Quartiles Calculator for ungrouped data with examples

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.

Quartiles Calculator for ungrouped Data

Use this calculator to find the Quartiles for ungrouped (raw) data.

How to calculate quartiles for ungrouped data?

Step 1 - Enter the $x$ values separated by commas

Step 2 - Click on "Calculate" button to get quartiles for ungrouped data

Step 3 - Gives the output as number of observations $n$

Step 4 - Gives the output as ascending order data

Step 5 - Gives all the quartiles $Q_1$, $Q_2$ and $Q_3$

Quartiles for Ungrouped Data Example 1

A random sample of 15 patients yielded the following data on the length of stay (in days) in the hospital.</p> <p>5, 6, 9, 10, 15, 10, 14, 12, 10, 13, 13, 9, 8, 10, 12.</p> <p>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:

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

Thus, lower $25$ % 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:

$$ \begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{2(15+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(8\big)^{th} \text{ observation}\\ &=10 \end{aligned} $$

Thus, lower $50$ % 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:

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

Quartiles for Ungrouped Data 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:

$$ \begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{1(20+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(5.25\big)^{th} \text{ observation}\\ &= \text{Value of }\big(5\big)^{th} \text{ obs.}\\ & \quad +0.25 \big(\text{Value of } \big(6\big)^{th}\text{ obs.}-\text{Value of }\big(5\big)^{th} \text{ obs.}\big)\\ &=75+0.25\big(75 -75\big)\\ &=75 \end{aligned} $$
Thus, lower $25$ % 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:

$$ \begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{2(20+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(10.5\big)^{th} \text{ observation}\\ &= \text{Value of }\big(10\big)^{th} \text{ obs.}\\ &\quad +0.5 \big(\text{Value of } \big(11\big)^{th}\text{ obs.}-\text{Value of }\big(10\big)^{th} \text{ obs.}\big)\\ &=78+0.5\big(79 -78\big)\\ &=78.5 \end{aligned} $$

Thus, lower $50$ % 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:

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

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

Quartiles for ungrouped data Example 3

The following data are the heights, correct to the nearest centimeters, for a group of children:

126, 129, 129, 132, 132, 133, 133, 135, 136, 137, 
137, 138, 141, 143, 144, 146, 147, 152, 154, 161 

Find quartiles for the above data.

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

126, 129, 129, 132, 132, 133, 133, 135, 136, 137, 137, 138, 141, 143, 144, 146, 147, 152, 154, 161

First Quartile $Q_1$

The first quartle $Q_1$ can be computed as follows:

$$ \begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{1(20+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(5.25\big)^{th} \text{ observation}\\ &= \text{Value of }\big(5\big)^{th} \text{ obs.}\\ &\quad +0.25 \big(\text{Value of } \big(6\big)^{th}\text{ obs.}-\text{Value of }\big(5\big)^{th} \text{ obs.}\big)\\ &=132+0.25\big(133 -132\big)\\ &=132.25 \text{ cm}. \end{aligned} $$

Thus, lower $25$ % of the children had height less than or equal to $132.25$ cm.

Second Quartile $Q_2$

The second quartile $Q_2$ can be computed as follows:

$$ \begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{2(20+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(10.5\big)^{th} \text{ observation}\\ &= \text{Value of }\big(10\big)^{th} \text{ obs.}\\ &\quad +0.5 \big(\text{Value of } \big(11\big)^{th}\text{ obs.}-\text{Value of }\big(10\big)^{th} \text{ obs.}\big)\\ &=137+0.5\big(137 -137\big)\\ &=137 \end{aligned} $$

Thus, lower $50$ % of the children had height less than or equal to $137$ cm.

Third Quartile $Q_3$

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

$$ \begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{3(20+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(15.75\big)^{th} \text{ observation}\\ &= \text{Value of }\big(15\big)^{th} \text{ obs.}\\ &\quad +0.75 \big(\text{Value of } \big(16\big)^{th}\text{ obs.}-\text{Value of }\big(15\big)^{th} \text{ obs.}\big)\\ &=144+0.75\big(146 -144\big)\\ &=145.5 \text{ cm}. \end{aligned} $$

Thus, lower $75$ % of the children had height less than or equal to $145.5$ cm.

Quartiles for ungrouped data Example 4

The following measurement were recorded for the drying time in hours, of a certain brand of latex paint.

3.4 2.5 4.8 2.9 3.6 2.8 3.3 5.6 
3.7 2.8 4.4 4.0 5.2 3.0 4.8.

Compute Quatiles for the above data.

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

2.5, 2.8, 2.8, 2.9, 3, 3.3, 3.4, 3.6, 3.7, 4, 4.4, 4.8, 4.8, 5.2, 5.6

First Quartile $Q_1$

The first quartle $Q_1$ can be computed as follows:

$$ \begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{1(15+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(4\big)^{th} \text{ observation}\\ &=2.9 \text{ hours}. \end{aligned} $$

Thus, lower $25$ % of the drying time is less than or equal to $2.9$ hours.

Second Quartile $Q_2$

The second quartile $Q_2$ can be computed as follows:

$$ \begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{2(15+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(8\big)^{th} \text{ observation}\\ &= \text{Value of }\big(8\big)^{th} \text{ obs.}\\ &=3.6\\ &=3.6 \end{aligned} $$

Thus, lower $50$ % of the drying time is less than or equal to $3.6$ hours.

Third Quartile $Q_3$

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

$$ \begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{3(15+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(12\big)^{th} \text{ observation}\\ &=4.8 \text{ hours}. \end{aligned} $$

Thus, lower $75$ % of the drying time is less than or equal to $4.8$ hours.

Quartiles for ungrouped data Example 5

The rice production (in Kg) of 10 acres is given as: 1120, 1240, 1320, 1040, 1080, 1720, 1600, 1470, 1750, and 1885. Find the quartiles for the given data.

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

1040, 1080, 1120, 1240, 1320, 1470, 1600, 1720, 1750, 1885

First Quartile $Q_1$

The first quartle $Q_1$ can be computed as follows:

$$ \begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{1(10+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(2.75\big)^{th} \text{ observation}\\ &= \text{Value of }\big(2\big)^{th} \text{ obs.}\\ &\quad +0.75 \big(\text{Value of } \big(3\big)^{th}\text{ obs.}-\text{Value of }\big(2\big)^{th} \text{ obs.}\big)\\ &=1080+0.75\big(1120 -1080\big)\\ &=1110 \text{ Kg}. \end{aligned} $$

Thus, lower $25$ % of the plots had rice production less than or equal to $1110$ Kg.

Second Quartile $Q_2$

The second quartile $Q_2$ can be computed as follows:

$$ \begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{2(10+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(5.5\big)^{th} \text{ observation}\\ &= \text{Value of }\big(5\big)^{th} \text{ obs.}\\ &\quad +0.5 \big(\text{Value of } \big(6\big)^{th}\text{ obs.}-\text{Value of }\big(5\big)^{th} \text{ obs.}\big)\\ &=1320+0.5\big(1470 -1320\big)\\ &=1395 \end{aligned} $$

Thus, lower $50$ % of the plots had rice production less than or equal to $1395$ Kg.

Third Quartile $Q_3$

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

$$ \begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{3(10+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(8.25\big)^{th} \text{ observation}\\ &= \text{Value of }\big(8\big)^{th} \text{ obs.}\\ &\quad +0.25 \big(\text{Value of } \big(9\big)^{th}\text{ obs.}-\text{Value of }\big(8\big)^{th} \text{ obs.}\big)\\ &=1720+0.25\big(1750 -1720\big)\\ &=1727.5 \text{ Kg}. \end{aligned} $$

Thus, lower $75$ % of the plots had rice production less than or equal to $1727.5$ Kg.

Conclusion

In this tutorial, you learned about formula for quartiles for ungrouped data and how to calculate quartiles for ungrouped data. You also learned about how to solve numerical problems based on quartiles for ungrouped data.

To learn more about other descriptive statistics measures, please refer to the following tutorials:

Descriptive Statistics

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