IB Analysis and Approaches HL · Descriptive statistics · Section 2 of 5

Graphical representations of data

3 worked examples · By Álvaro López

Key idea

A well-chosen graph shows at a glance how the data are spread out: where they cluster, whether they are symmetric and whether there are unusual values. Each type of data calls for its own graph, and in the IB exam you have to draw them, read them and get the median and quartiles from the cumulative frequency curve.

Choosing the graph according to the type of data

The general rule: the type of variable decides the representation.

Data	Usual graph
Qualitative or discrete with few values	Bar chart, pie chart
Continuous or grouped into intervals	Histogram
You want to read the median and quartiles	Cumulative frequencies, box-and-whisker diagram

Bar charts and pie charts

The pie chart is convenient for comparing parts with the whole, but it is no use if there are many categories or if what matters is the order of the values.

Histogram

If all the classes have the same width, the area is proportional to the height and it is enough to draw the frequency as the height. But if the widths are different, you have to correct, or the graph misleads: a very wide class would look enormous just because it covers more values.

In the figure, the classes have widths 10, 10, 20 and 20 years. The class from 20 to 40 years brings together 40 people, and its bar has height $40/20 = 2$ and area $2 \cdot 20 = 40$ .

Histogram of ages with intervals of different widths: heights 2, 3, 2 and 1 over the intervals 0 to 10, 10 to 20, 20 to 40 and 40 to 60

Cumulative frequencies

The cumulative frequency of a value is the number of data less than or equal to it. For grouped data it is calculated at the upper boundary of each class: it is the number of data that lie below that boundary.

It is the graph that answers questions such as "how many data are there below a given value?", and also the reverse: "below which value does a given amount of data lie?".

Median and quartiles on the cumulative curve

The quartiles divide the ordered data into four groups with the same number of values: $Q_{1}$ leaves a quarter below it, $Q_{2}$ (the median) half, and $Q_{3}$ three quarters.

Interpolation is equivalent to assuming that within each class the data are spread uniformly. That is why the result is an estimate: with grouped data the real values are not known.

The interquartile range is $IQR = Q_{3} - Q_{1}$ and measures how far the central half of the data extends.

Box-and-whisker diagram

The box contains the central 50% of the data and each whisker roughly the outer 25%. A box with an off-centre median or a whisker much longer than the other indicates skewness.

Outliers

In the figure there is an example with twelve values: 2, 4, 5, 6, 6, 7, 8, 9, 10, 11, 13 and 24. With $Q_{1} = 5.5$ , $Q_{2} = 7.5$ and $Q_{3} = 10.5$ we get $IQR = 5$ , so the upper fence is $10.5 + 1.5 \cdot 5 = 18$ . The 24 exceeds it and is drawn separately; the right whisker reaches 13. The lower fence, $5.5 - 7.5 = - 2$ , is not exceeded by any datum.

Box-and-whisker diagram: left whisker at 2, box from 5.5 to 10.5 with the median at 7.5, right whisker at 13 and an outlier at 24; the fence at 18 is marked with a dashed line

Worked examples

Example 1 Basic

Angles of a pie chart

At a school, 60 students are asked how they get to school. The answers are: 24 by bus, 9 by bicycle, 15 on foot and 12 by car.

a) Calculate the angle of the sector for each mode of transport, in the order given.

b) Find the percentage of students who walk to school.

Solution

a) The angle is $\frac{f}{60} \cdot 36 0^{\circ} = 6 f$ degrees:

Bus: $6 \cdot 24 = 14 4^{\circ}$ .
Bicycle: $6 \cdot 9 = 5 4^{\circ}$ .
On foot: $6 \cdot 15 = 9 0^{\circ}$ .
Car: $6 \cdot 12 = 7 2^{\circ}$ .

Check: $144 + 54 + 90 + 72 = 360$ .

b) $\frac{15}{60} = \frac{1}{4}$ , that is, $25 %$ (it is also seen in the graph: $9 0^{\circ}$ is a quarter of a turn).

Example 2 Medium

Histogram with classes of different widths

The waiting time, in minutes, of 80 patients at a health centre is recorded:

Waiting time $t$ (min)	$0 \leq t < 5$	$5 \leq t < 10$	$10 \leq t < 20$	$20 \leq t < 40$
Number of patients	12	18	30	20

a) Find the height of each rectangle of a histogram in which the area is proportional to the frequency (height = patients per minute), in the order of the table.

b) Find the proportion of patients who waited less than 20 minutes.

c) Find the proportion of patients who waited 10 minutes or more.

Solution

a) The widths are $5$ , $5$ , $10$ and $20$ minutes. Height $= \frac{frequency}{width}$ :

$\frac{12}{5} = \frac{12}{5}, \frac{18}{5}, \frac{30}{10} = 3, \frac{20}{20} = 1$

That is, $2.4$ ; $3.6$ ; $3$ and $1$ . If the frequency were drawn directly as the height, the class from 10 to 20 minutes (30 patients) would look much larger than the class from 5 to 10 (18 patients), although each minute of that class contains fewer patients.

b) Less than 20 minutes are the first three classes: $12 + 18 + 30 = 60$ patients, and $\frac{60}{80} = \frac{3}{4}$ .

c) Ten minutes or more are the last two classes: $30 + 20 = 50$ patients, and $\frac{50}{80} = \frac{5}{8}$ .

Example 3 Hard

Median, quartiles and outliers with grouped data

The loading time of a web page, in seconds, is measured for 100 connections:

Time $t$ (s)	$0 \leq t < 2$	$2 \leq t < 4$	$4 \leq t < 6$	$6 \leq t < 8$	$8 \leq t < 10$	$10 \leq t < 14$
Number of connections	10	20	40	20	6	4

a) Build the row of cumulative frequencies, in the order of the table.

b) Estimate the median.

c) Estimate $Q_{1}$ and $Q_{3}$ and calculate the interquartile range. Give the result as the tuple $(Q_{1}, Q_{3}, IQR)$ .

d) Find the upper fence for outliers and decide whether a connection of 12 seconds would be an outlier. Give the value of the fence.

Solution

a) The frequencies are added up: $10$ , $10 + 20 = 30$ , $30 + 40 = 70$ , $70 + 20 = 90$ , $90 + 6 = 96$ , $96 + 4 = 100$ .

b) Half of the data is $\frac{100}{2} = 50$ . The cumulative frequency goes from $30$ (at $t = 4$ ) to $70$ (at $t = 6$ ), so the median lies in the class $4 \leq t < 6$ . Interpolating:

$Q_{2} = 4 + 2 \cdot \frac{50 - 30}{70 - 30} = 4 + 2 \cdot \frac{20}{40} = 5$

c) For $Q_{1}$ we look for the height $\frac{100}{4} = 25$ , which falls between $10$ (at $t = 2$ ) and $30$ (at $t = 4$ ):

$Q_{1} = 2 + 2 \cdot \frac{25 - 10}{30 - 10} = 2 + 2 \cdot \frac{15}{20} = \frac{7}{2}$

For $Q_{3}$ we look for $\frac{3 \cdot 100}{4} = 75$ , which falls between $70$ (at $t = 6$ ) and $90$ (at $t = 8$ ):

$Q_{3} = 6 + 2 \cdot \frac{75 - 70}{90 - 70} = 6 + 2 \cdot \frac{5}{20} = \frac{13}{2}$

The interquartile range is $IQR = \frac{13}{2} - \frac{7}{2} = 3$ .

d) The upper fence is $Q_{3} + 1.5 \cdot IQR = \frac{13}{2} + 1.5 \cdot 3 = 6.5 + 4.5 = 11$ . A loading time of $12$ s exceeds $11$ , so it is an outlier. (The lower fence, $3.5 - 4.5 = - 1$ , cannot be reached with positive times.)

In the exam: when interpolating, write down the height you are looking for ( $\frac{n}{4}$ , $\frac{n}{2}$ , $\frac{3 n}{4}$ ) and the class where it falls; that is what earns the method marks. To say that a value is an outlier, it is not enough for it to "look large": you have to calculate the fence and compare.

Common mistakes

Summary

Bar charts and pie charts for qualitative or discrete data; histogram for grouped data. In the histogram there are no gaps and the area is the frequency.
With classes of different widths, height $= \frac{frequency}{width}$ .
Angle of a sector $= \frac{f}{n} \cdot 36 0^{\circ}$ .
The cumulative frequency is associated with the upper boundary of each class; the polygon or curve starts at $0$ and ends at $n$ .
To read $Q_{1}$ , $Q_{2}$ and $Q_{3}$ , look for the heights $\frac{n}{4}$ , $\frac{n}{2}$ and $\frac{3 n}{4}$ and go down to the horizontal axis; with a table, interpolate. $IQR = Q_{3} - Q_{1}$ .
Box-and-whisker diagram: minimum, $Q_{1}$ , median, $Q_{3}$ , maximum. Outliers are the data outside $Q_{1} - 1.5 IQR$ and $Q_{3} + 1.5 IQR$ .
In Paper 2 the GDC draws these graphs and gives the quartiles, but always write down how you obtained the result.

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