Presenting Data – A Level Maths Year 1

1 Types of Data and Choosing a Diagram

Type	Meaning	Example
Qualitative (categorical)	Non-numerical descriptions.	Eye colour, type of car
Quantitative, discrete	Numerical, can only take particular values.	Number of siblings, shoe size
Quantitative, continuous	Numerical, can take any value in a range.	Time, mass, height

Diagram	Best for	Shows well
Histogram	Grouped continuous data, especially with unequal classes	Shape of the distribution
Cumulative frequency graph	Grouped continuous data	Median, quartiles, percentiles
Box plot	Summarising a single variable; comparing two or more data sets	Median, spread, skew, outliers
Scatter diagram	Two quantitative variables measured on the same items	Correlation and relationships

You should already know bar charts, pie charts and stem-and-leaf diagrams from GCSE; they may still appear. At A Level the emphasis is on interpreting diagrams and critiquing the choice of diagram, not just drawing them.

2 Histograms

In a histogram the area of each bar is proportional to the frequency — not the height. When classes have different widths, the vertical axis shows frequency density.

Frequency density

$\text{frequency density} = \dfrac{\text{frequency}}{\text{class width}}$ so $\text{frequency} = \text{frequency density} \times \text{class width}$

Worked Example 1 — Drawing a histogram

The commuting times of $80$ workers are shown. Calculate the frequency densities and draw a histogram.

Time $t$ (min)	Frequency	Class width	Frequency density
$0 \leqslant t < 10$	$8$	$10$	$0.8$
$10 \leqslant t < 20$	$22$	$10$	$2.2$
$20 \leqslant t < 30$	$26$	$10$	$2.6$
$30 \leqslant t < 50$	$18$	$20$	$0.9$
$50 \leqslant t < 80$	$6$	$30$	$0.2$

Plotting frequency instead would make the $30 \leqslant t < 50$ bar look almost as important as the $20$–$30$ bar, even though its data are spread over twice the width. Frequency density corrects for that.

Worked Example 2 — Estimating from a histogram

Using the histogram above, estimate how many workers took between $25$ and $40$ minutes.

①$25$ to $30$: width $5$, density $2.6$, so $5 \times 2.6 = 13$.

②$30$ to $40$: width $10$, density $0.9$, so $10 \times 0.9 = 9$.

③Estimate: $13 + 9 = 22$ workers.

This assumes the values are spread evenly within each class — that is why it is only an estimate.

Worked Example 3 — Histograms drawn to an unknown scale

On a histogram, a bar $1.5$ cm wide and $8$ cm tall represents a frequency of $36$. Another bar is $3$ cm wide and $2$ cm tall. What frequency does it represent?

①First bar: area $1.5 \times 8 = 12$ cm$^2$ represents $36$, so $1$ cm$^2$ represents $3$.

②Second bar: area $3 \times 2 = 6$ cm$^2$.

③Frequency $= 6 \times 3 = 18$.

Exam histograms are often drawn without a labelled frequency density scale. Always work from area $\propto$ frequency.

3 Cumulative Frequency

Cumulative frequency is a running total. Plot each cumulative frequency at the upper class boundary, then join the points with a smooth curve or with straight lines.

Worked Example 4 — Median, quartiles and a percentile

Use the commuting data to estimate the median, the interquartile range and the $90$th percentile.

Time less than	$10$	$20$	$30$	$50$	$80$
Cumulative frequency	$8$	$30$	$56$	$74$	$80$

①$n = 80$. Median at the $\tfrac{80}{2} = 40$th position; $Q_1$ at the $20$th; $Q_3$ at the $60$th; $90$th percentile at the $72$nd.

②Median: $40$ lies in $20 \leqslant t < 30$ ($30$ before it, $26$ in it): $20 + \dfrac{40 - 30}{26} \times 10 = 23.8$ min.

③$Q_1$: $10 + \dfrac{20 - 8}{22} \times 10 = 15.5$ min. $Q_3$: $30 + \dfrac{60 - 56}{18} \times 20 = 34.4$ min.

④IQR $\approx 34.4 - 15.5 = 19.0$ min.

⑤$90$th percentile: $30 + \dfrac{72 - 56}{18} \times 20 = 47.8$ min.

Joining the points with straight lines and reading off is exactly the same as linear interpolation in the table — see Measures of Location and Spread. For grouped data use $\tfrac{n}{2}$, not $\tfrac{n+1}{2}$.

4 Box Plots and Comparing Distributions

A box plot shows the five-number summary: lowest value, $Q_1$, median, $Q_3$, highest value. Outliers are usually plotted as separate crosses, and the whiskers then stop at the most extreme values that are not outliers.

A common outlier rule

An outlier is a value below $Q_1 - 1.5 \times \text{IQR}$ or above $Q_3 + 1.5 \times \text{IQR}$
(the question will always tell you which rule to use — see Correlation and Outliers)

Worked Example 5 — Drawing and comparing box plots

Test marks for two groups are summarised below. In group B the highest mark, $95$, is followed by $64$. Identify any outliers, draw box plots and compare the groups.

	Lowest	$Q_1$	Median	$Q_3$	Highest
Group A	$12$	$31$	$42$	$50$	$71$
Group B	$20$	$38$	$47$	$54$	$95$

①A: IQR $= 19$, fences $31 - 28.5 = 2.5$ and $50 + 28.5 = 78.5$. No outliers.

②B: IQR $= 16$, fences $38 - 24 = 14$ and $54 + 24 = 78$. $95 > 78$, so $95$ is an outlier; the upper whisker ends at $64$.

③Location: B's median ($47$) is higher than A's ($42$), so on average group B scored higher.

④Spread: B's IQR ($16$) is smaller than A's ($19$), so B's marks were more consistent.

Compare in context. Full marks need one comparison of location and one of spread, each with numbers, each interpreted in the context ("group B scored higher on average"). Use median and IQR together, or mean and standard deviation together.

Skew. If $Q_3 - Q_2 > Q_2 - Q_1$ (a longer upper section) the data are positively skewed: a tail of high values. If $Q_3 - Q_2 < Q_2 - Q_1$, the data are negatively skewed. For group A, $Q_2 - Q_1 = 11$ and $Q_3 - Q_2 = 8$, so A is slightly negatively skewed. The commuting times in Section 2 are positively skewed — the histogram has a long tail to the right.

5 Scatter Diagrams

When two variables are measured on each item (bivariate data), plot one against the other. The explanatory (independent) variable goes on the horizontal axis and the response (dependent) variable on the vertical axis.

Worked Example 6 — Choosing a diagram

Suggest a suitable diagram for each: (a) the ages of $200$ people at a concert, grouped into unequal classes; (b) the reaction times of two groups, to compare them; (c) the arm span and height of $30$ students.

①(a) A histogram with frequency density — classes are unequal.

②(b) Two box plots on the same scale — designed for comparison.

③(c) A scatter diagram — two variables measured on each student.

Interpreting scatter diagrams, correlation and regression lines is covered in Correlation and Outliers.

6 Quick Reference

Frequency density

Frequency $\div$ class width.

Histogram

Area $\propto$ frequency; no gaps between bars.

Unknown scale

Find frequency per unit area from a known bar.

Cumulative frequency

Plot at upper class boundaries.

Grouped median

Read at $\tfrac{n}{2}$; quartiles at $\tfrac{n}{4}$, $\tfrac{3n}{4}$.

Box plot

Min, $Q_1$, median, $Q_3$, max; outliers as crosses.

Skew

Longer upper part $\Rightarrow$ positive skew.

Comparing

One location + one spread, with numbers, in context.

Scatter

Explanatory on $x$-axis, response on $y$-axis.

7 Practice Questions

Question 1

Calculate the frequency densities for this table of leaf lengths $l$ cm: $0 \leqslant l < 5$: $10$; $5 \leqslant l < 10$: $18$; $10 \leqslant l < 20$: $24$; $20 \leqslant l < 40$: $16$.

▶ Show solution

$10 \div 5 = 2$, $18 \div 5 = 3.6$, $24 \div 10 = 2.4$, $16 \div 20 = 0.8$.

Question 2

On a histogram, the bar for $15 \leqslant x < 25$ has frequency density $3.4$. How many values lie in this class?

▶ Show solution

Frequency $= 3.4 \times 10 = 34$.

Question 3

The heights $h$ cm of $50$ seedlings are: $0 \leqslant h < 4$: $7$; $4 \leqslant h < 8$: $13$; $8 \leqslant h < 12$: $18$; $12 \leqslant h < 20$: $12$. Use interpolation to estimate the median and interquartile range.

▶ Show solution

Cumulative frequencies: $7, 20, 38, 50$.

Median ($25$th): $8 + \dfrac{25 - 20}{18} \times 4 = 9.11$ cm.

$Q_1$ ($12.5$th): $4 + \dfrac{12.5 - 7}{13} \times 4 = 5.69$ cm.

$Q_3$ ($37.5$th): $8 + \dfrac{37.5 - 20}{18} \times 4 = 11.89$ cm.

IQR $\approx 11.89 - 5.69 = 6.20$ cm.

Question 4

A data set has $Q_1 = 22$, median $27$ and $Q_3 = 31$. Its two smallest values are $8$ and $12$, and its two largest are $40$ and $52$. Using the $1.5 \times \text{IQR}$ rule, identify any outliers and state where the whiskers of the box plot should end.

▶ Show solution

IQR $= 9$, so $1.5 \times 9 = 13.5$.

Lower fence $22 - 13.5 = 8.5$; upper fence $31 + 13.5 = 44.5$.

$8 < 8.5$ and $52 > 44.5$, so both $8$ and $52$ are outliers.

The whiskers end at $12$ and $40$, with $8$ and $52$ marked as crosses.

Question 5

Box plots of the times (in seconds) taken to complete a puzzle show: Class X median $58$, IQR $20$; Class Y median $64$, IQR $12$. Compare the two classes.

▶ Show solution

Class X has the lower median ($58 < 64$), so on average Class X completed the puzzle more quickly.

Class Y has the smaller IQR ($12 < 20$), so Class Y's times were more consistent.

Question 6

On a histogram, the class $20 \leqslant x < 30$ has frequency $15$ and is drawn with height $3$ cm. The class $30 \leqslant x < 50$ is drawn with height $1.2$ cm, on the same horizontal scale. Find its frequency.

▶ Show solution

Heights are proportional to frequency density. First class: density $15 \div 10 = 1.5$ is drawn as $3$ cm, so $1$ cm represents a density of $0.5$.

Second class: density $1.2 \times 0.5 = 0.6$, so frequency $= 0.6 \times 20 = 12$.

Question 7

Explain why a histogram with unequal class widths should use frequency density rather than frequency on the vertical axis.

▶ Show solution

Wider classes naturally contain more values. Using frequency would make wide classes look more important than they are. With frequency density, the area of each bar is proportional to its frequency, so the diagram gives a true picture of how the data are distributed.

Question 8

A student draws a pie chart to show the heights of $40$ trees, using classes $0$–$5$ m, $5$–$10$ m, $10$–$20$ m and $20$–$40$ m. Give a criticism and suggest a better diagram.

▶ Show solution

Height is continuous numerical data with unequal classes; a pie chart hides the order of the classes and their different widths, so the shape of the distribution cannot be seen.

A histogram using frequency density (or a cumulative frequency graph) would be better.

Question 9

On a histogram, a bar $2$ cm wide and $6$ cm tall represents $24$ people. How many people are represented by a bar $4$ cm wide and $1.5$ cm tall?

▶ Show solution

$12$ cm$^2$ represents $24$, so $1$ cm$^2$ represents $2$ people.

Second bar: $4 \times 1.5 = 6$ cm$^2$, so $12$ people.

Question 10

The weekly time $h$ hours spent on homework by $60$ students is shown.

$h$	$0 \leqslant h < 2$	$2 \leqslant h < 4$	$4 \leqslant h < 6$	$6 \leqslant h < 10$	$10 \leqslant h < 15$
Frequency	$6$	$15$	$20$	$14$	$5$

(a) Find the frequency density for each class. (b) Estimate the number of students who spent more than $8$ hours. (c) Estimate the median and interquartile range. (d) A second group of students has median $6.2$ hours and IQR $2.5$ hours. Compare the two groups. (e) Describe the skew of the first group, giving a reason.

▶ Show solution

(a) $6 \div 2 = 3$, $15 \div 2 = 7.5$, $20 \div 2 = 10$, $14 \div 4 = 3.5$, $5 \div 5 = 1$.

(b) From $8$ to $10$: $2 \times 3.5 = 7$; plus all $5$ in the last class. Estimate $12$ students.

(c) Cumulative frequencies: $6, 21, 41, 55, 60$.

Median ($30$th): $4 + \dfrac{30 - 21}{20} \times 2 = 4.9$ hours.

$Q_1$ ($15$th): $2 + \dfrac{15 - 6}{15} \times 2 = 3.2$ hours. $Q_3$ ($45$th): $6 + \dfrac{45 - 41}{14} \times 4 = 7.14$ hours.

IQR $\approx 7.14 - 3.2 = 3.94$ hours.

(d) The second group has a higher median ($6.2 > 4.9$), so on average they spent longer on homework. They also have a smaller IQR ($2.5 < 3.94$), so their times were more consistent.

(e) $Q_2 - Q_1 = 1.7$ and $Q_3 - Q_2 = 2.24$, so the upper half is more spread out: positive skew, consistent with the long tail of the histogram to the right.

Presenting Data · A Level Maths Year 1 · Created with MathJax