IB Applications and Interpretation SL · Descriptive statistics · Section 3 of 5

Measures of central tendency

3 worked examples · By Álvaro López

Key idea

Measures of central tendency summarise a set of data with a single "typical" value: the mean (the fair share), the median (the value at the centre) and the mode (the most repeated value). In the IB exam you have to calculate them from raw data, from frequency tables and from grouped data, and justify which one describes a particular set of data best.

The mean

The mean is the fair share: if all the data were made equal, the total would not change. That is why it uses all the data, and a very distant value drags it towards itself.

Mean of grouped data

When the data are in intervals, the exact value of each datum is not known. An approximation is made: all the data in a class are taken to be equal to the class midpoint.

The result is an estimate, not the exact value, because the information about where each datum falls within its class is lost. In the exam, write "an estimate of the mean".

The median

For grouped data the exact median cannot be given, but the median class can: the first class whose cumulative frequency reaches the position $\frac{n}{2}$ . An estimate of the value is read from the cumulative frequency graph, looking for the point at height $\frac{n}{2}$ .

The mode is the only measure that also works for qualitative data, such as the most popular colour. Its result is the value, not its frequency: if $5$ appears $9$ times, the mode is $5$ .

Which one to use?

Measure	What it uses	Effect of a very distant datum	When it is best
Mean	All the data	It shifts a lot	Symmetric data, or when the total matters
Median	The order	Hardly changes	Skewed data or data with outliers
Mode	The frequencies	Does not change	Qualitative data, or the most common value

Graphic display calculator

Worked examples

Example 1 Basic

Points per game

A basketball player has scored these points in five games: $3$ , $4$ , $4$ , $9$ and $10$ .

a) Find the mean.

b) Find the median.

c) Find the mode.

d) Due to a recording error, in the last game the player scored $40$ points, not $10$ . Find the new mean and the new median.

Solution

a) The sum is $3 + 4 + 4 + 9 + 10 = 30$ , and there are $5$ data: $x ˉ = \frac{30}{5} = 6$

b) The data are already ordered. The position of the median is $\frac{5 + 1}{2} = 3$ , which is a whole number. The third datum is $4$ , so the median is $4$ .

c) The $4$ appears twice and no other value is repeated as often: the mode is $4$ .

d) Now the sum is $3 + 4 + 4 + 9 + 40 = 60$ : $x ˉ = \frac{60}{5} = 12$ The median is still the third datum, $4$ , because the order has not changed. The mean has doubled and the median has not moved.

Example 2 Medium

Coffees per worker

In an office of $20$ people, the number of coffees each one drinks in a day has been recorded:

Coffees $x_{i}$	0	1	2	3	4	5
People $f_{i}$	2	3	4	1	6	4

a) Find the mean.

b) Find the median.

c) Find the mode.

Solution

First we check that $\sum f_{i} = 2 + 3 + 4 + 1 + 6 + 4 = 20 = n$ .

We complete the table with $f_{i} x_{i}$ and the cumulative frequencies $F_{i}$ :

$x_{i}$	$f_{i}$	$f_{i} x_{i}$	$F_{i}$
$0$	$2$	$0$	$2$
$1$	$3$	$3$	$5$
$2$	$4$	$8$	$9$
$3$	$1$	$3$	$10$
$4$	$6$	$24$	$16$
$5$	$4$	$20$	$20$
Total	$20$	$58$

a) The mean is $x ˉ = \frac{58}{20} = \frac{29}{10}$

b) Since $n = 20$ is even, the median is the mean of the data in positions $10$ and $11$ , because $\frac{20 + 1}{2} = 10.5$ . The cumulative frequency $F_{3} = 10$ shows that position $10$ holds the value $3$ , and $F_{4} = 16$ shows that position $11$ holds the value $4$ . Therefore $median = \frac{3 + 4}{2} = \frac{7}{2}$

c) The value with the highest frequency is $4$ , with $6$ people. The mode is $4$ .

Example 3 Hard

Minutes walking and a missing frequency

$20$ neighbours have been asked how many minutes a day they spend walking. The table is incomplete: the frequency of the class $[10, 20 [$ is not known, so we call it $k$ .

Minutes a day	Neighbours $f$
$[0, 10 [$	2
$[10, 20 [$	$k$
$[20, 30 [$	6
$[30, 40 [$	5

It is known that the estimated mean of the data is $22$ minutes.

a) Find $k$ .

b) State the modal class.

c) State the class that contains the median.

Solution

a) We use the class midpoints $5$ , $15$ , $25$ and $35$ . The number of data is $n = 2 + k + 6 + 5 = 13 + k$ , and the approximate sum is $2 \cdot 5 + 15 k + 6 \cdot 25 + 5 \cdot 35 = 335 + 15 k$ . Then $\frac{335 + 15 k}{13 + k} = 22$ Cross-multiplying: $335 + 15 k = 286 + 22 k \Rightarrow 49 = 7 k \Rightarrow k = 7$ Check: $n = 20$ and the estimated mean is $\frac{440}{20} = 22$ .

b) With $k = 7$ , the highest frequency is $7$ , and it corresponds to $[10, 20 [$ . That is the modal class.

c) The cumulative frequencies are $2$ , $9$ , $15$ and $20$ . The position $\frac{n}{2} = 10$ is not reached until $[20, 30 [$ , where the cumulative frequency goes from $9$ to $15$ . That is the median class.

Common mistakes

Summary

The mean is the sum of all the data divided by $n$ . From a table, $x ˉ = \frac{\sum f _{i} x _{i}}{n}$ . From grouped data, use the class midpoints and the result is an estimate.
The median is in position $\frac{n + 1}{2}$ . If $n$ is even, it is the mean of the two central data. With grouped data, the median class is identified.
The mode is the most frequent value. With grouped data one speaks of the modal class: the interval that contains the most data.
Choose the measure according to the data: the median if there are very distant values, the mean if the total matters.
On Paper 2, the GDC does the calculations, but always write the set-up and check that the number of data is $n$ .

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Measures of central tendency

The mean

Mean of grouped data

The median

The mode and the modal class

Which one to use?

Graphic display calculator

Worked examples

Common mistakes

Summary

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