Quartiles, spread and the cumulative frequency diagram – Statistics with Ungrouped Data (Senior 1 Mathematics) | Ganzaa

Statistics with Ungrouped Data

Lessons

Senior 1 · Mathematics

Statistics with Ungrouped Data

5 lessons

Lesson 5 of 5

Quartiles, spread and the cumulative frequency diagram

25 min

Same average, different story

Two S1 classes both have a median mark of 60 in mathematics. In the first class almost everybody scored between 55 and 65; in the second, marks ran from 20 to 95. The averages are the same, but a teacher would plan very differently for the two classes. To describe a data set fully you need its centre and its spread: how far the values are scattered.

Definition

Range: the difference between the largest and the smallest value.

\text{range} = \text{largest value} - \text{smallest value}

Midrange: the value halfway between the largest and the smallest value.

\text{midrange} = \frac{\text{largest value} + \text{smallest value}}{2}

Definition

Quartiles: the three values that divide the ordered data into four parts with equal numbers of values.

$Q_{1}$ , the lower quartile: a quarter of the values lie below it.
$Q_{2}$ , the median: half of the values lie below it.
$Q_{3}$ , the upper quartile: three quarters of the values lie below it.

For $n$ values in order, their positions are

Q_1: \ \frac{n+1}{4}\text{th value}, \qquad Q_2: \ \frac{n+1}{2}\text{th value}, \qquad Q_3: \ \frac{3(n+1)}{4}\text{th value}.

Definition

Inter-quartile range (IQR): the spread of the middle half of the data,

\text{IQR} = Q_3 - Q_1.

Unlike the range, it is not affected by one very large or very small value, because it ignores the lowest and highest quarters.

Worked example 1: eleven temperatures

The maximum temperatures in Kigali on eleven days in September were (in $^\circ\mathrm{C}$ ): 26, 28, 25, 27, 29, 26, 24, 27, 28, 26, 30. Find the quartiles, the inter-quartile range, the range and the midrange.

Solution First put the data in order:

The three quartiles cut the ordered data into four groups of equal size.

Here $n = 11$ , so $n + 1 = 12$ .

$Q_{1}$ is the $\frac{12}{4} = 3$ rd value: $Q_1 = 26^\circ\mathrm{C}$ .
$Q_{2}$ is the $\frac{12}{2} = 6$ th value: median $= 27^\circ\mathrm{C}$ .
$Q_{3}$ is the $\frac{3 \times 12}{4} = 9$ th value: $Q_3 = 28^\circ\mathrm{C}$ .
$\text{IQR} = 28 - 26 = 2^\circ\mathrm{C}$ ; range $= 30 - 24 = 6^\circ\mathrm{C}$ ; midrange $= \frac{30 + 24}{2} = 27^\circ\mathrm{C}$ .

Worked example 2: when the position is a half

Find the quartiles and the IQR of the nine quiz marks from Lesson 4: 9, 11, 12, 14, 15, 15, 15, 17, 18.

Solution $n = 9$ , so $n + 1 = 10$ .

$Q_{1}$ : position $\frac{10}{4} = 2.5$ , halfway between the 2nd and 3rd values: $\frac{11 + 12}{2} = 11.5$ .
$Q_{2}$ : position $\frac{10}{2} = 5$ : the median is 15.
$Q_{3}$ : position $\frac{30}{4} = 7.5$ , halfway between the 7th and 8th values: $\frac{15 + 17}{2} = 16$ .
$\text{IQR} = 16 - 11.5 = 4.5$ .

With an odd number of values, as in this lesson, each position is either a whole number or a half.

Worked example 3: quartiles from a frequency table

Thirty-one residents of a village recorded how many of last year's 12 monthly Umuganda sessions they attended. Find the median, the quartiles and the IQR.

Sessions attended $(x)$	6	7	8	9	10	11	12
Residents $(f)$	1	2	3	5	7	8	5
Cumulative frequency	1	3	6	11	18	26	31

Solution $n = 31$ , so $n + 1 = 32$ . Find each position, then read down the cumulative frequencies for the first value that reaches it.

$Q_{1}$ : position $\frac{32}{4} = 8$ . The cumulative frequency of 8 sessions is 6 (too small) and of 9 sessions is 11, so the 7th to 11th values are all 9: $Q_{1} = 9$ .
Median: position $\frac{32}{2} = 16$ . The cumulative frequencies are 11 at 9 and 18 at 10, so the 16th value is 10: median $= 10$ .
$Q_{3}$ : position $\frac{3 \times 32}{4} = 24$ . The cumulative frequencies are 18 at 10 and 26 at 11, so the 24th value is 11: $Q_{3} = 11$ .
$\text{IQR} = 11 - 9 = 2$ sessions; range $= 12 - 6 = 6$ sessions.

The middle half of the residents attended between 9 and 11 sessions.

Definition

Cumulative frequency diagram (ogive): a graph of cumulative frequency against value. Each point $(x, \text{cumulative frequency of } x)$ is plotted and the points are joined in order. The graph always rises from left to right and ends at $n$ .

Drawing and reading the ogive

Plot each value against its cumulative frequency from Worked example 3, start at 0 one step before the smallest value, and join the points with a ruler:

The graph starts at 0 one step below the smallest value, and ends at 31, the total number of residents.

To answer "how many residents attended 9 sessions or fewer?", go up from 9 to the graph and across: 11 residents. To find how many attended more than 10, read 18 at 10 and subtract from the total: $31 - 18 = 13$ . For data that are whole numbers, read the graph only at the plotted values; the quartiles come from the table, as in Worked example 3.

Worked example 4: from an ogive back to a frequency table

Books read this term by 25 S1 learners. Each small square is one learner high.

Make the frequency table for the number of books read.

Solution Read the cumulative frequency at each value, then subtract each one from the next:

Books $(x)$	Cumulative frequency (read from the graph)	Frequency
0	2	2
1	6	$6 - 2 = 4$
2	13	$13 - 6 = 7$
3	19	$19 - 13 = 6$
4	23	$23 - 19 = 4$
5	25	$25 - 23 = 2$

Check: $2 + 4 + 7 + 6 + 4 + 2 = 25$ , the last cumulative frequency. ✓

Warning

Two points about quartiles

Order the data first, exactly as for the median.
Books use slightly different rules. Some books find $Q_{1}$ as the median of the lower half of the data and $Q_{3}$ as the median of the upper half. For an odd number of values, as in every example here, that gives the same answers as the positions $\frac{n+1}{4}$ and $\frac{3(n+1)}{4}$ . For an even number of values the two rules can differ a little, so use the rule your teacher uses.

Tip

Compare data sets with two numbers: an average for the centre and a spread. The two classes at the start of this lesson have the same median, 60, but the second has a much larger IQR, so its marks are far less consistent.

Project: a statistical exploration of our school

Work in a group of four and use everything in this unit.

Ask a question that data can answer, such as "How long do S1 learners take to travel to school?" or "How many books do learners read in a month?"
Plan the data: what type it is, and how you will collect it (questionnaire, observation or measurement). Write fair questions.
Collect data from a random sample of 31 learners, with a tally chart.
Organise the data in a frequency table with cumulative frequencies.
Present the data in two suitable graphs, one of them a cumulative frequency diagram.
Summarise with the mean, median, mode, quartiles, IQR and range.
Report to the class: answer your question in three sentences, and say one way your sample might mislead.

Khan Academy: How to calculate interquartile range IQR (6 min). Sal finds the quartiles as the medians of the lower and upper halves of the data; compare his answer with the position rule in this lesson.

✦

Practice

Use an odd number of values, as in the lesson. Question 7 is harder.

Find the range and the midrange of: 14, 9, 21, 17, 12.
For the data 3, 5, 6, 8, 9, 11, 12, 15, 18, 20, 22, find $Q_{1}$ , the median, $Q_{3}$ and the IQR.
For the data 40, 45, 47, 52, 55, 58, 61, 64, 70, find $Q_{1}$ , $Q_{3}$ and the IQR.
Using the ogive of Umuganda sessions, how many residents attended (a) 8 sessions or fewer, (b) more than 11 sessions?
The cumulative frequencies of the marks 1 to 5 in a short test are 3, 8, 15, 21 and 23. Find the frequency of each mark, the median and the quartiles.
Explain why the IQR is often a better measure of spread than the range.
Class A's marks have median 62 and IQR 8; class B's have median 62 and IQR 21. Keza says the classes did equally well. Is she right? Explain.

Show answersHide answers

Range $21 - 9 = 12$ ; midrange $\frac{21 + 9}{2} = 15$ .
$n = 11$ : $Q_{1} =$ 3rd value $= 6$ , median $=$ 6th $= 11$ , $Q_{3} =$ 9th $= 18$ ; IQR $= 12$ .
$n = 9$ : $Q_{1}$ at 2.5, $\frac{45 + 47}{2} = 46$ ; $Q_{3}$ at 7.5, $\frac{61 + 64}{2} = 62.5$ ; IQR $= 16.5$ .
(a) 6 (b) $31 - 26 = 5$ .
Frequencies 3, 5, 7, 6, 2. With $n = 23$ : $Q_{1}$ is the 6th value $= 2$ , the median is the 12th $= 3$ , $Q_{3}$ is the 18th $= 4$ .
The range uses only the two most extreme values, so one unusual value changes it a lot. The IQR uses the middle half of the data and ignores the extremes.
Only partly. Their typical (median) marks are the same, but class B's marks are much more spread out: some learners in B did much better, and some much worse, than in A. Class A's results are more consistent.

1Same average, different story

2Drawing and reading the ogive

3Project: a statistical exploration of our school

Same average, different story

Drawing and reading the ogive

Project: a statistical exploration of our school