IB Applications and Interpretation SL · Descriptive statistics · Section 2 of 5

Graphical representations

3 worked examples · By Álvaro López

Key idea

A graph shows at a glance the shape of the data: where they cluster, whether they are symmetric and whether they have values far away from the rest. In the IB exam you have to choose the right graph for the type of variable, draw it with properly labelled axes and read specific values from it. The most common mistakes occur with the histogram and the cumulative frequency graph, so pay close attention to those two sections.

Which graph to use?

Before drawing, check what type of variable you have.

Type of data	Usual graph
Qualitative	Bar chart or pie chart
Discrete quantitative, with few values	Bar chart
Continuous, or grouped into intervals	Histogram and frequency polygon
Questions such as "how many at most"	Cumulative frequency graph

Bar chart

If you join the tops of the bars with line segments, you obtain the frequency polygon of the diagram.

Pie chart

A pie chart divides a circle into parts. It is used to compare the share that each value represents of the total, and it works best with a few categories.

Histogram

When all the classes have the same width, the height of each bar is simply its frequency. If the widths differ, the height must be corrected, and for that we use the frequency density.

Histogram of absolute frequencies of 30 waiting times in minutes, with class width 10: 4, 9, 12 and 5 data in the intervals 0 to 10, 10 to 20, 20 to 30 and 30 to 40

Frequency polygon

The frequency polygon is obtained by joining with line segments the points placed above the class midpoint of each bar, at the height of its frequency. So that the line reaches the axis, a point at height $0$ is added in the classes adjacent to the two ends.

Cumulative frequency graph

Cumulative frequency graph of 30 waiting times in minutes, with the points (0, 0), (10, 4), (20, 13), (30, 25) and (40, 30) joined by a line

Worked examples

Example 1 Basic

Calls received by a switchboard

A switchboard records, over 20 hours, the number of calls it receives in each hour. The results are in this table:

Calls $x$	0	1	2	3	4
Hours $f$	4	7	5	3	1

a) In a bar chart, what is the height of the bar for $x = 2$ ?

b) How many hours had at most $2$ calls?

c) In a pie chart, how many degrees does the sector for $x = 3$ measure?

Solution

There are $n = 20$ hours, which is the sum of the frequencies: $4 + 7 + 5 + 3 + 1 = 20$ .

a) The height of the bar is its absolute frequency, $f = 5$ .

b) "At most $2$ " means $x \leq 2$ , and that is the cumulative frequency $F$ of $2$ : $F = 4 + 7 + 5 = 16$

c) By the sector formula, $α = \frac{3}{20} \cdot 36 0^{\circ} = 5 4^{\circ}$

Example 2 Medium

Waiting time at the clinic with the cumulative graph

The waiting time, in minutes, of 40 patients at a health centre has been recorded. The grouped data are:

Waiting $t$ (min)	Patients $f$
$[0, 10 [$	4
$[10, 20 [$	10
$[20, 30 [$	14
$[30, 40 [$	8
$[40, 50 [$	4

A cumulative frequency graph is drawn.

a) How many patients waited at most $30$ minutes?

b) Estimate how many patients waited at most $25$ minutes. Use the line joining the two points of the graph that surround $25$ .

c) In which class is the patient who occupies place $30$ if the patients are ordered by waiting time?

Solution

First the cumulative frequencies are calculated: $4$ , $4 + 10 = 14$ , $14 + 14 = 28$ , $28 + 8 = 36$ and $36 + 4 = 40$ . The points of the graph are $(10, 4)$ , $(20, 14)$ , $(30, 28)$ , $(40, 36)$ and $(50, 40)$ , in addition to the origin.

a) These are the data with waiting time less than or equal to $30$ minutes. We read the point $(30, 28)$ directly: there are $28$ patients.

b) Between $(20, 14)$ and $(30, 28)$ the line rises $28 - 14 = 14$ patients in $10$ minutes. Up to $25$ it has risen half of that segment, so $14 + \frac{14}{2} = 21$ We estimate $21$ patients.

c) Up to the class $[20, 30 [$ there are $28$ patients. Places $29$ and $30$ do not fall there, so place $30$ is in the class $[30, 40 [$ .

Example 3 Hard

Histogram with unequal class widths

The distance covered by each of 52 couriers in one day has been recorded, in kilometres. The class widths are not equal:

Kilometres $[a, b [$	Couriers $f$
$[0, 4 [$	12
$[4, 6 [$	8
$[6, 10 [$	14
$[10, 16 [$	18

A histogram is drawn with the height of each bar equal to the frequency density.

a) What is the height of the bar for $[6, 10 [$ ?

b) Which class has the tallest bar in the histogram?

c) What proportion of couriers cover between $4$ and $10$ kilometres? Give the answer as an irreducible fraction.

Solution

We calculate the density of each class by dividing its frequency by its width:

Class	Width	$f$	Height $f / width$
$[0, 4 [$	$4$	$12$	$3$
$[4, 6 [$	$2$	$8$	$4$
$[6, 10 [$	$4$	$14$	$\frac{7}{2}$
$[10, 16 [$	$6$	$18$	$3$

a) The height of $[6, 10 [$ is $\frac{14}{4} = \frac{7}{2}$ .

b) The tallest bar is that of $[4, 6 [$ , with height $4$ . The class with the most couriers is $[10, 16 [$ (with $18$ ), but it is the widest, and that is why its bar is not the tallest. This is the trap in the exercise.

c) Between $4$ and $10$ kilometres lie the classes $[4, 6 [$ and $[6, 10 [$ , with $8 + 14 = 22$ couriers. As $n = 12 + 8 + 14 + 18 = 52$ : $\frac{22}{52} = \frac{11}{26}$

Common mistakes

Summary

A bar chart is for a discrete or qualitative variable with few values (separated bars). For proportions, use a pie chart, with angle $\frac{f _{i}}{n} \cdot 36 0^{\circ}$ .
A continuous or grouped variable calls for a histogram with joined bars. The height is $f / width$ ; with classes of the same width, the frequency alone is enough.
The frequency polygon joins the points at the class midpoints, at the height of their frequencies.
The cumulative frequency graph places each point at the upper limit of its class. It lets you read how many data lie up to a value, and estimate the median and the quartiles.
Always label the axes, use a uniform scale and check that the frequencies add up to $n$ .

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