Quartiles

GCSELevel 6-7Edexcel iGCSE

Quartiles

Where the median is the value that splits the data set evenly in two, the quartiles can be used to further separate the data into four equally sized sets.

The lower quartile can be thought of as the “median” of the lower half of the data.

Similarly the upper quartile can be thought of as the “median” of the higher half of the data.

Upper and Lower Quartiles

The lower quartile ($Q_1$) is the value in the ordered data such that $75\%$ of the data is larger than it and $25\%$ of the data is smaller

In a data set size $n$, $Q_1$ is in position $\dfrac{n+1}{4}$

The upper quartile ($Q_3$) is the value in the ordered data such that $75\%$ of the data is smaller than it and $25\%$ of the data is larger

In a data set size $n$, $Q_3$ is in position $\dfrac{3(n+1)}{4}$

$Q_2$ denotes the median, found at $\dfrac{n+1}{2}$

If the position of the upper or lower quartile is not a whole number, we must take an average of the data in the next position up, and the data in the next position down. For example if $n = 26$, then

$\dfrac{26+1}{4} = 6.75$ and $\dfrac{3(25+1)}{4} = 20.25$

In this case $Q_1$ will be the mean of the data in position $6$ and $7$, added together and divided by two. $Q_3$ will be the mean of the data in position $20$ and $21$.

Level 6-7GCSEEdexcel iGCSE

Interquartile Range

Interquartile range (IQR) is a measure of the distribution of the data.

$IQR = Q_3 - Q_1$

IQR can be more useful than range as extreme outliers will lie outside it and do not affect the IQR.

Level 6-7GCSEEdexcel iGCSE

Example 1: Finding the UQ

Considering the following data set

$2, 36, 48, 49, 58, 62, 92$

find the upper quartile.

[1 mark]

There are $7$ numbers in this data set. We can find the position of $Q_3$

$Q_3 \text{ position } = \dfrac{3(7+ 1)}{4} = 6$th,

The sixth value is $62$, so $Q_3 = 62$.

Level 6-7GCSEEdexcel iGCSE

Example 2: Finding the LQ

Considering the integers from $1$ to $159$, find their lower quartile.

[1 mark]

There are $159$ numbers in this data set. We can find the position of $Q_1$

$Q_1 \text{ position } = \dfrac{159+ 1}{4} = 40$th,

As the first value is $1$ and the second value is $2$ and so on, the $40^{\text{th}}$ value is $40$.

Hence, $Q_1 = 40$

Level 6-7GCSEEdexcel iGCSE

Example 3: Finding the IQR

Considering the following data set

$-45, 3, 5, 5, 5, 6, 7, 8, 10, 10, 118$

find the interquartile range.

[3 marks]

As the data set has $11$ elements, we can find the position of $Q_1$

$Q_1 \text{ position } = \dfrac{11+ 1}{4} = 3$ rd,

The third number is $5$, so $Q_1 = 5$. Now for $Q3$,

$Q_3 \text{ position } = \dfrac{3(11+ 1)}{4} = 9$th,

The ninth number is $10$, so $Q_3 = 10$. Now for the IQR.

$IQR = Q_3 - Q_1 = 10 - 5 = 5$

The IQR $= 5$.

Notice how this differs from the range of $163$ as outliers are ignored.

Level 6-7GCSEEdexcel iGCSE

Quartiles Example Questions

a) We can find the position of the lower quartile in the data set with the formula

$Q_1$ Position $= \dfrac{n+1}{4}$

There are $19$ entries in the data set and so $n = 19$

$Q_1$ Position $= \dfrac{19+1}{4}=\dfrac{20}{4}=5$,

The $5^{th}$ value in the data set is the lower quartile, which is $43$.

b) We can find the position of the upper quartile in the data set with the formula

$Q_3$ Position $= \dfrac{3(n+1)}{4}$

There are $19$ entries in the data set and so $n = 19$

$Q_3$ Position $= \dfrac{3(19+1)}{4}$

$Q_3$ Position $= \dfrac{3 \times 20}{4} = \dfrac{60}{4} = 15$

The $15^{th}$ value in the data set is the upper quartile, which is $168$

Gold Standard Education

Firstly we must order this data. Which when done gives us

$13, 27, 33, 37, 55, 57, 71, 77, 84, 94, 97$

The data set has $11$ elements. So we can find the position of $Q_1$ and $Q_3$

$Q_1$ Position $= \dfrac{11+1}{4} = 3$

The third element is $33$ and so it is the lower quartile.

$Q_3$ Position $= \dfrac{3(11+1)}{4} = 9$

The ninth element is $84$ and so it is the upper quartile.

Then the interquartile range

$IQR = Q_3 -Q_1$
$IQR = 84 - 33 = 51$

Gold Standard Education

Both classes 1 and 2 have $11$ students. So for both, the lower quartile is in the $3^{\text{rd}}$ position and the upper quartile is in the $9^{\text{th}}$ position.

Class 1:
$Q_3 =99$
$Q_1 = 72$
$IQR = 99 - 72 = 27$

Class 2:
$Q_3 =79$
$Q_1 = 71$
$IQR = 79 - 71 = 8$

While class 2 had the lower median grade, it did have the more consistent grades, despite outliers.