# Five (5) number summary calculator with examples

## Five number summary for ungrouped data

A five number summary is a quick and easy way to determine the the center, the spread and outliers (if any) of a data set.

5 number summary includes five values, namely,

• minimum value ($\min$),
• first quartile ($Q_1$),
• $\text{median }$ ($Q_2$),
• third quartile ($Q_3$),
• maximum value ($\max$).

The five number summary of a data gives you a rough idea about the range of the data, center of the data, variation in the data and symmetry of the data. A box and whisker plot is the visual representation of five number summary.

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.

## Five Number Summary Calculator for ungrouped data

Use this calculator to find the five number summary for ungrouped (raw) data.

Five (5) Number Summary Calculator
Enter the X Values (Separated by comma,)
Results
Number of Obs. (n):
Minimum of X:
First Quartile : ($Q_1$)
Median : ($Q_2$)
Third Quartile : ($Q_3$)
Maximum of X :

## How to calculate five number summary for ungrouped data?

Step 1 - Enter the (X) values seperated by comma (,)

Step 2 - Click on "Calculate" button to get mean, median and mode for ungrouped data

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

Step 4 - Calculate minimum value

Step 5 - Calculate first quartile $Q_1$

Step 6 - Calculate second quartile $Q_2$ (median)

Step 7 - Calculate third quartile $Q_3$

Step 8 - Calculate maximum value

## Example 1 - Find 5 number summary

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 the five number summary.

#### Solution

Arrange the data in ascending order

5, 6, 8, 9, 9, 10, 10, 10, 10, 12, 12, 13, 13, 14, 15.

Minimum Value

The minimum the length of stay in the hospital is $\min = 5$ days.

Maximum Value

The maximum the length of stay in the hospital is $\max = 15$ days.

Quartiles

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.

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}\\ &= \text{Value of }\big(4\big)^{th} \text{ obs.}\\ &\quad +0 \big(\text{Value of } \big(4\big)^{th}\text{ obs.}-\text{Value of }\big(4\big)^{th} \text{ obs.}\big)\\ &=9+0\big(9 -9\big)\\ &=9 \end{aligned}

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

Median (M) (i.e., Second Quartile $Q_2$)

The median ($M$) or second quartile $Q_2$ can be computed as follows:

 \begin{aligned} M &=\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.}\\ &\quad +0 \big(\text{Value of } \big(8\big)^{th}\text{ obs.}-\text{Value of }\big(8\big)^{th} \text{ obs.}\big)\\ &=10+0\big(10 -10\big)\\ &=10 \end{aligned}

Thus, lower $50$ % of patients had the 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}\\ &= \text{Value of }\big(12\big)^{th} \text{ obs.}\\ &\quad +0 \big(\text{Value of } \big(12\big)^{th}\text{ obs.}-\text{Value of }\big(12\big)^{th} \text{ obs.}\big)\\ &=13+0\big(13 -13\big)\\ &=13 \end{aligned}
Thus, lower $75$ % of patients had the length of stay in the hospital less than or equal to $13$ days.

Thus the five number summary of given data set is

$\min = 5$ days, $Q_1 = 9$ days, $\text{median }=10$ days, $Q_3=13$ days and $\max = 15$ days.

## Example 2 - Calculate five number summary

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 five number summary of the data.

#### Solution

Arrange the data in ascending order

72, 73, 73, 73, 75, 75, 76, 76, 78, 78, 79, 83, 84, 85, 86, 87, 87, 89, 97, 99.

Minimum Value

The minimum blood sugar level is $\min = 72$ mg/dl.

Maximum Value

The maximum blood sugar level is $\max = 99$ mg/dl.

Quartiles

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.

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 patients had blood sugar level less than or equal to $75$ mg/dl.

Median (M) (i.e., Second Quartile $Q_2$)

The median ($M$) or second quartile $Q_2$ can be computed as follows:

 \begin{aligned} M &=\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 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)\\ &=86+0.75\big(87 -86\big)\\ &=86.75 \end{aligned}
Thus, lower $75$ % of patients had blood sugar level less than or equal to $86.75$ mg/dl.

Thus the five number summary of given data set is

$\min = 72$ mg/dl, $Q_1 = 75$ mg/dl, $\text{median }=78.5$ mg/dl, $Q_3=86.75$ mg/dl and $\max = 99$ mg/dl.

## Example 3 - Get five number summary for data

The ages of 20 randomly selected students from a class are as follows:

23,22,21,27,19,21,18,25,26,25,29,28,18,22,20,17,19,21,24 and 20.

Find the five number summary of the data.

#### Solution

Arrange the data in ascending order

17, 18, 18, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 24, 25, 25, 26, 27, 28, 29.

Minimum Value

The minimum age of students is $\min = 17$ years.

Maximum Value

The maximum age of students is $\max = 29$ years.

Quartiles

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.

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)\\ &=19+0.25\big(20 -19\big)\\ &=19.25 \end{aligned}

Thus, lower $25$ % of the students had age less than or equal to $19.25$ years.

Median (M) (i.e., Second Quartile $Q_2$)

The median ($M$) or second quartile $Q_2$ can be computed as follows:

 \begin{aligned} M &=\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)\\ &=21+0.5\big(22 -21\big)\\ &=21.5 \end{aligned}

Thus, lower $50$ % of the students had age less than or equal to $21.5$ years.

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)\\ &=25+0.75\big(25 -25\big)\\ &=25 \end{aligned}
Thus, lower $75$ % of the students had age less than or equal to $25$ years.

Thus the five number summary of given data set is

$\min = 17$ years, $Q_1 = 19.25$ years, $\text{median }=21.5$ years, $Q_3=25$ years and $\max = 29$ years.

## Example 4 - Find Five (5) number summary

Following are the thorax lengths (in mm) of a sample of male fruit flies:

0.72,0.90, 0.84, 0.68,0.84, 0.90,0.92, 0.84, 0.64, 0.84,0.78

Compute five number summary.

#### Solution

Arrange the data in ascending order

17, 18, 18, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 24, 25, 25, 26, 27, 28, 29.

Minimum Value

The minimum thorax length is $\min = 17$ mm.

Maximum Value

The maximum thorax length is $\max = 29$ mm.

Quartiles

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.

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)\\ &=19+0.25\big(20 -19\big)\\ &=19.25 \end{aligned}

Thus, lower $25$ % of fruit flies thorax length less than or equal to $19.25$ mm.

Median (M) (i.e., Second Quartile $Q_2$)

The median ($M$) or second quartile $Q_2$ can be computed as follows:

 \begin{aligned} M &=\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)\\ &=21+0.5\big(22 -21\big)\\ &=21.5 \end{aligned}

Thus, lower $50$ % of fruit flies thorax length less than or equal to $21.5$ mm.

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)\\ &=25+0.75\big(25 -25\big)\\ &=25 \end{aligned}
Thus, lower $75$ % of fruit flies thorax length less than or equal to $25$ mm.

Thus the five number summary of given data set is

$\min = 17$ mm, $Q_1 = 19.25$ mm, $\text{median }=21.5$ mm, $Q_3=25$ mm and $\max = 29$ mm.

## Example 5 - Calculate five Number summary

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 five number summary.

#### Solution

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.

Minimum Value

The minimum drying time of latex paint is $\min = 2.5$ hours.

Maximum Value

The maximum drying time of latex paint is $\max = 5.6$ hours.

Quartiles

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.

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}\\ &= \text{Value of }\big(4\big)^{th} \text{ obs.}\\ &\quad +0 \big(\text{Value of } \big(4\big)^{th}\text{ obs.}-\text{Value of }\big(4\big)^{th} \text{ obs.}\big)\\ &=2.9+0\big(2.9 -2.9\big)\\ &=2.9 \end{aligned}

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

Median (M) (i.e., Second Quartile $Q_2$)

The median ($M$) or second quartile $Q_2$ can be computed as follows:

 \begin{aligned} M &=\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.}\\ &\quad +0 \big(\text{Value of } \big(8\big)^{th}\text{ obs.}-\text{Value of }\big(8\big)^{th} \text{ obs.}\big)\\ &=3.6+0\big(3.6 -3.6\big)\\ &=3.6 \end{aligned}

Thus, lower $50$ % of drying time of latex paint 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}\\ &= \text{Value of }\big(12\big)^{th} \text{ obs.}\\ &\quad +0 \big(\text{Value of } \big(12\big)^{th}\text{ obs.}-\text{Value of }\big(12\big)^{th} \text{ obs.}\big)\\ &=4.8+0\big(4.8 -4.8\big)\\ &=4.8 \end{aligned}
Thus, lower $75$ % of drying time of latex paint is less than or equal to $4.8$ hours.

Thus the five number summary of given data set is

$\min = 2.5$ hours, $Q_1 = 2.9$ hours, $\text{median }=3.6$ hours, $Q_3=4.8$ hours and $\max = 5.6$ hours.

## Conclusion

In this tutorial, you learned about formula for five nimber summary for ungrouped data and how to calculate five number summary for ungrouped data. You also learned about how to solve numerical problems based on five (5) number summary for ungrouped data. 