Quartiles Calculator for grouped data
Use this calculator to find the Quartiles for grouped (frequency distribution) data.
| Quartiles Calculator (Grouped Data) | |
|---|---|
| Type of Freq. Dist. | DiscreteContinuous |
| Enter the Classes for X (Separated by comma,) | |
| Enter the frequencies (f) (Separated by comma,) | |
| Results | |
| Number of Obs. (N): | |
| First Quartile : ($Q_1$) | |
| Second Quartile : ($Q_2$) | |
| Third Quartile : ($Q_3$) | |
How to calculate Quartiles for grouped data?
Step 1 – Select type of frequency distribution (Discrete or continuous)
Step 2 – Enter the Range or classes (X) seperated by comma (,)
Step 3 – Enter the Frequencies (f) seperated by comma
Step 4 – Click on "Calculate" for quartiles
Step 5 – Gives output as number of observation (N)
Step 6 – Calculate three quartiles
Quartiles for grouped data
Quartiles are the values which divide whole distribution 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.
For discrete frequency distribution, the formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where,
- $N$ is total number of observations.
For continuous frequency distribution, the formula for $i^{th}$ quartile is
$Q_i=l + \bigg(\dfrac{\dfrac{iN}{4} - F_<}{f}\bigg)\times h$; $i=1,2,\cdots,3$
where,
- $l$ is the lower limit of the $i^{th}$ quartile class
- $N=\sum f$ total number of observations
- $f$ frequency of the $i^{th}$ quartile class
- $F_<$ cumulative frequency of the class previous to $i^{th}$ quartile class
- $h$ is the class width
Quartiles for grouped data Example 1
A class teacher has the following data about the number of absences of 35 students of a class. Compute quartiles for the following frequency distribution.
| No.of days ($x$) | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|
| No. of Students ($f$) | 1 | 15 | 10 | 5 | 4 |
Solution
| $x_i$ | $f_i$ | $cf$ | |
|---|---|---|---|
| 2 | 1 | 1 | |
| 3 | 15 | 16 | |
| 4 | 10 | 26 | |
| 5 | 5 | 31 | |
| 6 | 4 | 35 | |
| Total | 35 |
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where $N$ is the total number of observations.
First Quartile $Q_1$
$$ \begin{aligned} Q_{1} &=\bigg(\dfrac{1(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(35)}{4}\bigg)^{th}\text{ value}\\ &=\big(8.75\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $8.75$ is $16$. The corresponding value of $X$ is the $1^{st}$ quartile. That is, $Q_1 =3$ days.
Thus, $25$ % of the students had absences less than or equal to $3$ days.
Second Quartile $Q_2$
$$ \begin{aligned} Q_{2} &=\bigg(\dfrac{2(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(35)}{4}\bigg)^{th}\text{ value}\\ &=\big(17.5\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $17.5$ is $26$. The corresponding value of $X$ is the $2^{nd}$ quartile. That is, $Q_2 =4$ days.
Thus, $50$ % of the students had absences less than or equal to $4$ days.
Third Quartile $Q_3$
$$ \begin{aligned} Q_{3} &=\bigg(\dfrac{3(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(35)}{4}\bigg)^{th}\text{ value}\\ &=\big(26.25\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $26.25$ is $31$. The corresponding value of $X$ is the $3^{rd}$ quartile. That is, $Q_3 =5$ days.
Thus, $75$ % of the students had absences less than or equal to $5$ days.
Quartiles for grouped data Example 2
The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students. Compute quartiles for the following frequency distribution.
| Time spent on Internet ($x$) | No. of Students ($f$) |
|---|---|
| 10-12 | 3 |
| 13-15 | 12 |
| 16-18 | 15 |
| 19-21 | 24 |
| 22-24 | 2 |
Solution
Let $X$ denote the amount of time (in minutes) spent on the internet.
Here the classes are inclusive. To make them exclusive type subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.
| Class Interval | Class Boundries | $f_i$ | $cf$ |
|---|---|---|---|
| 10-12 | 9.5-12.5 | 3 | 3 |
| 13-15 | 12.5-15.5 | 12 | 15 |
| 16-18 | 15.5-18.5 | 15 | 30 |
| 19-21 | 18.5-21.5 | 24 | 54 |
| 22-24 | 21.5-24.5 | 2 | 56 |
| Total | 56 |
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where $N$ is the total number of observations.
First Quartile $Q_1$
$$ \begin{aligned} Q_{1} &=\bigg(\dfrac{1(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(56)}{4}\bigg)^{th}\text{ value}\\ &=\big(14\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $14$ is $15$. The corresponding class $12.5-15.5$ is the $1^{st}$ quartile class.
Thus
- $l = 12.5$, the lower limit of the $1^{st}$ quartile class
- $N=56$, total number of observations
- $f =12$, frequency of the $1^{st}$ quartile class
- $F_< = 3$, cumulative frequency of the class previous to $1^{st}$ quartile class
- $h =3$, the class width
The first quartile $Q_1$ can be computed as follows:
$$ \begin{aligned} Q_1 &= l + \bigg(\frac{\frac{1(N)}{4} - F_<}{f}\bigg)\times h\\ &= 12.5 + \bigg(\frac{\frac{1*56}{4} - 3}{12}\bigg)\times 3\\ &= 12.5 + \bigg(\frac{14 - 3}{12}\bigg)\times 3\\ &= 12.5 + \big(0.9167\big)\times 3\\ &= 12.5 + 2.75\\ &= 15.25 \text{ minutes} \end{aligned} $$
Thus, $25$ % of the students spent less than or equal to $15.25$ minutes on the internet.
Second Quartile $Q_2$
$$ \begin{aligned} Q_{2} &=\bigg(\dfrac{2(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(56)}{4}\bigg)^{th}\text{ value}\\ &=\big(28\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $28$ is $30$. The corresponding class $15.5-18.5$ is the $2^{nd}$ quartile class.
Thus
- $l = 15.5$, the lower limit of the $2^{nd}$ quartile class
- $N=56$, total number of observations
- $f =15$, frequency of the $2^{nd}$ quartile class
- $F_< = 15$, cumulative frequency of the class previous to $2^{nd}$ quartile class
- $h =3$, the class width
The second quartile $Q_2$ can be computed as follows:
$$ \begin{aligned} Q_2 &= l + \bigg(\frac{\frac{2(N)}{4} - F_<}{f}\bigg)\times h\\ &= 15.5 + \bigg(\frac{\frac{2*56}{4} - 15}{15}\bigg)\times 3\\ &= 15.5 + \bigg(\frac{28 - 15}{15}\bigg)\times 3\\ &= 15.5 + \big(0.8667\big)\times 3\\ &= 15.5 + 2.6\\ &= 18.1 \text{ minutes} \end{aligned} $$
Thus, $50$ % of the students spent less than or equal to $18.1$ minutes on the internet.
Third Quartile $Q_3$
$$ \begin{aligned} Q_{3} &=\bigg(\dfrac{3(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(56)}{4}\bigg)^{th}\text{ value}\\ &=\big(42\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $42$ is $54$. The corresponding class $18.5-21.5$ is the $3^{rd}$ quartile class.
Thus
- $l = 18.5$, the lower limit of the $3^{rd}$ quartile class
- $N=56$, total number of observations
- $f =24$, frequency of the $3^{rd}$ quartile class
- $F_< = 30$, cumulative frequency of the class previous to $3^{rd}$ quartile class
- $h =3$, the class width
The third quartile $Q_3$ can be computed as follows:
$$ \begin{aligned} Q_3 &= l + \bigg(\frac{\frac{3(N)}{4} - F_<}{f}\bigg)\times h\\ &= 18.5 + \bigg(\frac{\frac{3*56}{4} - 30}{24}\bigg)\times 3\\ &= 18.5 + \bigg(\frac{42 - 30}{24}\bigg)\times 3\\ &= 18.5 + \big(0.5\big)\times 3\\ &= 18.5 + 1.5\\ &= 20 \text{ minutes} \end{aligned} $$
Thus, $75$ % of the students spent less than or equal to $20$ minutes on the internet.
Quartiles for grouped data Example 3
The Scores of students in a Math test is given in the table below :
| Class Interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
|---|---|---|---|---|---|---|
| Frequency ($f$) | 6 | 8 | 12 | 10 | 5 | 4 |
Find quartiles for the given grouped data.
Solution
| Class Interval | Class Boundries | $f_i$ | $cf$ | |
|---|---|---|---|---|
| 10-20 | 10-20 | 6 | 6 | |
| 20-30 | 20-30 | 8 | 14 | |
| 30-40 | 30-40 | 12 | 26 | |
| 40-50 | 40-50 | 10 | 36 | |
| 50-60 | 50-60 | 5 | 41 | |
| 60-70 | 60-70 | 4 | 45 | |
| Total | 45 |
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where $N$ is the total number of observations.
First Quartile $Q_1$
$$ \begin{aligned} Q_{1} &=\bigg(\dfrac{1(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(45)}{4}\bigg)^{th}\text{ value}\\ &=\big(11.25\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $11.25$ is $14$. The corresponding class $20-30$ is the $1^{st}$ quartile class.
Thus
- $l = 20$, the lower limit of the $1^{st}$ quartile class
- $N=45$, total number of observations
- $f =8$, frequency of the $1^{st}$ quartile class
- $F_< = 6$, cumulative frequency of the class previous to $1^{st}$ quartile class
- $h =10$, the class width
The first quartile $Q_1$ can be computed as follows:
$$ \begin{aligned} Q_1 &= l + \bigg(\frac{\frac{1(N)}{4} - F_<}{f}\bigg)\times h\\ &= 20 + \bigg(\frac{\frac{1*45}{4} - 6}{8}\bigg)\times 10\\ &= 20 + \bigg(\frac{11.25 - 6}{8}\bigg)\times 10\\ &= 20 + \big(0.6562\big)\times 10\\ &= 20 + 6.5625\\ &= 26.5625 \text{ Scores} \end{aligned} $$
Thus, $25$ % of the students scores less than or equal to $26.5625$ marks in Math test.
Second Quartile $Q_3$
$$ \begin{aligned} Q_{2} &=\bigg(\dfrac{2(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(45)}{4}\bigg)^{th}\text{ value}\\ &=\big(22.5\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $22.5$ is $26$. The corresponding class $30-40$ is the $2^{rd}$ quartile class.
Thus
- $l = 30$, the lower limit of the $2^{rd}$ quartile class
- $N=45$, total number of observations
- $f =12$, frequency of the $2^{rd}$ quartile class
- $F_< = 14$, cumulative frequency of the class previous to $2^{rd}$ quartile class
- $h =10$, the class width
The second quartile $Q_2$ can be computed as follows:
$$ \begin{aligned} Q_2 &= l + \bigg(\frac{\frac{2(N)}{4} - F_<}{f}\bigg)\times h\\ &= 30 + \bigg(\frac{\frac{2*45}{4} - 14}{12}\bigg)\times 10\\ &= 30 + \bigg(\frac{22.5 - 14}{12}\bigg)\times 10\\ &= 30 + \big(0.7083\big)\times 10\\ &= 30 + 7.0833\\ &= 37.0833 \text{ Scores} \end{aligned} $$
Thus, $50$ % of the students scores less than or equal to $37.0833$ marks in Math Test.
Third Quartile $Q_3$
$$ \begin{aligned} Q_{3} &=\bigg(\dfrac{3(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(45)}{4}\bigg)^{th}\text{ value}\\ &=\big(33.75\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $33.75$ is $36$. The corresponding class $40-50$ is the $3^{rd}$ quartile class.
Thus
- $l = 40$, the lower limit of the $3^{rd}$ quartile class
- $N=45$, total number of observations
- $f =10$, frequency of the $3^{rd}$ quartile class
- $F_< = 26$, cumulative frequency of the class previous to $3^{rd}$ quartile class
- $h =10$, the class width
The third quartile $Q_3$ can be computed as follows:
$$ \begin{aligned} Q_3 &= l + \bigg(\frac{\frac{3(N)}{4} - F_<}{f}\bigg)\times h\\ &= 40 + \bigg(\frac{\frac{3*45}{4} - 26}{10}\bigg)\times 10\\ &= 40 + \bigg(\frac{33.75 - 26}{10}\bigg)\times 10\\ &= 40 + \big(0.775\big)\times 10\\ &= 40 + 7.75\\ &= 47.75 \text{ Scores} \end{aligned} $$
Thus, $75$ % of the students scores less than or equal to $47.75$ marks in Math Test.
Quartiles for grouped data Example 4
The following data shows the distribution of maximum loads in short tons supported by certain cables produced by a company:
| Maximum load | No. of Cables |
|---|---|
| 9.25-9.75 | 2 |
| 9.75-10.25 | 5 |
| 10.25-10.75 | 12 |
| 10.75-11.25 | 17 |
| 11.25-11.75 | 14 |
| 11.75-12.25 | 6 |
| 12.25-12.75 | 3 |
| 12.75-13.25 | 1 |
Compute quartiles for the above frequency distribution.
Solution
| Class Interval | Class Boundries | $f_i$ | $cf$ | |
|---|---|---|---|---|
| 9.25-9.75 | 9.25-9.75 | 2 | 2 | |
| 9.75-10.25 | 9.75-10.25 | 5 | 7 | |
| 10.25-10.75 | 10.25-10.75 | 12 | 19 | |
| 10.75-11.25 | 10.75-11.25 | 17 | 36 | |
| 11.25-11.75 | 11.25-11.75 | 14 | 50 | |
| 11.75-12.25 | 11.75-12.25 | 6 | 56 | |
| 12.25-12.75 | 12.25-12.75 | 3 | 59 | |
| 12.75-13.25 | 12.75-13.25 | 1 | 60 | |
| Total | 60 |
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$
where $N$ is the total number of observations.
First Quartile $Q_1$
$$ \begin{aligned} Q_{1} &=\bigg(\dfrac{1(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(60)}{4}\bigg)^{th}\text{ value}\\ &=\big(15\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $15$ is $19$. The corresponding class $10.25-10.75$ is the $1^{st}$ quartile class.
Thus
- $l = 10.25$, the lower limit of the $1^{st}$ quartile class
- $N=60$, total number of observations
- $f =12$, frequency of the $1^{st}$ quartile class
- $F_< = 7$, cumulative frequency of the class previous to $1^{st}$ quartile class
- $h =0.5$, the class width
The first quartile $Q_1$ can be computed as follows:
$$ \begin{aligned} Q_1 &= l + \bigg(\frac{\frac{1(N)}{4} - F_<}{f}\bigg)\times h\\ &= 10.25 + \bigg(\frac{\frac{1*60}{4} - 7}{12}\bigg)\times 0.5\\ &= 10.25 + \bigg(\frac{15 - 7}{12}\bigg)\times 0.5\\ &= 10.25 + \big(0.6667\big)\times 0.5\\ &= 10.25 + 0.3333\\ &= 10.5833 \text{ tons} \end{aligned} $$
Thus, $25$ % of the cables less than or equal to $10.5833$ tons of maximum load.
Second Quartile $Q_2$
$$ \begin{aligned} Q_{2} &=\bigg(\dfrac{2(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(60)}{4}\bigg)^{th}\text{ value}\\ &=\big(30\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $30$ is $36$. The corresponding class $10.75-11.25$ is the $2^{rd}$ quartile class.
Thus
- $l = 10.75$, the lower limit of the $2^{rd}$ quartile class
- $N=60$, total number of observations
- $f =17$, frequency of the $2^{rd}$ quartile class
- $F_< = 19$, cumulative frequency of the class previous to $2^{rd}$ quartile class
- $h =0.5$, the class width
The second quartile $Q_2$ can be computed as follows:
$$ \begin{aligned} Q_2 &= l + \bigg(\frac{\frac{2(N)}{4} - F_<}{f}\bigg)\times h\\ &= 10.75 + \bigg(\frac{\frac{2*60}{4} - 19}{17}\bigg)\times 0.5\\ &= 10.75 + \bigg(\frac{30 - 19}{17}\bigg)\times 0.5\\ &= 10.75 + \big(0.6471\big)\times 0.5\\ &= 10.75 + 0.3235\\ &= 11.0735 \text{ tons} \end{aligned} $$
Thus, $50$ % of the cables less than or equal to $11.0735$ tons of maximum load.
Third Quartile $Q_3$
$$ \begin{aligned} Q_{3} &=\bigg(\dfrac{3(N)}{4}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(60)}{4}\bigg)^{th}\text{ value}\\ &=\big(45\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $45$ is $50$. The corresponding class $11.25-11.75$ is the $3^{rd}$ quartile class.
Thus
- $l = 11.25$, the lower limit of the $3^{rd}$ quartile class
- $N=60$, total number of observations
- $f =14$, frequency of the $3^{rd}$ quartile class
- $F_< = 36$, cumulative frequency of the class previous to $3^{rd}$ quartile class
- $h =0.5$, the class width
The third quartile $Q_3$ can be computed as follows:
$$ \begin{aligned} Q_3 &= l + \bigg(\frac{\frac{3(N)}{4} - F_<}{f}\bigg)\times h\\ &= 11.25 + \bigg(\frac{\frac{3*60}{4} - 36}{14}\bigg)\times 0.5\\ &= 11.25 + \bigg(\frac{45 - 36}{14}\bigg)\times 0.5\\ &= 11.25 + \big(0.6429\big)\times 0.5\\ &= 11.25 + 0.3214\\ &= 11.5714 \text{ tons} \end{aligned} $$
Thus, $75$ % of the cables less than or equal to $11.5714$ tons of maximum load.
Conclusion
In this tutorial, you learned about formula for quartiles for grouped data and how to calculate quartiles for grouped data. You also learned about how to solve numerical problems based on quartiles for grouped 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 Quartiles calculator for grouped data with examples and your thought on this article.
