Just like the median divides the set of observation into two equal parts when arranged in the numerical order, in the same way quartile divides the set of observation into 4 equal parts.

The value of the middle term, between the first term and median is known as*first or Lower Quartile* and is denoted as **Q**_{1}. The value of middle term between the last term and the median is known as *third or Upper Quartile *and is denoted as **Q**_{3}.The median itself is known as the Second Quartile and is denoted as **Q**_{2}

The value of the middle term, between the first term and median is known as

When the set of observation is arranged in an ascending order, then the Lower quartile is given as:

Q$_1$ = $(\frac{n+1}{4})^{th}$ term

If the solution is a decimal number then, Lower quartile Q_{1} is given by rounding it to the nearest whole integer.

The Second quartile, which is the median of the set of observation is given as:

The Second quartile, which is the median of the set of observation is given as:

Q$_2$ = $(\frac{n+1}{2})^{th}$ term

The Upper quartile is given as,

Q$_3$ = $(\frac{3(n+1)}{4})^{th})$ term

If the solution is a decimal number then, Upper quartile Q_{3} is given by rounding it to the nearest whole integer.

The lower and the upper quartile value helps us to find the measure of dispersion in the set of observation, which is called as '**inter-quartile range',** it is denoted as **IQR **and it is the difference between upper and lower quartile.

IQR = $Q_3 - Q_1$

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Below are the problems based on Quartile: ### Solved Examples

**Question 1: **Find the median, lower quartile, upper quartile and inter-quartile range of the following data set of scores:
19, 22, 24, 20, 24, 27, 25, 24, 30 ?

** Solution: **

First, lets arrange of the values in an ascending order:

19, 20, 22, 24, 24, 24, 25, 27, 30

Now lets calculate the Median,

Median = $\left(\frac{n + 1}{2} \right)^{th}$ term

= $\left(\frac{9 + 1}{2} \right)^{th}$ term

= 5^{th} term

= 24

Lower quartile = $\left(\frac{n + 1}{4} \right)^{th}$ term

= $\left(\frac{9 + 1}{4} \right)^{th}$ term

= $\left(\frac{10}{4} \right)^{th}$ term

= 2.5^{th}

Find the average of 2^{nd} and 3^{rd} term

= $\frac{20 + 22}{2}$

= $\frac{42}{2}$

= 21

Upper quartile = $\left(\frac{3(n + 1)}{4} \right)^{th}$

= $\left(\frac{3(9 + 1)}{4} \right)^{th}$

= $\left(\frac{3(10)}{4} \right)^{th}$

= $\left(\frac{30}{4} \right)^{th}$

= 7.5^{th}

(lets find the average of 7th and 8th term)

= $\frac{25 + 27}{2}$

= $\frac{52}{2}$

= 26

Inter - quartile = Upper quartile - lower quartile

= 26 - 21

= 5

**Question 2: **Find the first quartile, second quartile and third quartile of the given information of the following sequence
4, 77, 16, 59, 93, 88 ?

** Solution: **

First, lets arrange of the values in an ascending order :

4, 16, 59, 77, 88, 93

Given n = 6

$\therefore$ Lower quartile = $\left(\frac{n + 1}{4} \right)^{th}$ term

= $\left(\frac{6 + 1}{4} \right)^{th}$ term

= $\left(\frac{7}{4} \right)^{th}$ term

= 1.7^{th} term

Here we can consider the 2^{nd} term (rounding 1.7 to nearest whole integer) from the set of observation.

$\Rightarrow$ 2^{nd} term = 16

Lower quartile = 16

Upper quartile = $\left(\frac{3(n + 1)}{4} \right)^{th}$ term

= $\left(\frac{3(6 + 1)}{4} \right)^{th}$ term

= $\left(\frac{21}{4} \right)^{th}$ term

= 5.25^{th}

Here we can consider the 5^{th }term (rounding 5.25 to nearest whole integer) from the set of observation.

$\Rightarrow$ 5.25^{th} = 88

Upper quartile = 88

Inter - quartile = Upper quartile - lower quartile

= 88 - 16

= 72

First, lets arrange of the values in an ascending order:

19, 20, 22, 24, 24, 24, 25, 27, 30

Now lets calculate the Median,

Median = $\left(\frac{n + 1}{2} \right)^{th}$ term

= $\left(\frac{9 + 1}{2} \right)^{th}$ term

= 5

= 24

Lower quartile = $\left(\frac{n + 1}{4} \right)^{th}$ term

= $\left(\frac{9 + 1}{4} \right)^{th}$ term

= $\left(\frac{10}{4} \right)^{th}$ term

= 2.5

Find the average of 2

= $\frac{20 + 22}{2}$

= $\frac{42}{2}$

= 21

Upper quartile = $\left(\frac{3(n + 1)}{4} \right)^{th}$

= $\left(\frac{3(9 + 1)}{4} \right)^{th}$

= $\left(\frac{3(10)}{4} \right)^{th}$

= $\left(\frac{30}{4} \right)^{th}$

= 7.5

(lets find the average of 7th and 8th term)

= $\frac{25 + 27}{2}$

= $\frac{52}{2}$

= 26

Inter - quartile = Upper quartile - lower quartile

= 26 - 21

= 5

First, lets arrange of the values in an ascending order :

4, 16, 59, 77, 88, 93

Given n = 6

$\therefore$ Lower quartile = $\left(\frac{n + 1}{4} \right)^{th}$ term

= $\left(\frac{6 + 1}{4} \right)^{th}$ term

= $\left(\frac{7}{4} \right)^{th}$ term

= 1.7

Here we can consider the 2

$\Rightarrow$ 2

Lower quartile = 16

Upper quartile = $\left(\frac{3(n + 1)}{4} \right)^{th}$ term

= $\left(\frac{3(6 + 1)}{4} \right)^{th}$ term

= $\left(\frac{21}{4} \right)^{th}$ term

= 5.25

Here we can consider the 5

$\Rightarrow$ 5.25

Upper quartile = 88

Inter - quartile = Upper quartile - lower quartile

= 88 - 16

= 72

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Interquartile Range Formula | |