Quartiles and Interquartile Range

Quartiles and Interquartile Range exercise

• Explain how to find the interquartile range.

  Hints

  Here you see the list ordered.

  Usually the data gets ordered first.

  For an odd number of data points we get two sets, one left of of the median and one right of the median, with an equal number of data points.

  Solution

  To get the interquartile range, we have to proceed with the following steps:

  1. Order the data, as pictured to the right.
  2. Count the total number of data points, $19$.
  3. Split the list into two equal halves. Because the number of data points is odd, the median is right in the middle at position $10$ and we have that $Q_2=8$.
  4. Split the lower as well as the upper part of the list, each containing $9$ elements. So we get $Q_1=5$ as the median of the lower part and $Q_3=15$ as the median of the upper part.
  5. The interquartile range is defined as the difference of $Q_3$ and $Q_1$. So we get $IQR=Q_3-Q_1=15-5=10$.

• Determine the median, the quartiles, and the interquartile range.

  Hints

  If you've got an odd number of data points the median lies right in the middle of the list of data. For example, $4$ is the median of the list

  $\begin{array}{ccccccc} 1&2&3&4&5&6&7 \end{array}$

  For the list pictured, we have that:

  • $Q_2=4$
  • $Q_1=2$
  • $Q_3=6$
  • $IQR=6-2=4$

  Solution

  To get the interquartile range, we have to proceed with the following steps:

  1. Order the data, as pictured to the right.
  2. Count the total number of data points, $19$.
  3. Split the list into two equal halves. Because the number of data points is odd, the median is right in the middle at position $10$ and we have that $Q_2=8$.
  4. Split the lower as well as the upper part of the list, each containing $9$ elements. So we get $Q_1=5$ as the median of the lower part and $Q_3=15$ as the median of the upper part.
  5. The interquartile range is defined as the difference of $Q_3$ and $Q_1$. So we get $IQR=Q_3-Q_1=15-5=10$.

  We can also then see that:

  • The first quartile range starts with $0$ and ends with $4$.
  • The second quartile range starts with $6$ and ends with $7$.
  • The third quartile range starts with $9$ and ends with $14$.
  • The fourth quartile range starts with $17$ and ends with $20$.

• Determine the median, the quartiles, and the interquartile range.

  Hints

  To get the mean of two numbers add them and divide the sum by $2$.

  The number of data points in the list pictured is $6$.

  The median is the mean of the data points at position $\frac 62=3$ and $4$; i.e. the mean of $9$ and $11$, which is

  $\frac{9+11}2=\frac{20}2=10$.

  Solution

  Here we have an ordered list with $20$ data points, an even number.

  The median lies between the data points at position $\frac{20}2=10$ and position $11$; i.e. between the data points $7$ and $9$. We take the mean of these points as the median:

  $\frac{9+11}2=\frac{20}2=10$.

  Thus we can conclude that $Q_2=10$.

  We proceed in a similar fashion for the lower part of this list,

  $\begin{array}{cccccccccc} 1&1&3&4&5&5&6&7&8&9 \end{array}$

  $Q_1=\frac{5+7}2=\frac{12}2=6$,

  and for the upper one,

  $\begin{array}{cccccccccc} 11&12&12&13&14&16&16&17&18&20 \end{array}$

  $Q_3=\frac{14+16}2=\frac{30}2=15$.

  Finally we can determine the interquartile range as the difference of $Q_3$ and $Q_1$:

  $IQR=15-6=9$.

• Calculate which data points are in which quartile range.

  Hints

  First order the data list.

  This is the ordered list.

  Each quartile contains $4$ data points.

  Be careful: the medians should not be highlighted.

  Solution

  First you have to order the list of data. You can see the ordered list to the right.

  Next determine the medians $Q_1$, $Q_2$, and $Q_3$ which divide the list in the four quartile ranges.

  Let's start with $Q_2$; it's right in the middle of the starting list: $Q_2=17$.

  The middle of the lower part,

  $\begin{array}{ccccccccc} 2&5&7&7&9&10&10&13&16, \end{array}$

  is $Q_1=9$.

  The middle of the upper part,

  $\begin{array}{ccccccccc} 19&20&20&22&24&25&26&26&26, \end{array}$

  is $Q_3=24$.

  Thus we have the following quartiles:

  • The first one: $\begin{array}{cccc}2&5&7&7\end{array}$
  • The second one: $\begin{array}{cccc}10&10&13&16\end{array}$
  • The third one: $\begin{array}{cccc}19&20&20&22\end{array}$
  • The fourth one: $\begin{array}{cccc}25&26&26&26\end{array}$

• Identify how to find the median.

  Hints

  For this list the median is $7$.

  For this list the median is the mean of $7$ and $9$.

  For the example above, the mean is $\frac{7+9}2=\frac{16}2=8$.

  Solution

  How can we find the median of a given ordered list?

  We have to differentiate between two cases:

  • The number of data points is odd.
  • The number of data points is even.

  Let's illustrate what we do in each case with an example:

  Let's start with an odd number of data points:

  $\begin{array}{ccccc} 2&5&7&8&12 \end{array}$

  The median is the data point lying right in the middle of the list, in this example at position $3$. So we have that the median is $7$. The median belongs to the list of data.

  Now let's see what we do when we have an even number of data points:

  $\begin{array}{cccccc} 2&5&7&9&12&13 \end{array}$

  The median lies between $7$ and $9$. So we take the mean of those values as the median:

  $\frac{7+9}2=\frac{16}2=8$.

  This median could belong to the list of data, but not always.

  Independent of even or odd numbers, we can conclude that the median splits the list into two parts containing the same number of data points.

• Determine the median, the quartiles, and the interquartile range.

  Hints

  First order the given list.

  This is the ordered list.

  The number of data points is $13$ and thus odd. So the median is right in the middle of the ordered list.

  The number of data points in the lower as well as the upper part of the list is $6$ and thus even. So the median is the mean of both data points lying in the middle.

  Solution

  First determine $Q_1$, $Q_2$, and $Q_3$.

  $Q_2$ is the median of the ordered list. Because the number of data points is odd, namely $13$, the median is right in the middle. Thus we have $Q_2=26$.

  The lower as well as the upper part of the list contains $6$ data points. So we have to determine the corresponding median as the mean of both of the data points lying in the middle.

  Let's start with $Q_1$:

  $\begin{array}{cccccc} 9&12&\mathbf{15}&\mathbf{16}&21&23 \end{array}$

  So $Q_1=\frac{15+16}2=\frac{31}2=15.5$.

  Now, to determine $Q_3$:

  $\begin{array}{cccccc} 28&33&\mathbf{35}&\mathbf{37}&41&44 \end{array}$

  So $Q_3=\frac{35+37}2=\frac{72}2=36$.

  With $Q_3$ and $Q_1$ known, we can calculate the interquartile range:

  $IQR=Q_3-Q_1=36-15.5=20.5$

  We can now establish the quartile ranges:

  • The first one: $\begin{array}{ccc}9&12&15\end{array}$
  • The second one: $\begin{array}{ccc}16&21&23\end{array}$
  • The third one: $\begin{array}{ccc}28&33&35\end{array}$
  • The fourth one: $\begin{array}{ccc}37&41&44\end{array}$