Inter Quartile Range for Grouped Data Calculator
Use this calculator to find the Inter Quartile Range for grouped (frequency distribution) data.
Calculator
Inter Quartile Range Calculator (Grouped Data) | |
---|---|
Type of Frequency Distribution | DiscreteContinuous |
Enter the Classes for X (Separated by comma,) | |
Enter the frequencies (f) (Separated by comma,) | |
Results | |
Number of Observation (N): | |
First Quartile : ($Q_1$) | |
Second Quartile : ($Q_2$) | |
Third Quartile : ($Q_3$) | |
Inter Quartile Range : $IQR$ | |
How to find Inter Quartile Range (IQR) 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 Inter quartile range
Step 5 - Gives output as number of observation (N)
Step 6 - Calculate three quartiles $Q_1$, $Q_2$ and $Q_3$
Step 7 - Calculate Inter quartile range (IQR)
Inter Quartile Range for grouped data
Inter quartile range is the difference between the third quartile $Q_3$ and first quartile $Q_1$. It is a good measure of spread to use for skewed distribution. Inter quartile range (IQR) 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
$$ \begin{aligned} Q_i=l + \bigg(\frac{\frac{iN}{4} - F_<}{f}\bigg)\times h; \quad i=1,2,3 \end{aligned} $$
where,
- $l :$ 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
- $h :$ the class width
Inter Quartile Range Example 1
A class teacher has the following data about the number of absences of 35 students of a class. find inter quartile range 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 |
Inter quartile range (IQR)
The inter quartile range is given by $IQR= Q_3-Q_1$.
The formula for $i^{th}$ quartile range for grouped data 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.
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.
Inter quartile range
The inter quartile range is
$$ \begin{aligned} IQR & = Q_3 - Q_1\\ &= 5 - 3\\ & = 2. \end{aligned} $$
Inter Quartile Range Example 2
The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students.
Time spent on Internet ($x$) | No. of students ($f$) |
---|---|
10-12 | 3 |
13-15 | 12 |
16-18 | 15 |
19-21 | 24 |
22-24 | 2 |
Calculate Inter quartile range for the frequency distribution.
Solution
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 range for grouped data 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.
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.
Inter quartile range
The inter quartile range is
$$ \begin{aligned} IQR & = Q_3 - Q_1\\ &= 20 - 15.25\\ & = 4.75 \text{ minutes}. \end{aligned} $$
Inter Quartile Range 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 inter quartile range 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 for grouped data 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.
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.
Inter quartile range
The inter quartile range is
$$ \begin{aligned} IQR & = Q_3 - Q_1\\ &= 47.75 - 26.5625\\ & = 21.1875 \text{ Scores}. \end{aligned} $$
Inter Quartile Range 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 |
Calculate inter quartile range 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 for grouped data 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.
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.
Inter quartile range
The inter quartile range is
$$ \begin{aligned} IQR & = Q_3 - Q_1\\ &= 11.5714 - 10.5833\\ & = 0.9881 \text{ tons}. \end{aligned} $$
Conclusion
Hope you like article on how to calculate Inter Quartile for grouped data and step by step procedure to solve numerical problems based on IQR 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 Inter Quartile Range calculator for grouped data with examples and your thought on this article.
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