5.1.2 Stem and Leaf Diagrams - GCEALevelMaths9709

In this topic we will learn how to:

draw and interpret stem and leaf diagrams

A stem and leaf diagram is used to represent discrete data. It consists of a stem, which defines the scale for the data, and a leaf on which the data is plotted. It also consists of a key which explains how to read the data.

We can calculate the median from a stem and leaf plot. To do that 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.

We can calculate the lower quartile from a stem and leaf plot. To do that we use the formula,
$q_{1} = \frac{1}{4}(n + 1)$ Where $q_{1}$ represents the lower quartile.

We can calculate the upper quartile from a stem and leaf plot. To do that we use the formula,
$q_{3} = \frac{3}{4}(n + 1)$ Where $q_{3}$ represents the upper quartile.

We can calculate the interquartile range from a stem and leaf plot. To do that 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 following table gives the weekly snowfall, in centimetres, for $11$ weeks in $2018$ at two ski resorts, Dados and Linva.

Dados	6	8	12	15	10	36	42	28	10	22	16
Linva	2	11	15	16	0	32	36	40	10	12	9

(9709/52/O/N/20 number 5)

(a) Represent the information in a back-to-back stem-and-leaf diagram.

$\colorbox{white!45}{\begin{tabular}{lllll|c|lllll}\multicolumn{5}{r|}{\textbf{Dados}} & \textbf{} & \multicolumn{5}{l}{\textbf{Linva}}\\[0.024cm]\hline& & & 8 & 6 & 0 & 0 & 2 & 9 & & \\[0.124cm]6 & 5 & 2 & 0 & 0 & 1 & 0 & 1 & 2 & 5 & 6 \\[0.124cm]& & & 8 & 2 & 2 & & & & & \\[0.124cm]& & & & 6 & 3 & 2 & 6 & & & \\[0.124cm]& & & & 2 & 4 & 0 & & & & \\[0.124cm]\end{tabular}}$

To determine the scale in the stem, look at the lowest and highest values in the dataset. In our case, that is $0$ and $42$ respectively. This means our scale should range from $0$ to $50$ to encompass all the data, hence the reason for the scale above. Then plot the data, for Davos on the left hand side, from right to left i.e numbers increase from right to left. Plot data for Linva on the right hand side, from left hand side to right hand side i.e numbers increase going to the right. Don’t forget to insert a key to help read your stem-and-leaf diagram.

(b) Find the median and interquartile range for the weekly snowfall in Dados.

To find the median we use the formula,
$q_{2} = \frac{n + 1}{2}$ Substitute into the formula,
$q_{2} = \frac{11 + 1}{2}$ $q_{2} = 6$ Note: Remember that this formula gives us the position of the median.

Find the data point in position number $6$ for Dados, remember to count from right to left,

$\colorbox{white!45}{\begin{tabular}{lllll|c|lllll}\multicolumn{5}{r|}{\textbf{Dados}} & \textbf{} & \multicolumn{5}{l}{\textbf{Linva}}\\[0.024cm]\hline& & & 8 & 6 & 0 & 0 & 2 & 9 & & \\[0.124cm]6 & \textcolor{blue}{5} & 2 & 0 & 0 & 1 & 0 & 1 & 2 & 5 & 6 \\[0.124cm]& & & 8 & 2 & 2 & & & & & \\[0.124cm]& & & & 6 & 3 & 2 & 6 & & & \\[0.124cm]& & & & 2 & 4 & 0 & & & & \\[0.124cm]\end{tabular}}$

$q_{2} = \textcolor{#2192ff}{15}$ To find the interquartile range we use the formula,
$IQR = q_{3} - q_{1}$ Let’s find $q_{3}$ ,
$q_{3} = \frac{3}{4}(n + 1)$ $q_{3} = \frac{3}{4}(11 + 1)$ $q_{3} = 9$ Find the data point in position $9$ ,

$\colorbox{white!45}{\begin{tabular}{lllll|c|lllll}\multicolumn{5}{r|}{\textbf{Dados}} & \textbf{} & \multicolumn{5}{l}{\textbf{Linva}}\\[0.024cm]\hline& & & 8 & 6 & 0 & 0 & 2 & 9 & & \\[0.124cm]6 & 5 & 2 & 0 & 0 & 1 & 0 & 1 & 2 & 5 & 6 \\[0.124cm]& & & \textcolor{blue}{8} & 2 & 2 & & & & & \\[0.124cm]& & & & 6 & 3 & 2 & 6 & & & \\[0.124cm]& & & & 2 & 4 & 0 & & & & \\[0.124cm]\end{tabular}}$

$q_{3} = \textcolor{#2192ff}{28}$ Let’s find $q_{1}$ ,
$q_{1} = \frac{1}{4}(n + 1)$ $q_{1} = \frac{1}{4}(11 + 1)$ $q_{1} = 3$ Find the data point in position $3$ ,

$\colorbox{white!45}{\begin{tabular}{lllll|c|lllll}\multicolumn{5}{r|}{\textbf{Dados}} & \textbf{} & \multicolumn{5}{l}{\textbf{Linva}}\\[0.024cm]\hline& & & 8 & 6 & 0 & 0 & 2 & 9 & & \\[0.124cm]6 & 5 & 2 & 0 & \textcolor{blue}{0} & 1 & 0 & 1 & 2 & 5 & 6 \\[0.124cm]& & & 8 & 2 & 2 & & & & & \\[0.124cm]& & & & 6 & 3 & 2 & 6 & & & \\[0.124cm]& & & & 2 & 4 & 0 & & & & \\[0.124cm]\end{tabular}}$

$q_{1} = \textcolor{#2192ff}{10}$ Let’s substitute into the formula for $IQR$ ,
$IQR = q_{3} - q_{1}$ $IQR = 28 - 10$ $IQR = 18$ Therefore, the final answer is,
$q_{2} = 15 \ \ \ \ \ \ IQR = 18$ (c) The median, lower quartile and upper quartile of the weekly snowfall for Linva are $12$ , $9$ , and $32$ cm, respectively. Use this information and your answers to part $(b)$ to compare the central tendency and the spread of the weekly snowfall in Dados and Linva.

The median of Dados, $15$ , is higher than that of Linva, $12$ . This means that on average, the snowfall in Dados is higher than in Linva.
Let’s calculate, $IQR$ for Linva,
$IQR = q_{3} - q_{1}$ $IQR = 32 - 9$ $IQR = 23$
The interquartile range of Dados, $18$ , is lower than that of Linva, $23$ . This means that the amount of snowfall in Linva varies more than that of Dados.

2. The weights, in kg, of $15$ rugby players in the Rebels club and $15$ soccer players in the Sharks Club are shown below

Rebels	75	78	79	80	82	82	83	84	85	86	89	93	95	99	102
Sharks	66	68	71	72	74	75	75	76	78	83	83	84	85	86	92

(9709/51/O/N/21 number 6)

(a) Represent the data by drawing a back-to-back stem-and-leaf diagram with Rebels on the left-hand side of the diagram.

$\colorbox{white!45}{\begin{tabular}{llllllll|c|lllllll}\multicolumn{8}{r|}{\textbf{Rebels}} & \textbf{} & \multicolumn{7}{l}{\textbf{Sharks}}\\[0.024cm]\hline& & & & & & & & 6 & 6 & 8 & & & & & \\[0.124cm]& & & & & 9 & 8 & 5 & 7 & 1 & 2 & 4 & 5 & 5 & 6 & 8 \\[0.124cm]9 & 6 & 5 & 4 & 3 & 2 & 2 & 0 & 8 & 3 & 3 & 4 & 5 & 6 & & \\[0.124cm]& & & & & 9 & 5 & 3 & 9 & 2 & & & & & & \\[0.124cm]& & & & & & 2 & 10 & & & & & & & \\[0.124cm]\end{tabular}}$

To determine the scale in the stem, look at the lowest and highest values in the dataset. In our case, that is $66$ and $102$ respectively. This means our scale should range from $60$ to $110$ to encompass all the data, hence the reason for the scale above. Then plot the data, for Rebels on the left hand side, from right to left i.e numbers increase from right to left. Plot data for Sharks on the right hand side, from left hand side to right hand side i.e numbers increase going to the right. Don’t forget to insert a key to help read your stem-and-leaf diagram.

(b) Find the median and interquartile range for the Rebels.

To find the median we use the formula,
$q_{2} = \frac{n + 1}{2}$ Substitute into the formula,
$q_{2} = \frac{15 + 1}{2}$ $q_{2} = 8$ Note: Remember that this formula gives us the position of the median.

Find the data point in position number $8$ for Rebels, remember to count from right to left,
$q_{2} = 84$ To find the interquartile range we use the formula,
$IQR = q_{3} - q_{1}$ Let’s find $q_{3}$ ,
$q_{3} = \frac{3}{4}(n + 1)$ $q_{3} = \frac{3}{4}(15 + 1)$ $q_{3} = 12$ Find the data point in position $12$ ,
$q_{3} = 93$ Let’s find $q_{1}$ ,
$q_{3} = \frac{1}{4}(n + 1)$ $q_{3} = \frac{1}{4}(15 + 1)$ $q_{3} = 4$ Find the data point in position $4$ ,
$q_{1} = 80$ Let’s substitute into the formula for $IQR$ ,
$IQR = q_{3} - q_{1}$ $IQR = 93 - 80$ $IQR = 13$ Therefore, the final answer is,
$q_{2} = 84 \ \ \ \ \ \ IQR = 13$