Quartiles for ungrouped data
Quartiles are the values of arranged data which divide whole data into four equal parts. They are 3 in numbers namely $Q_1$, $Q_2$ and $Q_3$. Here $Q_1$ is first quartile, $Q_2$ is second quartile and $Q_3$ is third quartile.
The formula for $i^{th}$ quartile is
$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$
where $n$ is the total number of observations.
Example 1
A random sample of 15 patients yielded the following data on the length of stay (in days) in the hospital.
5, 6, 9, 10, 15, 10, 14, 12, 10, 13, 13, 9, 8, 10, 12.
Find quartiles.
Solution
The formula for $i^{th}$ quartile is
$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$
where $n$ is the total number of observations.
Arrange the data in ascending order
5, 6, 8, 9, 9, 10, 10, 10, 10, 12, 12, 13, 13, 14, 15
First Quartile $Q_1$
The first quartle $Q_1$ can be computed as follows:
$$ \begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{1(15+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(4\big)^{th} \text{ obs.}\\ &=9 \end{aligned} $$
Thus, $25$ % of the patients had length of stay in the hospital less than or equal to $9$ days.
Second Quartile $Q_2$
The second quartile $Q_2$ can be computed as follows:
$$ \begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{2(15+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(8\big)^{th} \text{ obs.}\\ &=10 \end{aligned} $$
Thus, $50$ % of the patients had length of stay in the hospital less than or equal to $10$ days.
Third Quartile $Q_3$
The third quartile $Q_3$ can be computed as follows:
$$ \begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{3(15+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(12\big)^{th} \text{ obs.}\\ &=13 \end{aligned} $$
Thus, $75$ % of the patients had length of stay in the hospital less than or equal to $13$ days.
Example 2
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 the value of $Q_1$, $Q_2$ and $Q_3$.
Solution
The formula for $i^{th}$ quartile is
$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$
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, 80, 82, 83, 84, 85, 86, 87, 97, 99
First Quartile $Q_1$
The first quartle $Q_1$ can be computed as follows:
$$ \begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{1(20+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(5.25\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(5\big)^{th} \text{ obs.}+0.25 \big(\text{Value of } \big(6\big)^{th}\text{ obs.}-\text{Value of }\big(5\big)^{th} \text{ obs.}\big)\\ &=75+0.25\big(75 -75\big)\\ &=75 \end{aligned} $$
Thus, $25$ % of the patients had blood sugar level less than or equal to $75$ mg/dl.
Second Quartile $Q_2$
The second quartile $Q_2$ can be computed as follows:
$$ \begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{2(20+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(10.5\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(10\big)^{th} \text{ obs.}+0.5 \big(\text{Value of } \big(11\big)^{th}\text{ obs.}-\text{Value of }\big(10\big)^{th} \text{ obs.}\big)\\ &=78+0.5\big(79 -78\big)\\ &=78.5 \end{aligned} $$
Thus, $50$ % of the patients had blood sugar level less than or equal to $78.5$ mg/dl.
Third Quartile $Q_3$
The third quartile $Q_3$ can be computed as follows:
$$ \begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{3(20+1)}{4}\bigg)^{th} \text{ obs.}\\ &= \text{ Value of }\big(15.75\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(15\big)^{th} \text{ obs.}+0.75 \big(\text{Value of } \big(16\big)^{th}\text{ obs.}-\text{Value of }\big(15\big)^{th} \text{ obs.}\big)\\ &=84+0.75\big(85 -84\big)\\ &=84.75 \end{aligned} $$
Thus, $75$ % of the patients had blood sugar level less than or equal to $84.75$ mg/dl.
