IB Analysis and Approaches HL · Descriptive statistics · Section 4 of 5

Measures of dispersion

3 worked examples · By Álvaro López

Key idea

The mean or the median tell you where the centre of the data is, but not how far the data lie from it. The measures of dispersion (range, interquartile range, variance and standard deviation) measure that spread and allow you to compare two sets with the same centre. In the IB exam you calculate them with the GDC in Paper 2 and by hand, using a table, in Paper 1.

Range and quartiles

Two classes can have the same average mark and be very different: in one everyone scores around 6, and in the other there are zeros and tens. What distinguishes them is the dispersion.

It is easy to calculate, but it depends on only two data and a single extreme value sets it off. That is why the quartiles, which ignore the extremes, are also used.

A way to locate the quartiles by hand with $n$ ordered data is to look at the positions $\frac{n + 1}{4}$ , $\frac{n + 1}{2}$ and $\frac{3 ( n + 1 )}{4}$ ; if the position is not an integer, the two neighbouring data are averaged. With data in a table, look for that position in the cumulative frequency column.

The five numbers $x_{m i n}$ , $Q_{1}$ , $Q_{2}$ , $Q_{3}$ and $x_{m a x}$ are the ones that draw the box-and-whisker diagram you saw in the section on graphs: the box measures exactly the IQR.

Outliers

A datum very far from the rest distorts the mean and the range. The IB sets a rule for deciding when a datum is considered an outlier.

On the box-and-whisker diagram, outliers are marked separately with a cross, and the whisker reaches only to the most extreme datum that is not an outlier.

Box-and-whisker diagram with an outlier to the right of the upper fence

Variance and standard deviation

The range and the IQR look at only some of the data. The standard deviation uses all of them, measuring how far each datum lies from the mean.

For each datum, $x_{i} - x ˉ$ is its deviation from the mean. If the deviations were simply added, positive and negative ones would cancel out and the result would be $0$ . To avoid this, they are squared and then averaged.

The second expression (the mean of the squares minus the square of the mean) is the most convenient by hand: you only need $\sum f_{i} x_{i}$ and $\sum f_{i} x_{i}^{2}$ . The standard deviation is measured in the same units as the data; the variance is not (it is in squared units).

Interpreting the standard deviation

A large standard deviation means data very spread out around the mean; a small one means data closely grouped. A single number says little: its value is interpreted by comparing it with another set measured in the same unit, or with the mean itself.

When two sets have similar means, the one with the smaller standard deviation is the more regular or homogeneous. When comparing, always write a complete sentence that mentions the context, for example "the pieces made by machine A are more uniform because their standard deviation is smaller". A number without a sentence loses the interpretation mark.

Some properties worth remembering: $σ \geq 0$ , and it is $0$ only if all the data are equal; and, unlike the range, it uses all the data, so an outlier increases it greatly. That is why, if there are outliers, the IQR is a more reliable measure of spread than $σ$ .

Worked examples

Example 1 Basic

Range, quartiles and outliers

A town council records the waiting time, in minutes, of eleven patients at a health centre:

Patient	1	2	3	4	5	6	7	8	9	10	11
Waiting time (min)	4	6	7	9	10	12	13	15	16	18	40

The data are already ordered.

a) Find the range.

b) Find $Q_{1}$ , $Q_{3}$ and the interquartile range.

c) Determine whether there is any outlier.

Solution

a) Range $= 40 - 4 = 36$ minutes.

b) There are $n = 11$ data. The positions are $\frac{n + 1}{4} = 3$ and $\frac{3 ( n + 1 )}{4} = 9$ , so $Q_{1}$ is the third datum and $Q_{3}$ the ninth: $Q_{1} = 7$ and $Q_{3} = 16$ . Then

$IQR = 16 - 7 = 9$

c) We calculate $1.5 \cdot IQR = 13.5$ . The limits are

$Q_{1} - 13.5 = 7 - 13.5 = - 6.5 Q_{3} + 13.5 = 16 + 13.5 = 29.5$

No datum lies below $- 6.5$ . Above $29.5$ lies the $40$ . The only outlier is $40$ , and it is the one in the figure above.

Example 2 Medium

Standard deviation from a frequency table

20 people are asked how many books they have read this month. The results are:

Books read, $x_{i}$	0	1	2	3	4
People, $f_{i}$	3	5	7	4	1

a) Find the mean.

b) Find the standard deviation of these 20 values, in exact form.

c) Now consider the 20 people to be a sample of the readers in a neighbourhood. Find $s_{n - 1}$ in exact form.

Solution

The table is extended with the columns $f_{i} x_{i}$ and $f_{i} x_{i}^{2}$ :

$x_{i}$	$f_{i}$	$f_{i} x_{i}$	$f_{i} x_{i}^{2}$
0	3	0	0
1	5	5	5
2	7	14	28
3	4	12	36
4	1	4	16
Sum	20	35	85

a) $x ˉ = \frac{35}{20} = \frac{7}{4} = 1.75$ books.

b) The variance is

$σ^{2} = \frac{85}{20} - (\frac{7}{4})^{2} = \frac{17}{4} - \frac{49}{16} = \frac{68 - 49}{16} = \frac{19}{16}$

and the standard deviation is $σ = \frac{19}{4} \approx 1.09$ books.

c) With $n = 20$ :

$s_{n - 1}^{2} = \frac{20}{19} \cdot \frac{19}{16} = \frac{5}{4} s_{n - 1} = \frac{5}{2} \approx 1.12 books$

As expected, $s_{n - 1}$ is slightly larger than $σ$ .

Example 3 Hard

Comparing two machines

A cereal factory has two machines that fill 50 g bags. Eight bags from each machine are weighed (in grams):

Machine	Weights
A	48, 50, 49, 51, 50, 52, 49, 51
B	45, 55, 50, 47, 53, 50, 46, 54

a) Find the mean of each machine.

b) Find the standard deviation $σ$ of each machine, in exact form.

c) State which machine is more regular. Answer with the letter "A" or "B".

d) For machine A, $Q_{1} = 49$ and $Q_{3} = 51$ . Find the limits beyond which a weight is an outlier, and decide whether a weight of $56$ g would be one. Answer as the tuple (lower limit, upper limit, is it an outlier?).

e) If the eight bags from B are treated as a sample of everything that machine produces, find $s_{n - 1}$ in exact form.

Solution

a) Machine A: $\frac{48 + 50 + 49 + 51 + 50 + 52 + 49 + 51}{8} = \frac{400}{8} = 50$ . Machine B: $\frac{45 + 55 + 50 + 47 + 53 + 50 + 46 + 54}{8} = \frac{400}{8} = 50$ . Both means are $50$ g.

b) As the means equal $50$ , the deviations are $x_{i} - 50$ .

A: $- 2, 0, - 1, 1, 0, 2, - 1, 1$ . Their squares add up to $4 + 0 + 1 + 1 + 0 + 4 + 1 + 1 = 12$ , so $σ_{A}^{2} = \frac{12}{8} = \frac{3}{2}$ and $σ_{A} = \frac{3}{2} = \frac{6}{2} \approx 1.22$ g.
B: $- 5, 5, 0, - 3, 3, 0, - 4, 4$ . Their squares add up to $25 + 25 + 0 + 9 + 9 + 0 + 16 + 16 = 100$ , so $σ_{B}^{2} = \frac{100}{8} = \frac{25}{2}$ and $σ_{B} = \frac{5 2}{2} \approx 3.54$ g.

c) The means coincide, so the spread decides: $σ_{A} \approx 1.22 < σ_{B} \approx 3.54$ . Machine A is more regular, because its bags are closer to 50 g.

d) $IQR = 51 - 49 = 2$ and $1.5 \cdot IQR = 3$ . The limits are $49 - 3 = 46$ and $51 + 3 = 54$ . As $56 > 54$ , a weight of $56$ g would be an outlier for machine A.

e) With $n = 8$ : $s_{n - 1}^{2} = \frac{100}{7}$ , and therefore $s_{n - 1} = \frac{10}{7} = \frac{10 7}{7} \approx 3.78$ g.

In the exam: when comparing, give the conclusion in a sentence in context ("machine A is more regular because its standard deviation is smaller"); a loose number does not earn the interpretation mark. For "outlier", show both limits and compare against them. And state whether you use $σ$ or $s_{n - 1}$ .

Common mistakes

Summary

Range $= x_{m a x} - x_{m i n}$ . Quartiles: $Q_{1}$ , $Q_{2}$ (median) and $Q_{3}$ ; $IQR = Q_{3} - Q_{1}$ is the width of the box. In the exam, the quartile given by the GDC is valid.
Outlier: a datum below $Q_{1} - 1.5 \cdot IQR$ or above $Q_{3} + 1.5 \cdot IQR$ .
Population variance: $σ^{2} = \frac{\sum f _{i} x _{i}^{2}}{n} - x ˉ^{2}$ ; standard deviation $σ = σ^{2}$ , in the same units as the data.
Sample: $s_{n - 1}^{2} = \frac{n}{n - 1} σ^{2}$ and $s_{n - 1} > σ$ . Use $σ$ if the data are the whole population, $s_{n - 1}$ if they are a sample used to estimate the population.
A smaller standard deviation means more tightly grouped and more regular data; always interpret it by comparing, with a sentence in context.
An outlier makes the range and the standard deviation grow a lot, but barely affects the IQR.
With the GDC (Paper 2) you obtain the mean, $σ$ , $s_{n - 1}$ , $Q_{1}$ , the median and $Q_{3}$ ; with grouped data, use the class midpoints.

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