Quartiles for grouped data

Quartiles are the values which divide whole distriution 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.

Formula

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

Example 1

A class teacher has the following data about the number of absences of 35 students of a class. Compute five number summary 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.

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 five number summary for the following frequency distribution.

Time spent on Internet ($x$) 10-12 13-15 16-18 19-21 22-24
No. of students ($f$) 3 12 15 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.

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