IB Applications and Interpretation SL · Descriptive statistics · Section 1 of 5

Data and frequency tables

3 worked examples · By Álvaro López

Key idea

Before drawing or calculating anything, you need to sort the data. A frequency table counts how many times each value, or each interval, appears, and from it you get the graphs, the mean, the median and the quartiles. In the IB exam you will be asked to construct the table, complete the missing cells and read counts and proportions from it. That is why it is worth mastering before moving on.

Population, sample and types of data

Studying every member of a group, the population, is often expensive or impossible. That is why we work with a sample, a part of the population, and the conclusions are extended to the whole.

The way the sample is chosen decides whether it is representative. In a random sample every individual has the same probability of being chosen, for example by drawing lots. If you only pick whoever is most convenient to reach, the sample is biased and its conclusions drift away from reality, even if it has plenty of data. A survey carried out outside a single supermarket, for example, does not represent a whole neighbourhood.

Everything measured on each individual is a variable. There are three types.

The type of variable decides how the data are organised. If the variable is qualitative, or discrete with few values, individual values are used. If it is continuous, or there are very many different values, the data are grouped into intervals.

Frequencies

Let $n$ be the total number of data values and $x_{1}, x_{2}, \dots, x_{k}$ the $k$ distinct values that appear.

There are three checks you can always make on a table.

Cumulative frequencies answer questions such as "how many have at most...?" or "what proportion has less than...?". For example, $F_{2}$ is the number of data values that are $x_{2}$ or less.

Grouped data in intervals

When there are many different values, a table of individual values would be almost as long as the list of data. The data are then distributed into classes: intervals that cover all the data without overlapping, so that each datum falls into exactly one class.

The notation $[a, b [$ means that $a$ is included in the class and $b$ is not. Thus, a datum equal to $b$ goes into the next class, and no datum is counted twice. The last class may be closed on both sides so that the largest datum is not left out.

Worked examples

Example 1 Basic

Pets per family

A survey asks 16 families how many pets they have. The answers are:

0, 1, 2, 1, 0, 3, 2, 1, 1, 2, 0, 4, 1, 2, 3, 1

a) Find the absolute frequency of the value $1$ .

b) Find the cumulative absolute frequency of the value $2$ .

c) Find the relative frequency of the value $3$ . Give the answer as a fraction.

Solution

There are $n = 16$ data values. We count how many times each value appears and complete the table:

$x_{i}$	$f_{i}$	$F_{i}$	$f_{i} / n$
$0$	$3$	$3$	$\frac{3}{16}$
$1$	$6$	$9$	$\frac{6}{16}$
$2$	$4$	$13$	$\frac{4}{16}$
$3$	$2$	$15$	$\frac{2}{16}$
$4$	$1$	$16$	$\frac{1}{16}$

Check: $3 + 6 + 4 + 2 + 1 = 16$ .

a) The absolute frequency of $1$ is $6$ .

b) The cumulative frequency of $2$ is $F_{3} = 13$ : these are the families with $0$ , $1$ or $2$ pets, that is, $3 + 6 + 4$ .

c) The relative frequency of $3$ is $\frac{f}{n} = \frac{2}{16} = \frac{1}{8}$

Example 2 Medium

Grouped arrival times

A record has been kept of the time, in minutes, that each of 20 students took to arrive at school:

3.4 · 4.1 · 5.8 · 6.2 · 6.9 · 7.5 · 8.3 · 9 · 9.6 · 10.2 · 11.7 · 12.4 · 13.1 · 14.8 · 15.5 · 16.3 · 17.9 · 19.2 · 21.6 · 26.8

You want to group the data into 5 classes of the same width, and the width must be a whole number.

a) Calculate the minimum width that allows all the data to be covered with 5 classes.

b) Starting at $3$ and with that width, write the classes. How many students are in the class $[8, 13 [$ ?

c) Find the class midpoint of the class $[13, 18 [$ .

Solution

a) The smallest datum is $3.4$ and the largest is $26.8$ , so the range is $26.8 - 3.4 = 23.4$ With 5 classes, $\frac{23.4}{5} = 4.68$ . Rounded up, the width is $5$ .

b) With width $5$ and starting at $3$ , the classes are $[3, 8 [$ , $[8, 13 [$ , $[13, 18 [$ , $[18, 23 [$ and $[23, 28 [$ . The last one reaches $28$ , which exceeds $26.8$ , so it covers all the data.

Class	$f_{i}$	$F_{i}$
$[3, 8 [$	$6$	$6$
$[8, 13 [$	$6$	$12$
$[13, 18 [$	$5$	$17$
$[18, 23 [$	$2$	$19$
$[23, 28 [$	$1$	$20$

In $[8, 13 [$ fall $8.3$ ; $9$ ; $9.6$ ; $10.2$ ; $11.7$ and $12.4$ . That makes $6$ students. The $13.1$ already falls into the next class.

c) The class midpoint is the midpoint: $\frac{13 + 18}{2} = \frac{31}{2} = 15.5$

Example 3 Hard

A table with gaps

60 people have been asked how many minutes a day they spend reading the press. The data are in a table of classes with width $10$ in which some values are missing:

Minutes a day	Absolute frequency
$[0, 10 [$
$[10, 20 [$	relative frequency $0.3$
$[20, 30 [$
$[30, 40 [$	$9$

In addition, the class $[20, 30 [$ has twice as many people as the class $[0, 10 [$ .

a) Find the absolute frequency of the classes $[0, 10 [$ and $[20, 30 [$ .

b) Find the proportion of people who read for less than $30$ minutes. Give the answer as an irreducible fraction.

c) Find the proportion of people who read for less than $20$ minutes. Give the answer as an irreducible fraction.

Solution

The frequency of $[10, 20 [$ is $0.3 \cdot 60 = 18$ . Let $a$ be the frequency of $[0, 10 [$ . Then the frequency of $[20, 30 [$ is $2 a$ , and since the total is $60$ : $a + 18 + 2 a + 9 = 60 \Rightarrow 3 a = 33 \Rightarrow a = 11$

a) The class $[0, 10 [$ has $11$ people, and the class $[20, 30 [$ has $2 \cdot 11 = 22$ . Check: $11 + 18 + 22 + 9 = 60$ .

b) Less than $30$ minutes covers the first three classes: $11 + 18 + 22 = 51$ people. The proportion is $\frac{51}{60} = \frac{17}{20}$

c) Less than $20$ minutes covers the first two classes: $11 + 18 = 29$ people. The proportion is $\frac{29}{60}$

Common mistakes

Summary

A frequency table organises the data with $f_{i}$ (absolute), $F_{i}$ (cumulative) and $\frac{f _{i}}{n}$ (relative). It always holds that $\sum f_{i} = n$ , $F_{k} = n$ and that the relative frequencies add up to $1$ .
With many values or a continuous variable, the data are grouped into classes $[a, b [$ of the same width, which do not overlap. The width is rounded up.
The class midpoint is $\frac{a + b}{2}$ and is used for calculations with grouped data.
Before accepting a table, check that the frequencies add up to $n$ .

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