

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 $9.3 - 9.7$ $2$ $9.25 - 9.75$ $2$ $9.8 - 10.2$ $5$ $9.75 - 10.25$ $2 + 3 = 7$ $10.3 - 10.7$ $12$ $10.25 - 10.75$ $7 + 12 = 19$ $10.8 - 11.2$ $17$ $10.75 - 11.25$ $19 + 17 = 36$ $11.3 - 11.7$ $14$ $11.25 - 11.75$ $36 + 14 = 50$ $11.8 - 12.2$ $6$ $11.75 - 12.25$ $50 + 6 = 56$ $12.3 - 12.7$ $3$ $12.25 - 12.75$ $56 + 3 = 59$ $12.8 - 13.2$ $1$ $12.75 - 13.25$ $59 + 1 = 60$

${Q_1} = {\text{Value of }}\left( {\frac{n}{4}} \right)th{\text{ item }} = {\text{Value of }}\left( {\frac{{60}}{4}} \right)th{\text{ item = }}15th{\text{ item}}$
${Q_1}$ lies in the class $10.25 - 10.75$
$\therefore$
${Q_1} = l + \frac{h}{f}\left( {\frac{n}{4} - c} \right)$

Here $l = 10.25$, $h = 0.5$, $f = 12$, $n/4 = 15$ 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 $11.25 - 11.75$
$\therefore$
${Q_3} = l + \frac{h}{f}\left( {\frac{{3n}}{4} - c} \right)$

Here $l = 11.25$, $h = 0.5$, $f = 14$, $3n/4 = 45$ and $c = 36$
$\therefore$
${Q_1} = 11.25 + \frac{{0.5}}{{14}}\left( {45 - 36} \right) = 11.25 + 0.32 = 11.57$