Higher: construct unequal-interval histograms
Use continuous class widths and frequency densities to draw adjoining rectangles whose areas recover the original frequencies.
Section 6.1 · Graphical representation of data
Exam scope and assessment notes
The first six lessons are shared Foundation work inherited by Higher; the last four are additional Higher histograms and cumulative-frequency work. Construct actual labelled diagrams with their scales, units and keys. Unequal continuous intervals need frequency density and area; partial-class and straight-guide readings are estimates, and a flat inverse can be an interval. Numerical agreement does not award marks or establish mastery; written explanations need human review. Statistical measures and probability remain separate. Linear November 2026/June 2027 entries and regional-centre eligibility differ from modular awards. Publisher map/contents are navigation only: F6.1Aii is a modular subpart, not the whole linear F6.1A row. Book1 Handling Data3 map398–428 differs from proof397/summary426/next428; Book2 Handling Data6 map208–222 differs from proof207/summary221/next222. Relevant chapter and Hub bodies remain unavailable; specialist and formal-assessment approval remain pending.
Before you begin
- Read continuous class limits and subtract signed boundaries.
- Use labelled Cartesian scales and calculate rectangle area.
- Complete Foundation frequency tables and interpret ordinary bar charts.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A continuous class runs from 2 to 9 and has frequency 28. What density is required?
What you will learn
- Find positive widths for continuous unequal intervals and calculate frequency density.
- Draw actual variable-width adjoining bars on labelled measurement and density axes.
- Recover frequencies as rectangle areas and distinguish height from count.
A complete sample of continuous time has these classes and frequencies: 0 < time ≤ 5: 12; 5 < time ≤ 15: 18; 15 < time ≤ 30: 24; 30 < time ≤ 50: 16. The measurement unit is min. Tabulate widths and frequency densities, then construct a labelled histogram with actual numerical class boundaries and adjoining variable-width rectangles. Check every class area and the total, and compare bar heights with their frequencies.
Choose your representation, enter its actual categories, counts, geometry and labels, and explicitly record it. No table margin, scale, missing sector, density, cumulative point or reading trace is filled for you.
Original supplied data
Only literal data are supplied. A missing entry is displayed as not supplied; requested totals, frequencies, angles, density calibration, cumulative constructions and read-off traces are not generated on this board.
All metric panels in this board share one camera and equal Cartesian drawing units. Equal full widths, heights and pie radii therefore stay visibly comparable; no individual panel is independently fitted. Histogram numerical frequency density is its entered vertical calibration. Tables and tally strokes are symbolic displays.
Blank construction plane
unready actual work. No target values or construction points are supplied.
Blue is supplied geometry, amber is my actual work, purple is dirty retained work and green is a separately gated reference. Rendering never repairs a wrong count, overlap, radius, calibration, cumulative decrease or read-off elbow. Tool coordinates are bounded to±1000000; finite unsupported portions are not silently clamped. Up to1000 explicit pictogram pieces and100 tally bundles per displayed row are rendered. Screen size is not physical instrument calibration. Readiness and fixed numerical comparison do not mark reasoning, award marks, prove mastery or provide teacher/specialist/formal-assessment approval.
Observed and whole-class counts stay distinct from within-class histogram or cumulative-guide estimates. A zero-frequency plateau can give an interval of inverse readings. No median, quartile, midpoint-mean or probability teaching credit is supplied by this display.
My actual complete rectangles and calibrated density axis
No actual representation is recorded for this action.
The board places only the coordinates you explicitly choose into the selected slot. It generates no other corner, bar, sector or reading elbow. Interval-area bounds are two measurements, not a planar point.
My quantities, conclusions and explanation
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Enter an exact fraction or finite decimal, with the precision and quantity unit explicitly requested in the question.
Explain the actual construction and its conditions. Text presence does not establish a correct argument.
These original fixed cases support self-checking. They do not mark proofs or written explanations, award marks or mastery, certify instrument accuracy, or provide specialist or formal assessment approval. Within-class histogram and cumulative-guide readings are estimates, while supplied whole-class counts remain exact. A plateau may give an interval rather than one inverse reading.
Keep continuous class boundaries
A histogram represents grouped continuous measurements. Put the class boundaries at their actual numerical positions on a uniformly calibrated measurement axis, including negative values where supplied. The width of a class is upper boundary minus lower boundary. Equal horizontal distances must represent equal changes in the measured quantity; unequal intervals therefore need genuinely different bar widths.
Use the stated interval convention consistently. In these tasks a class is lower < x ≤ upper, so adjacent classes share a boundary without counting a measurement in both. The complete sample lies in the supplied classes. Their bars adjoin because the measurement intervals are consecutive; separated equal-width category bars would not represent this grouped scale.
Choose height so area means frequency
When widths differ, bar height cannot be frequency. Define frequency density as frequency divided by class width. Label the vertical axis Frequency density and include observations per measurement unit where units are supplied. The numerical rectangle area, width multiplied by density, then equals the class frequency.
Calculate every width and density before drawing. Choose one consistent vertical scale high enough for all densities, mark actual boundary positions horizontally, and draw each rectangle across its whole class width. A fraction or decimal density is legitimate even though the class frequency counts whole observations. Keep exact working where possible so the area check remains clear.
Check areas before comparing bars
Multiply each drawn width by its intended density to recover the frequency, then add those frequencies to check the sample total. A taller narrow rectangle can contain fewer observations than a shorter wide rectangle. Explain a comparison by area or by the recovered frequencies, rather than by the heights alone.
The density rectangle summarises the class total; it does not reveal all individual positions within that class. A negative measurement boundary is allowed, but a negative width, frequency or density is not. Construct the actual labelled histogram as well as the calculation table. A plausible-looking outline without consistent units and scales cannot establish the required area relationship.
Sources for this lesson
- Pearson Mathematics A linear specification · Issue 2 · Foundation printed 24 / physical 30; Higher printed 39 / physical 45; inheritance printed 27 / physical 33
- Pearson 2026/2027 entry manual · centre regions physical 3; Mathematics A physical 25
- Pearson second-edition modular book/Hub map · literal navigation only · physical 7, 9, 11
Pause and explain
Which histogram rectangle contains more observations: width 3, density 6, or width 10, density 2?
Worked example
A complete sample of continuous time has these classes and frequencies: 0 < time ≤ 5: 12; 5 < time ≤ 15: 18; 15 < time ≤ 30: 24; 30 < time ≤ 50: 16. The measurement unit is min. Tabulate widths and frequency densities, then construct a labelled histogram with actual numerical class boundaries and adjoining variable-width rectangles. Check every class area and the total, and compare bar heights with their frequencies.
Show the worked solution
- Subtract consecutive numerical boundaries to get widths 5, 10, 15, 20 min; in particular the first width is 5−(0)=5. Place every boundary at its actual position on one uniform horizontal scale, including negative boundaries if present.
- Divide each frequency by its width: 12/5=12/5; 18/10=9/5; 24/15=8/5; 16/20=4/5. These are heights in observations per min, not the frequencies themselves.
- Draw and label the time (min) axis and a common zero-based Frequency density axis. Construct adjoining rectangles across 0, 5, 15, 30, 50 at heights 12/5, 9/5, 8/5, 4/5, keeping their different horizontal widths rather than drawing category bars.
- Check the rectangle areas: 5×12/5=12; 10×9/5=18; 15×8/5=24; 20×4/5=16. Their sum is 70. The tallest density is 12/5 in 0 < time ≤ 5, but the largest frequency 24 occurs in 15 < time ≤ 30; height alone does not compare class counts.
Answer Widths 5, 10, 15, 20 min; densities 12/5, 9/5, 8/5, 4/5 observations per min; rectangle areas 12, 18, 24, 16 observations; total 70. Construct the actual adjoining rectangles at the stated boundaries on uniform labelled axes. The tallest bar has frequency 12, whereas the largest class frequency is 24.
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A complete sample of continuous mass has these classes and frequencies: 20 < mass ≤ 25: 15; 25 < mass ≤ 35: 25; 35 < mass ≤ 55: 30; 55 < mass ≤ 80: 25. The measurement unit is g. Tabulate widths and frequency densities, then construct a labelled histogram with actual numerical class boundaries and adjoining variable-width rectangles. Check every class area and the total, and compare bar heights with their frequencies.
Give me a hint
Calculate upper minus lower for every class, then choose a height whose product with that width recovers its count.
Compare my reasoning
- Subtract consecutive numerical boundaries to get widths 5, 10, 20, 25 g; in particular the first width is 25−(20)=5. Place every boundary at its actual position on one uniform horizontal scale, including negative boundaries if present.
- Divide each frequency by its width: 15/5=3; 25/10=5/2; 30/20=3/2; 25/25=1. These are heights in observations per g, not the frequencies themselves.
- Draw and label the mass (g) axis and a common zero-based Frequency density axis. Construct adjoining rectangles across 20, 25, 35, 55, 80 at heights 3, 5/2, 3/2, 1, keeping their different horizontal widths rather than drawing category bars.
- Check the rectangle areas: 5×3=15; 10×5/2=25; 20×3/2=30; 25×1=25. Their sum is 95. The tallest density is 3 in 20 < mass ≤ 25, but the largest frequency 30 occurs in 35 < mass ≤ 55; height alone does not compare class counts.
Widths 5, 10, 20, 25 g; densities 3, 5/2, 3/2, 1 observations per g; rectangle areas 15, 25, 30, 25 observations; total 95. Construct the actual adjoining rectangles at the stated boundaries on uniform labelled axes. The tallest bar has frequency 15, whereas the largest class frequency is 30.
Look for these in your work
- Retain continuous class order, actual unequal widths and the stated units.
- Construct adjoining rectangles with frequency-density heights on calibrated axes.
- Verify every frequency by area and explain why height alone is insufficient.
A complete sample of continuous temperature has these classes and frequencies: -10 < temperature ≤ -5: 8; -5 < temperature ≤ 5: 24; 5 < temperature ≤ 20: 30; 20 < temperature ≤ 40: 12. The measurement unit is °C. Tabulate widths and frequency densities, then construct a labelled histogram with actual numerical class boundaries and adjoining variable-width rectangles. Check every class area and the total, and compare bar heights with their frequencies.
Give me a hint
Calculate upper minus lower for every class, then choose a height whose product with that width recovers its count.
Compare my reasoning
- Subtract consecutive numerical boundaries to get widths 5, 10, 15, 20 °C; in particular the first width is -5−(-10)=5. Place every boundary at its actual position on one uniform horizontal scale, including negative boundaries if present.
- Divide each frequency by its width: 8/5=8/5; 24/10=12/5; 30/15=2; 12/20=3/5. These are heights in observations per °C, not the frequencies themselves.
- Draw and label the temperature (°C) axis and a common zero-based Frequency density axis. Construct adjoining rectangles across -10, -5, 5, 20, 40 at heights 8/5, 12/5, 2, 3/5, keeping their different horizontal widths rather than drawing category bars.
- Check the rectangle areas: 5×8/5=8; 10×12/5=24; 15×2=30; 20×3/5=12. Their sum is 74. The tallest density is 12/5 in -5 < temperature ≤ 5, but the largest frequency 30 occurs in 5 < temperature ≤ 20; height alone does not compare class counts.
Widths 5, 10, 15, 20 °C; densities 8/5, 12/5, 2, 3/5 observations per °C; rectangle areas 8, 24, 30, 12 observations; total 74. Construct the actual adjoining rectangles at the stated boundaries on uniform labelled axes. The tallest bar has frequency 24, whereas the largest class frequency is 30.
Look for these in your work
- Retain continuous class order, actual unequal widths and the stated units.
- Construct adjoining rectangles with frequency-density heights on calibrated axes.
- Verify every frequency by area and explain why height alone is insufficient.
A complete sample of continuous length has these classes and frequencies: 0 < length ≤ 2: 10; 2 < length ≤ 6: 14; 6 < length ≤ 12: 15; 12 < length ≤ 20: 8. The measurement unit is cm. Tabulate widths and frequency densities, then construct a labelled histogram with actual numerical class boundaries and adjoining variable-width rectangles. Check every class area and the total, and compare bar heights with their frequencies.
Give me a hint
Calculate upper minus lower for every class, then choose a height whose product with that width recovers its count.
Compare my reasoning
- Subtract consecutive numerical boundaries to get widths 2, 4, 6, 8 cm; in particular the first width is 2−(0)=2. Place every boundary at its actual position on one uniform horizontal scale, including negative boundaries if present.
- Divide each frequency by its width: 10/2=5; 14/4=7/2; 15/6=5/2; 8/8=1. These are heights in observations per cm, not the frequencies themselves.
- Draw and label the length (cm) axis and a common zero-based Frequency density axis. Construct adjoining rectangles across 0, 2, 6, 12, 20 at heights 5, 7/2, 5/2, 1, keeping their different horizontal widths rather than drawing category bars.
- Check the rectangle areas: 2×5=10; 4×7/2=14; 6×5/2=15; 8×1=8. Their sum is 47. The tallest density is 5 in 0 < length ≤ 2, but the largest frequency 15 occurs in 6 < length ≤ 12; height alone does not compare class counts.
Widths 2, 4, 6, 8 cm; densities 5, 7/2, 5/2, 1 observations per cm; rectangle areas 10, 14, 15, 8 observations; total 47. Construct the actual adjoining rectangles at the stated boundaries on uniform labelled axes. The tallest bar has frequency 10, whereas the largest class frequency is 15.
Look for these in your work
- Retain continuous class order, actual unequal widths and the stated units.
- Construct adjoining rectangles with frequency-density heights on calibrated axes.
- Verify every frequency by area and explain why height alone is insufficient.
Common mistakes
- Using frequency as height when class widths are unequal.
- Drawing equal-width bars on a nonuniform class-boundary scale.
- Treating a tall bar as necessarily the class with the largest frequency.
What can you explain now?
Can a histogram of temperature have a negative horizontal boundary but only nonnegative density heights?
Compare with the explanation
Yes. Temperature can be negative; widths are positive differences between ordered boundaries, and frequencies and densities remain nonnegative.
The sign of a measured value is different from the sign of a count. Ordered upper-minus-lower boundaries give a positive class width even when a boundary is negative.
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Make it stick
Explain why density is frequency divided by class width, then describe the scales and area checks needed for an unequal-interval histogram.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.