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AQA A-level Mathematics (7357) · Data presentation and interpretation
Mini-Lesson

Data presentation and interpretation

This mini-lesson covers AQA section L: histograms (area = frequency), cumulative frequency and box plots, outliers (the 1.5 × IQR and mean ± 2 sd rules), standard deviation from summary statistics, and scatter diagrams with regression lines — including why correlation does not imply causation.

01020304050 min Q1 = 12 median Q3 = 20 whisker end outlier IQR = Q3 − Q1 = 8 · fences at Q1 − 1.5×IQR = 0 and Q3 + 1.5×IQR = 32
Box plot: min · Q1 · median · Q3 · max, with points beyond the 1.5 × IQR fences plotted separately as outliers.

Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. AQA note: you interpret and use a given regression line — calculating one is excluded. Press Start when you're ready.

Section L1 · histograms

Histograms: area = frequency

In a histogram the classes can have different widths, so the height is not frequency. Instead:

frequency density = frequency ÷ class width⇒ frequency = frequency density × class width = the AREA of the bar
0–55–1010–2020–35 freq density area = frequency
Unequal widths — a tall narrow bar and a short wide bar can hold the same frequency.
Worked example

Class 20 ≤ x < 35 has width 15 and frequency 54.

frequency density = 54 ÷ 15 = 3.6. A bar of height 3.6 and width 15 has area 54 ✓

Calculate

Frequency density

1In a histogram, the class 20 ≤ x < 30 has a frequency of 24. Calculate the frequency density.
Hint: Class width = 30 − 20 = 10. Frequency density = 24 ÷ 10.
Calculate

Reading a frequency

2On the same histogram, the bar for 0 ≤ x < 5 has a frequency density of 3.6. Calculate the frequency of that class.
Hint: frequency = frequency density × class width = 3.6 × 5.
Quick check

Area in a histogram

?In a histogram, what does the area of a bar represent?
Section L3–L4 · spread

Standard deviation and outliers

The mean and standard deviation summarise centre and spread; the median and IQR do the same but are resistant to outliers.

σ = √( Σx² / n − x̄² )variance = σ² = Σx²/n − x̄²  ·  you can also use Sxx: σ = √(Sxx / n), Sxx = Σx² − (Σx)²/n
Worked example — from summary statistics

n = 10, Σx = 50, Σx² = 300.

x̄ = 50 ÷ 10 = 5. Variance = 300/10 − 5² = 30 − 25 = 5.

σ = √5 = 2.24 (3 s.f.)

AQA accepts two outlier rules — the question will tell you which to use:

  • 1.5 × IQR rule: an outlier is any value below Q1 − 1.5 × IQR or above Q3 + 1.5 × IQR.
  • Mean ± 2 sd rule: an outlier is any value more than 2 standard deviations from the mean.

Cleaning data: an outlier is not automatically wrong. Only remove it if you can justify it (a typing error, a broken sensor). A genuine extreme value should be kept and commented on.

Calculate

Outlier fence

3A data set has Q1 = 12 and Q3 = 20. Using the 1.5 × IQR rule, calculate the upper boundary above which a value is an outlier.
Hint: IQR = 20 − 12 = 8. Upper fence = Q3 + 1.5 × IQR = 20 + 1.5 × 8 = 20 + 12.
Calculate

Standard deviation

4For a sample, n = 10, Σx = 50 and Σx² = 300. Calculate the standard deviation to 3 significant figures.
Hint: x̄ = 5. Variance = Σx²/n − x̄² = 30 − 25 = 5. σ = √5.
Sort it

Which diagram?

Tap a statement, then tap the diagram it describes.

📶 Histogram

📦 Box plot

✳️ Scatter diagram

Section L2 · bivariate data

Scatter, regression, and causation

A scatter diagram plots paired data (x, y). If the points lie close to a straight line there is linear correlation — positive, negative, or none.

  • The regression line of y on x, y = a + bx, is the line of best fit used to predict y from x. Here x is the explanatory (independent) variable and y the response (dependent) variable.
  • Interpolation — predicting inside the range of the data. Reasonably reliable.
  • Extrapolation — predicting outside the range. Unreliable: there is no evidence the linear pattern continues.
  • Watch for scatter diagrams containing distinct sections of the population (e.g. two separate clusters) — a single line of best fit is then misleading.
Worked example — using a given regression line

For data with x between 2 and 10, the regression line is y = 3.2 + 1.5x.

At x = 8 (inside the range): y = 3.2 + 1.5 × 8 = 3.2 + 12 = 15.2 — an interpolation, so reasonable.

At x = 40: y = 63.2 — an extrapolation, far outside the data, so not reliable.

Correlation ≠ causation. Ice-cream sales and drowning deaths correlate — because both rise with temperature. That third factor is a confounding variable.

Calculate

Using a regression line

5The regression line of y on x for data with x from 2 to 10 is y = 3.2 + 1.5x. Use it to estimate y when x = 8.
Hint: y = 3.2 + 1.5 × 8 = 3.2 + 12.
Quick check

Spot the outlier

?A data set has mean 62 and standard deviation 5. Using the mean ± 2 sd rule, which value is an outlier?
Quick check

Interpolate or extrapolate?

?The same regression line y = 3.2 + 1.5x is used to predict y when x = 40, well outside the data range of 2 to 10. What is the correct comment?
Match it

Statistics vocabulary

Tap a description on the left, then its term on the right.

Description
Term
Quick check

Correlation and causation

?Ice-cream sales and the number of drownings at the coast are strongly positively correlated. What can you conclude?
Section L1 · cumulative frequency

Cumulative frequency curves

A cumulative frequency diagram plots the running total of frequency against the upper class boundary of each class. The curve is then used to estimate the median and quartiles.

  • Median: read across at cumulative frequency n ÷ 2.
  • Lower quartile Q1: read across at n ÷ 4.   Upper quartile Q3: read across at 3n ÷ 4.
  • The IQR = Q3 − Q1 then feeds straight into the 1.5 × IQR outlier rule and into a box plot.
Worked example

A cumulative frequency curve is drawn for n = 80 values.

Median: read at 80 ÷ 2 = 40.   Q1: read at 80 ÷ 4 = 20.   Q3: read at 3 × 80 ÷ 4 = 60.

Why "estimate"? Grouped data has lost the individual values, so every statistic read from a cumulative frequency curve or calculated from a grouped table is an estimate. Say so in the exam.

Calculate

Reading a cumulative frequency curve

6A cumulative frequency diagram is drawn for 80 values. At which cumulative frequency do you read across to estimate the upper quartile?
Hint: Q3 is the 3n/4 value: 3 × 80 ÷ 4.
Quick check

Median from a curve

?For the same cumulative frequency diagram of 80 values, at which cumulative frequency do you read across to estimate the median?
Recap

The big ideas to know

Histogram: frequency density = frequency ÷ class width; area = frequency

Box plot: min · Q1 · median · Q3 · max; IQR = Q3 − Q1

Outliers: beyond Q1 − 1.5×IQR / Q3 + 1.5×IQR, or more than 2 sd from the mean

Standard deviation: σ = √(Σx²/n − x̄²) — mean & sd are affected by outliers; median & IQR are not

Regression: interpolation (inside the data) is fair; extrapolation (outside) is unreliable

Correlation does not imply causation — look for a confounding variable

That is the whole of section L — Data presentation and interpretation for AQA A-level Mathematics (7357). Press Finish to see your score.

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