IB Analysis and Approaches HL · 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 one single "typical" number: the mean (the fair share), the median (the value in the middle) and the mode (the most frequent value). In the IB exam you have to calculate them by hand from raw data, from tables and from grouped data, and reason about which of the three best describes a particular set of data.

The mean

The mean $x ˉ$ is the sum of all the data divided by how many there are. If the value $x_{i}$ appears $f_{i}$ times in a table with $k$ different values, summing "all the data" means adding each value as many times as it is repeated.

When working with a table, add a column $f_{i} x_{i}$ : the sum of that column is the numerator and the sum of the frequency column is $n$ .

The mean uses all the data, so one value far away pulls it towards itself. It is the natural measure when the data are symmetric or when the total (the sum) matters.

Mean of grouped data

When the data come in intervals, the exact value of each one is not known. An approximation is made: all the data in an interval are assumed to be worth the same as its midpoint, the class midpoint.

The result is an estimate, not the exact value, because the information about where each datum falls within its interval has been lost. Say so in the exam: "an estimate of the mean".

The median

The median is the value that leaves half of the data below it and the other half above, once they are ordered.

For grouped data the exact value of the median cannot be given with this recipe, but the median class can: the interval in which the cumulative frequency passes the position $\frac{n + 1}{2}$ . A finer estimate is read from the cumulative frequency graph, as you saw in the previous section.

The median depends on the order, not on the size of the extreme data. Changing the value of the largest datum, as long as it remains the largest, does not move it.

The mode is the only measure that also works for qualitative data (the most common colour, the best-selling brand). Be careful: the mode is the value of the datum, not its frequency. If the value $5$ appears $9$ times, the mode is $5$ , not $9$ .

Which to choose? Outliers

An outlier is a datum that is very different from the rest of the set. Each measure reacts differently.

Measure	Uses	Effect of an outlier	When it is best
Mean	All the data	Shifts a lot towards the outlier	Symmetric data, without unusual extremes; when the total matters
Median	Only the order	Almost unchanged	Skewed data or data with outliers (salaries, house prices)
Mode	Only the frequencies	Unchanged	Qualitative data, or when the most common value matters

A quick hint: if the mean is much greater than the median, there are probably very high data pulling it upwards.

Worked examples

Example 1 Basic

Three measures of waiting times

The waiting times, in minutes, of eight customers at a counter are

$12, 7, 9, 15, 9, 11, 9, 14$

a) Find the mean.

b) Find the median.

c) Find the mode.

d) Because of a recording error, the $15$ was actually $150$ . Find the new mean and the new median.

Solution

a) The sum is $12 + 7 + 9 + 15 + 9 + 11 + 9 + 14 = 86$ , and there are $8$ data:

$x ˉ = \frac{86}{8} = \frac{43}{4} = 10.75$

b) Ordering them: $7, 9, 9, 9, 11, 12, 14, 15$ . As $n = 8$ is even, the median is the mean of the data in positions $4$ and $5$ (the position is $\frac{8 + 1}{2} = 4.5$ ):

$\frac{9 + 11}{2} = 10$

c) The $9$ appears three times and no other value is repeated as often: the mode is $9$ .

d) The sum becomes $86 - 15 + 150 = 221$ :

$x ˉ = \frac{221}{8} = 27.625$

When the data are ordered, $150$ is still the largest datum and the rest do not change, so positions $4$ and $5$ have the same values and the median is still $10$ . The mean has almost tripled, and the median has not moved.

Example 2 Medium

Trip durations with grouped data

Forty trips on an urban bus have been timed. The duration $t$ , in minutes, is recorded in the table.

Duration $t$ (min)	Frequency
$0 \leq t < 10$	$5$
$10 \leq t < 20$	$13$
$20 \leq t < 30$	$12$
$30 \leq t < 40$	$7$
$40 \leq t < 50$	$3$

a) State the modal class.

b) Find an estimate of the mean duration.

c) Find the class that contains the median.

Solution

a) The highest frequency is $13$ , which corresponds to $10 \leq t < 20$ .

b) The class midpoint $x_{i}$ and the column $f_{i} x_{i}$ are added:

Interval	$f_{i}$	$x_{i}$	$f_{i} x_{i}$
$0 \leq t < 10$	$5$	$5$	$25$
$10 \leq t < 20$	$13$	$15$	$195$
$20 \leq t < 30$	$12$	$25$	$300$
$30 \leq t < 40$	$7$	$35$	$245$
$40 \leq t < 50$	$3$	$45$	$135$
Total	$40$		$900$

$x ˉ \approx \frac{900}{40} = 22.5 minutes$

c) The median is in position $\frac{40 + 1}{2} = 20.5$ , between the 20th and 21st data. The cumulative frequencies are $5, 18, 30, 37, 40$ . Up to the second class there are $18$ data, fewer than $20$ , and on reaching the third there are $30$ . The 20th and 21st data are therefore in $20 \leq t < 30$ .

Example 3 Hard

Missing frequency and an extreme value

In a group, the number of books each student read during the term has been recorded. It is known that the mean is $1.9$ .

Books $x$	$0$	$1$	$2$	$3$	$4$
Frequency	$4$	$6$	$k$	$5$	$3$

a) Find $k$ .

b) Find the median and the mode.

c) A new student who has read $20$ books joins the group. Find the mean and the median of the new group.

Solution

a) The sum of the data is $\sum f_{i} x_{i} = 0 \cdot 4 + 1 \cdot 6 + 2 k + 3 \cdot 5 + 4 \cdot 3 = 33 + 2 k$ , and the number of students is $n = 4 + 6 + k + 5 + 3 = 18 + k$ . Setting the mean equal to $1.9$ :

$\frac{33 + 2 k}{18 + k} = 1.9 ⟹ 33 + 2 k = 34.2 + 1.9 k ⟹ 0.1 k = 1.2 ⟹ k = 12$

b) With $k = 12$ there are $n = 30$ students and the median is in position $\frac{30 + 1}{2} = 15.5$ , between the 15th and 16th data. The cumulative frequencies are $4, 10, 22, 27, 30$ : both data fall on the value $2$ . Median $= 2$ .

The highest frequency is $12$ , which corresponds to the value $2$ : the mode is $2$ .

c) The sum of the data was $33 + 2 \cdot 12 = 57$ (this matches $1.9 \cdot 30$ ). Now there are $31$ data and the sum is $57 + 20 = 77$ :

$x ˉ = \frac{77}{31} \approx 2.48$

The median is in position $\frac{31 + 1}{2} = 16$ . The new datum, $20$ , goes at the end of the ordered list, so the first $16$ data are the old ones and the 16th still has the value $2$ . Median $= 2$ .

The mean has gone up from $1.9$ to almost $2.5$ because of a single student, while the median and the mode have stayed at $2$ .

In the exam: when the mean is given and a value is missing, set up the equation with "sum $= n x ˉ$ " and write all the steps. If you are asked to compare measures, reason in one sentence: the mean changes because it uses all the data, and the median does not because it depends only on the central position.

Common mistakes

Summary

Mean: $x ˉ = \frac{\sum f _{i} x _{i}}{n}$ . Total sum $= n x ˉ$ : use it for data that are missing or added.
Grouped data: use the class midpoint of each interval; the mean is then an estimate.
Median: the datum in position $\frac{n + 1}{2}$ of the ordered list (mean of the two central data if $n$ is even). With grouped data, identify the median class using the cumulative frequency.
Mode: the value with the highest frequency (there may be several); with grouped data, the modal class.
An outlier moves the mean a lot, barely affects the median and does not change the mode: with skewed data or outliers, the median describes what is typical better.
In Paper 2 the GDC gives the mean and the median; with grouped data, the mean of the class midpoints is an estimate. Always write down the set-up.

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Measures of central tendency

The mean

Mean of grouped data

The median

The mode and the modal class

Which to choose? Outliers

Worked examples

Common mistakes

Summary

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