Inter Quartile Range for ungrouped data

Inter Quartile Range for ungrouped data

Inter quartile range is given by

$IQR = Q_3-Q_1$

where

• $Q_1$ is the first quartile
• $Q_3$ is the 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:

$$\begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{1(15+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(4\big)^{th} \text{ observation}\\ &=9 \text{ days}. \end{aligned}$$

Thus, $25$ % of the patients had length of stay in the hospital less than or equal to $9$ days.

Third Quartile $Q_3$

The third quartile $Q_3$ can be computed as follows:

$$\begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{3(15+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(12\big)^{th} \text{ observation}\\ &=13 \text{ days}. \end{aligned}$$ Thus, $75$ % of the patients had length of stay in the hospital less than or equal to $13$ days.

Inter-Quartile Range

The inter-quartile range is

$$\begin{aligned} IQR &= Q_3 - Q_1\\ &= 13 - 9\\ &= 4\text{ days}. \end{aligned}$$

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:

$$\begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{1(20+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(5.25\big)^{th} \text{ observation}\\ &= \text{Value of }\big(5\big)^{th} \text{ obs.}+0.25 \big(\text{Value of } \big(6\big)^{th}\text{ obs.}-\text{Value of }\big(5\big)^{th} \text{ obs.}\big)\\ &=75+0.25\big(75 -75\big)\\ &=75 \text{ mg/dl}. \end{aligned}$$ Thus, $25$ % of the patients had blood sugar level less than or equal to $75$ mg/dl.

Third Quartile $Q_3$

The third quartile $Q_3$ can be computed as follows:

$$\begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{3(20+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(15.75\big)^{th} \text{ observation}\\ &= \text{Value of }\big(15\big)^{th} \text{ obs.}+0.75 \big(\text{Value of } \big(16\big)^{th}\text{ obs.}-\text{Value of }\big(15\big)^{th} \text{ obs.}\big)\\ &=84+0.75\big(85 -84\big)\\ &=84.75 \text{ mg/dl}. \end{aligned}$$

Thus, $75$ % of the patients had blood sugar level less than or equal to $84.75$ mg/dl.

Inter-Quartile Range

The inter-quartile range is

$$\begin{aligned} IQR &= Q_3 - Q_1\\ &= 84.75 - 75\\ &= 9.75 \text{ mg/dl}. \end{aligned}$$

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