AQA A-Level Mathematics: L: Data presentation and interpretation (7357)

This guide teaches Section L: Data presentation and interpretation (content references L1 to L4) of the AQA A-level Mathematics (7357) specification, version 1.3, for A-level exams from June 2018 onwards. The specification lists Section L under Paper 3, which also assesses any Paper 1 content. A calculator is required in every 7357 paper, and the specification says you must be able to use calculator technology to compute summary statistics for Sections K to O.

Use it with the Data presentation revision notes and the Data presentation practice questions. The 7357 course hub lists every topic, the printable checklist lets you tick off outcomes, and the free 10-minute diagnostics show where to start.

What Section L covers

Ref	What you must be able to do
L1	Interpret diagrams for single-variable data, understanding that area in a histogram represents frequency; connect to probability distributions
L2	Interpret scatter diagrams and regression lines for bivariate data, including diagrams with distinct sections of the population; informal interpretation of correlation; correlation does not imply causation (calculations involving regression lines are excluded)
L3	Interpret measures of central tendency and variation, extending to standard deviation; calculate standard deviation, including from summary statistics
L4	Recognise and interpret possible outliers; select or critique data presentation techniques; clean data, including dealing with missing data, errors and outliers

L1: Diagrams for single-variable data

Histograms: area represents frequency

A histogram shows grouped continuous data. The classes can have different widths, so the height of a bar is frequency density, not frequency:

frequency density = frequency ÷ class width
frequency         = frequency density × class width  (bar area)

If the vertical axis is labelled frequency density, area equals frequency exactly. If a diagram is drawn to some other scale, area is still proportional to frequency, so you find the scale factor first.

Worked example 1

The times, t minutes, that 70 people took to finish a puzzle are shown.

Time t (min)	0 ≤ t < 10	10 ≤ t < 15	15 ≤ t < 20	20 ≤ t < 30	30 ≤ t < 50
Frequency	8	14	20	18	10

(a) Find the frequency density of each class. (b) Estimate how many people took between 12 and 25 minutes.

(a) Divide each frequency by its class width:

8 ÷ 10 = 0.8    14 ÷ 5 = 2.8    20 ÷ 5 = 4.0    18 ÷ 10 = 1.8    10 ÷ 20 = 0.5

(b) Take the area of the histogram between 12 and 25. This assumes values are spread evenly within each class.

12 to 15:  3 × 2.8 = 8.4
15 to 20:  5 × 4.0 = 20
20 to 25:  5 × 1.8 = 9
Total = 37.4, so about 37 people

Worked example 2: a histogram drawn to a scale

On a histogram of the same data, the bar for 15 ≤ t < 20 is 2 cm wide and 8 cm tall. The bar for 20 ≤ t < 30 is 3.6 cm tall. Find the frequency of that class without using the table.

Bar for 15-20: area = 2 × 8 = 16 cm², representing 20 people
So 1 cm² represents 20 ÷ 16 = 1.25 people
A 5-minute class is 2 cm wide, so the 10-minute class is 4 cm wide
Bar for 20-30: area = 4 × 3.6 = 14.4 cm²
Frequency = 14.4 × 1.25 = 18

This matches the table.

Box plots and cumulative frequency diagrams

A box plot shows the minimum, Q₁, median, Q₃ and maximum, with possible outliers often plotted as crosses. Read off the median and IQR = Q₃ − Q₁, and judge skew:

• median closer to Q₁ and a long upper whisker: positive skew
• median central, whiskers similar: roughly symmetrical
• median closer to Q₃ and a long lower whisker: negative skew

A cumulative frequency diagram plots running totals against upper class boundaries. Read the median at half the total frequency and the quartiles at a quarter and three-quarters.

Connecting histograms to probability distributions

Divide every frequency by the total and the bar areas become relative frequencies that add to 1. Area over an interval then estimates the probability that a randomly chosen value lies in it. From Worked example 1:

P(15 ≤ t < 20) ≈ 20/70 = 0.286 (3 s.f.)
P(t ≥ 20)     ≈ (18 + 10)/70 = 0.4

A roughly symmetrical, bell-shaped histogram suggests a Normal model; Section N of the specification links the Normal distribution back to histograms, mean and standard deviation.

L2: Scatter diagrams, correlation and regression lines

A scatter diagram plots pairs (x, y). Describe what you see in words:

• direction: positive correlation (y tends to rise as x rises), negative correlation (y tends to fall), or no correlation
• strength: strong if points lie close to a straight line, weak if widely scattered
• anything unusual: a point far from the pattern or separate clusters

A regression line of y on x is used to predict y from x. Calculations involving regression lines are excluded, so the skill is interpretation:

• The gradient is the estimated change in y for each increase of 1 in x, in context.
• Predictions are more reliable within the range of the x data (interpolation) than outside it (extrapolation), and more reliable when the correlation is strong.
• Use a y on x line to predict y from x, not the other way round.

Worked example 3

For 15 days at a garden centre, a manager records the hours of sunshine, h, and the number of bags of compost sold, c. The values of h range from 2 to 11. The regression line of c on h is c = 14 + 3.5h, and the scatter diagram shows strong positive correlation.

(a) Interpret the 3.5. (b) Comment on using the line for a day with 15 hours of sunshine. (c) The manager says sunshine causes people to buy compost. Comment.

(a) For each extra hour of sunshine, about 3.5 more bags of compost are sold.

(b) 15 hours is outside the observed range of 2 to 11 hours. That is extrapolation, so the linear pattern may not continue and the prediction is unreliable.

(c) Correlation does not imply causation. A third factor, such as the season, could raise both: spring brings more sunshine and more gardening.

Distinct sections of the population

A scatter diagram may contain separate clusters, for example data from two regions or two species. Then:

• one line fitted to all the points can show a correlation that does not exist within either group, or hide one that does
• it is usually better to describe, or model, each group separately

L3: Measures of central tendency and variation

Central tendency: mean, median and mode. Variation: range, interquartile range, variance and standard deviation.

Appendix B of the specification lists the mean formula among those you must recall:

x̄ = Σx / n = Σfx / Σf

The variance is the mean of the squared deviations from the mean. The standard deviation is its square root:

σ² = Σ(x − x̄)² / n = Σx²/n − x̄²        σ = √(σ²)

The specification’s notation uses σ for a population standard deviation and s for a sample standard deviation. Calculators usually show both σx (divisor n) and sx (divisor n − 1), which are close for large n. This page uses the divisor n; follow the convention each question uses.

Worked example 4: standard deviation from summary statistics

The lengths, x cm, of 20 leaves give n = 20, Σx = 340 and Σx² = 6100.

x̄  = 340 / 20 = 17 cm
σ² = 6100/20 − 17² = 305 − 289 = 16
σ  = 4 cm

With the divisor n − 1: s² = (6100 − 20 × 17²)/19 = 320/19, so s = 4.10 cm (3 s.f.).

Add a 21st leaf of length 23 cm by updating the sums:

n = 21,  Σx = 363,  Σx² = 6100 + 23² = 6629
x̄ = 363/21 = 17.3 cm (3 s.f.)
σ² = 6629/21 − (363/21)² = 16.87...
σ  = 4.11 cm (3 s.f.)

Worked example 5: grouped data

Estimate the mean and standard deviation of the puzzle times in Worked example 1. Use class midpoints: 5, 12.5, 17.5, 25 and 40.

Σf   = 70
Σfx  = 8(5) + 14(12.5) + 20(17.5) + 18(25) + 10(40) = 1415
Σfx² = 8(25) + 14(156.25) + 20(306.25) + 18(625) + 10(1600) = 35 762.5
x̄   = 1415/70 = 20.2 minutes (3 s.f.)
σ²  = 35 762.5/70 − (1415/70)² = 102.27...
σ   = 10.1 minutes (3 s.f.)

These are estimates because midpoints replace the actual values. Calculator statistics mode, with midpoints as x and a frequency column, gives the same values.

Interpreting the measures

• Mean and standard deviation use every value: best for roughly symmetrical data.
• Median and IQR resist outliers and skew: best for skewed data.
• When you compare two data sets, compare one measure of location and one measure of spread, each in the context of the question: “Group A took longer on average, and their times were less consistent.”

L4: Outliers, data cleaning and choosing diagrams

Possible outliers

An outlier is a value that lies far from the rest of the data. Use the rule a question gives. Two common rules are:

• below Q₁ − 1.5 × IQR or above Q₃ + 1.5 × IQR
• more than 2 (or 3) standard deviations from the mean

Worked example 6

The masses, in grams, of 12 apples from one orchard are:

138, 142, 144, 146, 147, 149, 150, 151, 153, 155, 160, 210

(a) Show that 210 is an outlier using the 1.5 × IQR rule. (b) The 210 g apple turns out to be a different variety that was mixed into the batch. Clean the data and find the new mean and standard deviation.

(a) Median = (149 + 150)/2 = 149.5. Taking the median of each half, Q₁ = (144 + 146)/2 = 145 and Q₃ = (153 + 155)/2 = 154, so IQR = 9.

Upper limit = 154 + 1.5 × 9 = 167.5
210 > 167.5, so 210 is an outlier

Other quartile methods change Q₁ and Q₃ slightly, but 210 is still well above the limit.

(b) The apple is not from the population being studied, so remove it. For the other 11 values, Σx = 1635 and Σx² = 243 405.

x̄ = 1635/11 = 149 g (3 s.f.)
σ = √(243 405/11 − (1635/11)²) = 5.91 g (3 s.f.)

With 210 included, σ = 17.9 g: one extreme value tripled the standard deviation.

Cleaning data

Real data sets often need cleaning before you analyse them.

• Missing data: a blank or a code such as “n/a”. Leave it out of calculations, say so, and never replace it with 0.
• Errors: impossible values (a negative mass) or obvious typing slips. Correct them if the true value is known; otherwise remove them.
• Outliers: investigate first. Remove one only if it is an error or does not belong to the population; a genuine extreme value stays.

Selecting or critiquing a presentation technique

Data	Suitable diagrams
Continuous, grouped, unequal classes	Histogram (frequency density), cumulative frequency diagram
Comparing two or more distributions	Box plots on the same scale
Bivariate (pairs of values)	Scatter diagram
Categorical	Bar chart, pie chart

When you critique a diagram, look for frequency plotted instead of frequency density, different scales on diagrams being compared, cumulative frequency plotted at midpoints, and one trend line drawn through separate groups.

Common errors

• Reading the height of a histogram bar as a frequency when class widths differ.
• Skipping the area scale factor on a histogram not labelled frequency density.
• Trusting a prediction without checking that x lies inside the data range.
• Saying one variable causes another because they are correlated.
• Confusing Σx² with (Σx)², or forgetting to subtract x̄².
• Recomputing a standard deviation from the old σ after adding or removing a value, instead of updating n, Σx and Σx².
• Removing an outlier just because it is large.

Where to go next

Test yourself with the practice questions and recap with the revision notes. Section K, where your data come from, is in the statistical sampling guide.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams June 2018 onwards, published by AQA. Section 3.13, L: Data presentation and interpretation.