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.
To learn more about other descriptive statistics measures, please refer to the following tutorials:
Let me know in the comments if you have any questions on Five number summary calculator for ungrouped data with examples and your thought on this article.
