AQA A-Level Mathematics Paper 3, June 2025: Question 16

7 marks · Medium difficulty · Multi-step Problem

Use a histogram showing travel distances for 340 individuals to estimate the frequency within a given interval, justify why it is an estimate, identify the skewness, and explain the use of frequency density.

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Question

A histogram on a grid representing distance in kilometres travelled from home to work by 340 individuals. The horizontal axis is labelled 'Distance (km)' with markings from 0 to 6, and the vertical axis is labelled 'Frequency density' without numerical values. Five bars are shown across intervals [0.5, 2.5], [2.5, 3.5], [3.5, 4.0], [4.0, 5.5], and [5.5, 6.0]. Four sub-questions follow: (a)(i) asks to estimate the number of individuals travelling between 3 km and 5 km with full justification, (a)(ii) asks for a reason why it is only an estimate, (b) asks to state the type of skewness shown, and (c) asks to explain why frequency density is used.

Mark scheme

Show the mark scheme The mark scheme details marks for parts 16(a)(i) to 16(c). For 16(a)(i), 4 marks are available: M1 for finding total area or square count, M1 for calculating the scale factor (e.g., 340 / 17 = 20), M1 for setting up the calculation for the area between 3 and 5 km, and A1 for 190. For 16(a)(ii), 1 mark (E1) is given for stating that the actual distance for each person is unknown or distances are not evenly distributed. For 16(b), 1 mark (B1) for stating 'negative'. For 16(c), 1 mark (E1) for stating class widths or intervals are different.

How to answer it

Analysing Histograms with Unequal Class Widths

AQA A-Level Mathematics • Statistics • 7 Marks Total

What this question tests

This question evaluates your understanding of continuous data representation using histograms with unlabelled frequency density axes:

  • Determining the constant of proportionality (scale factor) linking histogram area to total frequency.
  • Applying linear interpolation across split intervals to estimate partial class frequencies.
  • Understanding the fundamental modeling assumptions behind grouped continuous data.
  • Identifying distribution skewness visually from graphical shapes.
  • Justifying the statistical purpose of plotting frequency density instead of raw frequency.

Part (a)(i): Estimating Frequencies in an Interval

4 Marks • AO 3.1a (2 marks), AO 1.1a (1 mark), AO 1.1b (1 mark)

📐 Step-by-Step Calculation

  1. Identify bar dimensions (Width × Height units):
    • 0.5 to 2.5: width = 2.0, height = 1 → Area = 2.0
    • 2.5 to 3.5: width = 1.0, height = 2 → Area = 2.0
    • 3.5 to 4.0: width = 0.5, height = 3 → Area = 1.5
    • 4.0 to 5.5: width = 1.5, height = 7 → Area = 10.5
    • 5.5 to 6.0: width = 0.5, height = 2 → Area = 1.0
  2. Calculate total area:
    Total Area = 2.0 + 2.0 + 1.5 + 10.5 + 1.0 = 17 area units
  3. Find the scale factor (individuals per unit area):
    Scale factor = 340 individuals ÷ 17 units = 20 individuals per area unit
  4. Calculate the required area between 3 km and 5 km:
    • 3.0 to 3.5 km: width = 0.5, height = 2 → Area = 0.5 × 2 = 1.0
    • 3.5 to 4.0 km: width = 0.5, height = 3 → Area = 0.5 × 3 = 1.5
    • 4.0 to 5.0 km: width = 1.0, height = 7 → Area = 1.0 × 7 = 7.0
    Area (3 to 5 km) = 1.0 + 1.5 + 7.0 = 9.5 area units
  5. Estimate the number of individuals:
    Frequency = 9.5 × 20 = 190

✅ Correct Answer

190 individuals

Fully supported by scale calculations showing either unit areas (17 units → 20/unit), large 1 cm squares (34 squares → 10/square), or 2 mm grid squares (850 squares → 0.4/square).

🧠 Exam Technique: Counting Units

The vertical axis has no numbers! You can define your own scale:

  • Let 1 major grid division vertically = 1 unit of frequency density.
  • Always sum all bars first to set up the equation: Total Area × k = Total Frequency .
  • Clearly split bars that lie partly inside the target interval (e.g. 3 to 3.5 km and 4 to 5 km).

❌ Common Errors to Avoid

  • Reading the height as frequency: Remembering that for histograms, Frequency ∝ Area (not height) is critical.
  • Taking the entire bar width: Adding the whole bar from 4.0 to 5.5 instead of truncating it at 5.0 km.
  • Arithmetic slips on class widths: Double-check interval boundaries (e.g. 5.5 − 4.0 = 1.5, not 1.0).
Mark Scheme Breakdown:
• M1 (AO 3.1a): Summing total area across all 5 bars correctly (e.g., 17 units, 34 large squares, or 850 small squares).
• M1 (AO 1.1a): Correct method to find the ratio (e.g., 340 ÷ 17 = 20 or 1 cm = 20 seen on vertical scale).
• M1 (AO 3.1a): Correct expression for required area multiplied by scale (e.g., 340 ÷ 17 × 9.5).
• A1 (AO 1.1b): Correct final value of 190.

Part (a)(ii): Why the Result is an Estimate

1 Mark • AO 3.5b

✅ Acceptable Explanations (Any ONE)

  • The actual distance travelled by each person is unknown.
  • The data is grouped into intervals.
  • Distances within each class are not evenly distributed (assumed uniform distribution / linear interpolation).

💡 Key Knowledge: The Continuity Assumption

When calculating frequencies across parts of an interval in a histogram, we assume that data points are uniformly distributed across each class. In reality, commuters might cluster around specific round values (e.g., exactly 4 km or 5 km).

❌ Common Errors

Vague answers such as "Because you cannot read the graph precisely" or "Human error in measurement" score 0. You must refer specifically to the grouped nature of the data or the unknown individual values.

Mark Scheme Breakdown:
• E1 (AO 3.5b): Explaining that raw values are grouped/unknown or that distances may not be uniformly distributed.

Part (b): Identifying Skewness

1 Mark • AO 2.2b

✅ Correct Answer

Negative skewness (or Negative)

💡 How to Identify Skewness from a Histogram

  • Negative skew: Long tail stretches to the left (lower values); the bulk/peak of data is on the right (higher values). Here, the peak is between 4 and 5.5 km with a long tail back to 0.5 km.
  • Positive skew: Long tail stretches to the right (higher values); the peak is on the left.
  • Symmetrical: Peak in the center with balanced tails on both sides.

❌ Common Student Trap

Students frequently confuse where the peak is with where the tail is. Skewness describes the direction of the tail, not the peak!

Mark Scheme Breakdown:
• B1 (AO 2.2b): States 'negative' clearly.

Part (c): Purpose of Frequency Density

1 Mark • AO 2.4

✅ Acceptable Explanations (Any ONE)

  • The class widths (intervals) are unequal / different.
  • It gives an accurate / fair representation of the distribution when interval sizes vary.

🧠 Exam Technique

Always inspect the widths on the horizontal axis: [0.5–2.5] has width 2, [3.5–4.0] has width 0.5, and [4.0–5.5] has width 1.5. Because the widths vary, plotting raw frequency as height would misrepresent the data by exaggerating wider classes.

Mark Scheme Breakdown:
• E1 (AO 2.4): States that class widths are unequal/different or that it ensures a fair/correct representation.

Topics

Statistics · L: Data presentation and interpretation

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.