Four displays come up again and again on Paper 5. Each one answers a different question about a data set, and each has a mark-losing trap: frequency density, class boundaries, the key, and reading a percentile.
In a histogram the AREA of a bar is the frequency, so the height is the frequency density: \(\text{frequency density} = \dfrac{\text{frequency}}{\text{class width}}\). On a cumulative frequency graph, read across from a position (such as \(\tfrac{n}{2}\) for the median) and down to the value.
Fig. 1: a histogram with unequal class widths. Each bar's area (width × frequency density) is its frequency: 8, 12, 18, 16 and 6, total 60
Section 01
Histograms with unequal widths.
Open the histogram tab of the lab and widen one class. The frequency stays the same, so the bar must get shorter: that is frequency density at work. Then read Fig. 1, where the widest class (30 to 50 s) holds only 6 values even though its bar is the longest.
Continuous data only. Histograms have no gaps between bars; the horizontal axis is a number line.
Class boundaries. For data recorded to the nearest whole number, the class 40-44 runs from 39.5 to 44.5, so its width is 5, not 4.
Height = frequency density = frequency ÷ class width. Area = frequency.
Estimating inside a class: assume the values are spread evenly, so part of a bar gives that fraction of its frequency.
Worked example Frequency densities
The masses, to the nearest kilogram, of 57 sacks are grouped as 40-44 (15 sacks), 45-54 (30 sacks) and 55-59 (12 sacks). Find the frequency density for each class.
Class boundaries: 39.5 to 44.5, 44.5 to 54.5 and 54.5 to 59.5.
Use Fig. 1 to estimate how many of the 60 times are between 12 and 25 seconds.
From 12 to 15 s: width 3, density 2.4, so \(3 \times 2.4 = 7.2\).
From 15 to 20 s: the whole bar, \(5 \times 3.6 = 18\).
From 20 to 25 s: half of the 20-30 class, \(5 \times 1.6 = 8\).
Total \(7.2 + 18 + 8 = 33.2\).
Answer: About 33 times
Your turn
A class \(10 \le x < 25\) has frequency 36. Its bar is drawn 4.8 cm high. The bar for \(25 \le x < 30\) is 5.6 cm high. Find the frequency of \(25 \le x < 30\).
Density of the first class \(= 36 \div 15 = 2.4\), so 1 cm stands for 0.5. The second bar has density \(5.6 \times 0.5 = 2.8\), so its frequency is \(2.8 \times 5 = 14\).
Examiner trap
Plotting frequency (not frequency density) on the vertical axis when the class widths differ. It makes wide classes look far too important and scores no marks for the heights.
Section 02
Stem-and-leaf and box plots.
A stem-and-leaf diagram keeps every raw value while showing the shape. A back-to-back diagram puts two data sets on one shared stem so they can be compared at a glance; the left-hand leaves are read from right to left. Always give a key for BOTH sides.
Class A leaves
Stem
Class B leaves
8 5 2
1
7 3 3 1
2
5 8
8 6 5 2 0
3
1 3 4 6 7 9
4 1
4
0 2 3 5 7 9
2
5
5
Key: 2 | 3 | 1 means 32 for Class A and 31 for Class B.
Worked example Quartiles from a back-to-back diagram
The diagram shows the marks of 15 students in each of Classes A and B. Find the median and interquartile range for each class and compare the two classes.
With 15 values the median is the 8th value; the lower quartile is the median of the lower 7 values (the 4th value) and the upper quartile is the 12th value.
Class A, in order: 12, 15, 18, 21, 23, 23, 27, 30, 32, 35, 36, 38, 41, 44, 52. So \(Q_1 = 21\), median 30, \(Q_3 = 38\), IQR \(= 17\).
Class B: 25, 28, 31, 33, 34, 36, 37, 39, 40, 42, 43, 45, 47, 49, 55. So \(Q_1 = 33\), median 39, \(Q_3 = 45\), IQR \(= 12\).
Compare in context: Class B scored higher on average (median 39 against 30) and their marks were more consistent (IQR 12 against 17).
Answer: A: median 30, IQR 17. B: median 39, IQR 12. B did better on average and was less spread out.
Worked example Box plot and skew
Draw a box-and-whisker plot for Class A and describe its shape.
The five values are: minimum 12, \(Q_1 = 21\), median 30, \(Q_3 = 38\), maximum 52.
Draw a scale, a box from 21 to 38 with a line at 30, and whiskers out to 12 and 52.
\(Q_3 - Q_2 = 8\) and \(Q_2 - Q_1 = 9\) are nearly equal, but the upper whisker (14) is longer than the lower one (9).
Answer: The box is roughly symmetrical with a slightly longer upper tail, so the data are close to symmetrical, at most slightly positively skewed.
Examiner trap
Forgetting to reverse the left-hand leaves. In the row \(8\ 6\ 5\ 2\ 0 \mid 3\), the smallest Class A value is 30 (the leaf next to the stem), not 38.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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A class \(20 \le x < 35\) has frequency 45. What is its frequency density?
Quick check+10 XP
Lengths are recorded to the nearest centimetre. What are the class boundaries of the class 10-19?
Quick check+10 XP
On a back-to-back stem-and-leaf diagram the key reads 4 | 2 | 7 means 24 for set P and 27 for set Q. What does the row \(9\ 1 \mid 3 \mid 0\ 5\) show for set P?
Quick check+10 XP
Where are cumulative frequencies plotted?
Section 03
Cumulative frequency graphs.
Plot each cumulative frequency at the UPPER class boundary and join the points (with a curve or straight lines). For \(n\) values, read across at \(\tfrac{n}{2}\) for the median, \(\tfrac{n}{4}\) and \(\tfrac{3n}{4}\) for the quartiles and \(\tfrac{k}{100}n\) for the \(k\)th percentile.
Height \(h\) (cm)
\(h \le 150\)
\(h \le 160\)
\(h \le 170\)
\(h \le 180\)
\(h \le 190\)
Cumulative frequency
12
50
120
176
200
The heights of 200 students; the shortest is 140 cm. The estimates below join the points with straight lines, which is what you read from a carefully drawn graph.
Worked example Median, quartiles and a percentile
Estimate the median, the interquartile range and the 90th percentile of the heights.
Median: the 100th student. Between 160 cm (50) and 170 cm (120): \(160 + \dfrac{100 - 50}{70} \times 10 = 167.1\) cm.
Lower quartile: the 50th student, which is exactly 160 cm.
Upper quartile: the 150th student. \(170 + \dfrac{150 - 120}{56} \times 10 = 175.4\) cm.
IQR \(\approx 175.4 - 160 = 15.4\) cm.
90th percentile: the 180th student. \(180 + \dfrac{180 - 176}{24} \times 10 = 181.7\) cm.
Answer: Median about 167 cm, IQR about 15.4 cm, 90th percentile about 182 cm
Worked example Proportion above a value
Estimate the number of students taller than 175 cm.
Read up from 175 cm: halfway between 120 and 176, so the cumulative frequency is \(120 + 28 = 148\).
Taller than 175 cm: \(200 - 148 = 52\).
Answer: About 52 students (26%)
Your turn
Using the same table, estimate the number of students with heights between 155 cm and 175 cm.
At 155 cm the cumulative frequency is \(12 + 0.5 \times 38 = 31\); at 175 cm it is 148. So about \(148 - 31 = 117\) students.
Examiner trap
Plotting cumulative frequencies at the class midpoints. A cumulative frequency counts everything UP TO the end of a class, so it belongs at the upper boundary.
Section 04
Choosing a display.
Paper 5 asks you to choose a representation and to give an advantage or disadvantage. Tie your reason to the data in the question.
Display
Good for
Drawback
Stem-and-leaf
Small data sets; keeps every raw value; shows shape; back-to-back compares two sets
Unwieldy for large data sets
Box-and-whisker
Comparing several sets side by side; shows median, quartiles, spread and skew
Raw values and the detailed shape are lost
Histogram
Large sets of continuous grouped data; shows the shape of the distribution
Raw values lost; median and quartiles not visible
Cumulative frequency
Estimating median, quartiles, percentiles and how many lie above or below a value
Raw values lost; values are only estimates
Stage 2 · Exam
Exam-style questions
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In a histogram the class \(10 \le x < 25\) has frequency 36 and its bar is 4.8 cm high. The bar for \(25 \le x < 30\) is 5.6 cm high. What is the frequency of \(25 \le x < 30\)?
A histogram has bars \(0 \le x < 10\) with density 1.5, \(10 \le x < 15\) with density 4 and \(15 \le x < 30\) with density 2. Estimate the number of values with \(8 \le x < 20\).
Exam style+20 XP
The ordered data are 4, 7, 9, 12, 15, 18, 20, 25, 31, 40, 52. What is the interquartile range?
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Data displays, sorted.
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