5.1.3 Box and Whisker Plots

In this topic we will learn how to:

• draw and interpret box and whisker plots

A box and whisker plot is used to represent discrete data. However, it does not show all the data points. It consists of the box and the whiskers. It helps to give a general understanding of how the data is spread.

To draw a box and whisker plot you need the following information:

• minimum value
• lower quartile
• median
• upper quartile
• maximum value

To calculate the median, we use the formula,
$$q_{2} = \frac{n + 1}{2}$$The formula above, gives us the position of the median, which is denoted by $q_{2}$ and $n$ represents the sample size.

To calculate the lower quartile, we use the formula,
$$q_{1} = \frac{1}{4}(n + 1)$$Where $q_{1}$ represents the lower quartile.

To calculate the upper quartile, we use the formula,
$$q_{3} = \frac{3}{4}(n + 1)$$Where $q_{3}$ represents the upper quartile.

To calculate the interquartile range, we use the formula,
$$IQR = q_{3} - q_{1}$$Where $IQR$ represents the interquartile range, $q_{3}$ represents the upper quartile, $q_{1}$ represents the lower quartile.

Let’s look at some past paper questions.

1. The times in minutes taken to run a marathon were recorded for a group of $33$ marathon runners and were found to be as follows:

(9709/62/M/J/19 number 6)

(a) Draw a box and whisker plot to illustrate the times for these $33$ people.

To be able to draw a box and whisker plot. We need the minimum value, lower quartile, median, upper quartile and the maximum value. The minimum value is,
$$190$$The maximum value is,
$$375$$Let’s find the lower quartile,
$$q_{1} = \frac{1}{4}(n + 1)$$$$q_{1} = \frac{1}{4}(33 + 1)$$$$q_{1} = 8.5$$Notice how, there is no data point in position $8.5$, so we get the average of data points on position $\textcolor{#2192ff}{8}$ and $\textcolor{#2192ff}{9}$,

$$q_{1} = \frac{\textcolor{#2192ff}{254} + \textcolor{#2192ff}{258}}{2}$$$$q_{1} = 256$$Let’s find the upper quartile,
$$q_{3} = \frac{3}{4}(n + 1)$$$$q_{3} = \frac{3}{4}(33 + 1)$$$$q_{3} = 25.5$$Notice how, there is no data point in position $25.5$, so we get the average of the data points on position $\textcolor{#2192ff}{25}$ and $\textcolor{#2192ff}{26}$,

$$q_{3} = \frac{\textcolor{#2192ff}{327} + \textcolor{#2192ff}{331}}{2}$$$$q_{3} = 329$$Let’s find the median,
$$q_{2} = \frac{1}{2}(n + 1)$$$$q_{2} = \frac{1}{2}(33 + 1)$$$$q_{2} = 17$$Find the data point at position $\textcolor{#2192ff}{17}$,

$$q_{2} = \textcolor{#2192ff}{280}$$Now that we have the necessary requirements, we can now draw the box and whisker plot. Mark the minimum and maximum values with a line each, these will represent the whiskers. To draw the box, mark the lower quartile and the upper quartile, these will be the edges of the box. Inside the box, label the median with a line. Finally, join the whiskers to the box. Don’t forget to label the $x$-axis, with the appropriate label. The box represents the interquartile range, the whiskers represent the extreme values.

(b) Find the interquartile range of these times.

The formula for interquartile range is,
$$IQR = q_{3} - q_{1}$$Substitute the lower and upper quartiles,
$$IQR = 329 - 256$$$$IQR = 73$$Therefore, the final answer is,
$$IQR = 73$$2. Two machines $A$ and $B$, produce metal rods of a certain type. The lengths, in metres, of $19$ rods produced by machine $A$ and $19$ rods produced by machine $B$ are measured. It is given that, the minimum value of $A$ is $0.220$, maximum value is $0.254$, lower quartile is $0.231$, upper quartile is $0.245$, median is $0.238$. It is given that, the minimum value of $B$ is $0.211$, maximum value is $0.256$, lower quartile is $0.224$, upper quartile is $0.243$, median is $0.232$.(9709/52/M/J/20 number 3)

(a) Draw box and whisker plots for $A$ and $B$.

Sketch the box and whisker plots, with $A$ above $B$, and label them. Ensure that the extreme values are the whiskers, and the box is marked by the lower and upper quartiles. Insert the lines for the median inside their respective boxes. Don’t forget to label the $x$-axis.

(b) Hence make two comparisons between the lengths of the rods produced by machine $A$ and those produced by machine $B$.

Since $B$ has a larger interquartile range (box) than $A$, the length of rods it produces is more spread.
Since the median of $B$ is smaller than that of $A$, the lengths of $B$ are, on average smaller than those of $A$.