Higher: recover counts and missing histogram scales
Read frequencies from areas, recover a common density calibration or missing bar, and make explicitly conditional partial-class estimates.
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
- Construct and interpret continuous unequal-width histograms.
- Calculate rectangle areas and recover a missing part of a total.
- Split a numerical interval at stated class boundaries.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
An 8-unit-wide histogram bar is 5 grid divisions high and contains 20 observations. What does one vertical grid division represent?
What you will learn
- Recover frequencies and a missing common density calibration from actual rectangle areas.
- Use a complete total to recover an absent class count and density without pre-drawing its target bar.
- Estimate partial-class counts under a stated uniform-density assumption and explain their limits.
A supplied continuous histogram has 0 < x ≤ 4: density 3; 4 < x ≤ 10: density 2; 10 < x ≤ 20: density 3/2. Recover the full frequency table and keep the density scale labelled in observations per unit of x. Estimate the count with 8 < x ≤ 14, explicitly assuming uniform density within each class. Show the split overlap areas. Draw or annotate the actual rectangles, check the total and compare at least two bar heights and frequencies without confusing them.
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.
Supplied known histogram bars; a missing bar stays absent
Supplied data only. This is a display state, not a correctness or marking result.
Entered density per vertical drawing unit: 1. Tick positions and labels are retained exactly as entered; mismatched calibration is not repaired.
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.
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.
Read an area with its calibration
For a histogram whose density axis is already calibrated, multiply each actual class width by its density height. Write the measurement intervals and recovered frequencies in a table, then check the total. A count is represented by area, not by the height label on its own. When no physical measurement unit is supplied, label the horizontal variable x and the density as observations per unit of x.
If the vertical calibration is missing but equal grid divisions are shown, first recover one density from a known class frequency divided by its width. Divide that density by the known bar’s number of grid divisions to find the density represented by one division. Apply that one common calibration to all heights; choosing a different calibration for each bar would destroy the area interpretation.
Recover an absent bar from a complete sample
A complete sample total can determine one missing class count. Calculate the frequencies of the known rectangles, add them, and subtract that sum from the stated total. Divide the remaining count by the missing class width to obtain its density, then draw the recovered rectangle at the actual boundaries on the same scale.
This recovery needs sufficient information. An unknown total and an entirely absent class do not generally determine a unique frequency. Negative remaining counts expose inconsistent data. Keep the missing rectangle and its numerical height as unknown work until they have been recovered; a supplied empty interval is not permission to invent an attractive height.
State what a partial-class reading assumes
For an interval cutting through classes, divide it at every class boundary. Multiply the overlap width within each class by that class’s density and add the resulting areas. This uses a uniform-within-class guide: it treats the class density as constant across that rectangle. The resulting partial count is an estimate of the grouped observations, even when its numerical value happens to be a whole number.
Whole-class areas recover the supplied grouped frequencies, but partial areas do not locate the raw individuals exactly. Label the requested overlap on the histogram and explain the assumption rather than announcing an exact observed count. For a complete supplied sample, intervals outside its class range contribute no additional observations; do not extend a bar beyond its actual class domain.
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
A partial-class area is 7.5 under an explicitly uniform within-class guide. Which statement is justified?
Worked example
A supplied continuous histogram has 0 < x ≤ 4: density 3; 4 < x ≤ 10: density 2; 10 < x ≤ 20: density 3/2. Recover the full frequency table and keep the density scale labelled in observations per unit of x. Estimate the count with 8 < x ≤ 14, explicitly assuming uniform density within each class. Show the split overlap areas. Draw or annotate the actual rectangles, check the total and compare at least two bar heights and frequencies without confusing them.
Show the worked solution
- The widths are 4, 6, 10. The supplied heights are already calibrated densities 3, 2, 3/2, so multiply each width by its stated density rather than treating a height as a frequency.
- Multiply widths by calibrated densities: 4×3=12; 6×2=12; 10×3/2=15. The frequencies are 12, 12, 15 and total 39.
- Mark 8 < x ≤ 14 on the histogram and split at its internal class boundaries. Overlap widths 0, 2, 4 give contributions 0, 4, 6; their sum is 10. This is an estimate under the stated uniform-within-class guide, not an exact raw-location count.
- Keep the original given bars and their actual boundaries. Compare area, not just height: the second and third bars have frequencies 12 and 15, even though their densities are 2 and 3/2. Whole-class areas match the grouped counts; the guide does not reveal individual positions within a class.
Answer Widths 4, 6, 10; densities 3, 2, 3/2; frequencies 12, 12, 15; total 39. The requested partial-interval count is estimated as 10, conditional on uniform within-class density. Retain actual class widths and use rectangle areas to compare counts.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
The supplied histogram has boundaries 10, 15, 25, 45, 60, and consecutive bar heights 6, 5, 4, 2 equal vertical grid divisions. Its first class contains 18 observations; the numerical density calibration is missing. Recover the density per grid division, all density heights, every class frequency and the total. Draw the calibrated density-axis labels on the supplied histogram and compare heights with class counts.
Give me a hint
Separate horizontal width, displayed grid height and calibrated density; use a known area or complete total before drawing an unknown target.
Compare my reasoning
- The first width is 5. Its density is 18÷5=18/5. Since that bar is 6 grid divisions high, each division represents (18/5)÷6=3/5 frequency-density units.
- Multiply widths by calibrated densities: 5×18/5=18; 10×3=30; 20×12/5=48; 15×6/5=18. The frequencies are 18, 30, 48, 18 and total 114.
- Label equal vertical grid divisions with increments of 3/5; the four actual heights correspond to densities 18/5, 3, 12/5, 6/5. Preserve the supplied widths 5, 10, 20, 15. The first bar is taller than the third but contains 18 rather than 48 observations, so compare their areas.
- Keep the original given bars and their actual boundaries. Compare area, not just height: the second and third bars have frequencies 30 and 48, even though their densities are 3 and 12/5. Whole-class areas match the grouped counts; the guide does not reveal individual positions within a class.
Widths 5, 10, 20, 15; densities 18/5, 3, 12/5, 6/5; frequencies 18, 30, 48, 18; total 114. One vertical grid division represents density 3/5; label that calibration on the supplied graph. Retain actual class widths and use rectangle areas to compare counts.
Look for these in your work
- Recover the common density calibration or missing class information from the supplied data.
- Show actual bar areas, a labelled density scale and the complete frequency table.
- Split requested partial overlaps at boundaries and state the uniform-density estimation limit.
A supplied continuous histogram has 0 < x ≤ 3: density 2; 3 < x ≤ 9: density 5/3; 9 < x ≤ 15: final bar and height missing. The complete sample contains 28 observations. Recover the full frequency table and keep the density scale labelled in observations per unit of x. Estimate the count with 3/2 < x ≤ 12, explicitly assuming uniform density within each class. Show the split overlap areas. Draw or annotate the actual rectangles, check the total and compare at least two bar heights and frequencies without confusing them.
Give me a hint
Separate horizontal width, displayed grid height and calibrated density; use a known area or complete total before drawing an unknown target.
Compare my reasoning
- The class widths are 3, 6, 6. Known areas are 6, 10; their sum is 16. The complete total leaves 28−16=12 observations for the missing class, giving density 12/6=2.
- Multiply widths by calibrated densities: 3×2=6; 6×5/3=10; 6×2=12. The frequencies are 6, 10, 12 and total 28.
- Mark 3/2 < x ≤ 12 on the histogram and split at its internal class boundaries. Overlap widths 3/2, 6, 3 give contributions 3, 10, 6; their sum is 19. This is an estimate under the stated uniform-within-class guide, not an exact raw-location count.
- Keep the original given bars and their actual boundaries. Draw the recovered missing rectangle only after calculating its height. Compare area, not just height: the second and third bars have frequencies 10 and 12, even though their densities are 5/3 and 2. Whole-class areas match the grouped counts; the guide does not reveal individual positions within a class.
Widths 3, 6, 6; densities 2, 5/3, 2; frequencies 6, 10, 12; total 28. The requested partial-interval count is estimated as 19, conditional on uniform within-class density. The missing final bar has density 2. Retain actual class widths and use rectangle areas to compare counts.
Look for these in your work
- Recover the common density calibration or missing class information from the supplied data.
- Show actual bar areas, a labelled density scale and the complete frequency table.
- Split requested partial overlaps at boundaries and state the uniform-density estimation limit.
A supplied continuous histogram has 0 < x ≤ 5: density 4; 5 < x ≤ 20: density 2; 20 < x ≤ 50: density 1. Recover the full frequency table and keep the density scale labelled in observations per unit of x. Estimate the count with 0 < x ≤ 25/2, explicitly assuming uniform density within each class. Show the split overlap areas. Draw or annotate the actual rectangles, check the total and compare at least two bar heights and frequencies without confusing them.
Give me a hint
Separate horizontal width, displayed grid height and calibrated density; use a known area or complete total before drawing an unknown target.
Compare my reasoning
- The widths are 5, 15, 30. The supplied heights are already calibrated densities 4, 2, 1, so multiply each width by its stated density rather than treating a height as a frequency.
- Multiply widths by calibrated densities: 5×4=20; 15×2=30; 30×1=30. The frequencies are 20, 30, 30 and total 80.
- Mark 0 < x ≤ 25/2 on the histogram and split at its internal class boundaries. Overlap widths 5, 15/2, 0 give contributions 20, 15, 0; their sum is 35. This is an estimate under the stated uniform-within-class guide, not an exact raw-location count.
- Keep the original given bars and their actual boundaries. Compare area, not just height: the second and third bars have frequencies 30 and 30, even though their densities are 2 and 1. Whole-class areas match the grouped counts; the guide does not reveal individual positions within a class.
Widths 5, 15, 30; densities 4, 2, 1; frequencies 20, 30, 30; total 80. The requested partial-interval count is estimated as 35, conditional on uniform within-class density. Retain actual class widths and use rectangle areas to compare counts.
Look for these in your work
- Recover the common density calibration or missing class information from the supplied data.
- Show actual bar areas, a labelled density scale and the complete frequency table.
- Split requested partial overlaps at boundaries and state the uniform-density estimation limit.
Common mistakes
- Reading an uncalibrated grid height directly as frequency density.
- Pre-drawing a missing target bar instead of recovering it from the total.
- Calling a partial-class estimate an exact raw count because its result is an integer.
What can you explain now?
Why is one known class frequency enough to calibrate all the equal vertical grid divisions of a consistent histogram?
Compare with the explanation
Its width gives the known class density; dividing that by its grid height gives density per division, which applies to every bar on the same axis.
The shared axis scale is common to the whole chart. A different scale for each bar would not be one consistent histogram.
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Make it stick
Describe recovery from a known bar and from a complete total, then explain how overlap areas yield a conditional partial-class estimate.
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.