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Histograms and cumulative frequency - Mathematics Year 10

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Introduction

Histogram area represents frequency; bar height needs the class width. A cumulative graph answers a different question: how many observations have accumulated up to a boundary?

Construct both representations for one complete grouped data set, then estimate quartiles with explicit interpolation assumptions. This is specification- and tier-sensitive GCSE progression; check your chosen route before treating every extension as core assessment.

Build unequal-width histograms from density

The fictional data describe 40 positive completion times x in minutes. The intervals below are open at the lower boundary and closed at the upper boundary. Thus 5 belongs to the first class and 10 to the second; no observation is counted twice. The model states that all 40 times lie within these classes.

Time class in minutes	Class width in minutes	Frequency	Frequency density per minute	Cumulative frequency
0 < x ≤ 5	5	20	4	20
5 < x ≤ 10	5	10	2	30
10 < x ≤ 20	10	4	0.4	34
20 < x ≤ 40	20	2	0.1	36
40 < x ≤ 60	20	4	0.2	40

For a frequency-density histogram, $density = \frac{frequency}{class width} .$ The source width 5/frequency 20 example gives height 4. Its rectangle area 5×4=20 matches its frequency. The wider 20-to 40 class has frequency 2 and width 20, so its height is 0.1 and area 20×0.1=2.

Draw a numerical horizontal time axis with boundaries 0, 5, 10, 20, 40, 60 at their true distances. Draw a vertical axis labelled “frequency density per minute”. The rectangle heights are 4, 2, 0.4, 0.1, 0.2. Adjacent intervals share boundaries, so their rectangles touch; do not give every class the same arbitrary width.

An independent check multiplies each height by its actual width: 5×4=20, 5×2=10, 10×0.4=4, 20×0.1=2, 20×0.2=4. Their areas sum to 40. Using ordinary frequencies 20, 10, 4, 2, 4 as heights with these unequal widths would give rectangle areas 100, 50, 40, 40, 80, which no longer preserve the intended relative frequencies.

Compare the 10-to 20 and 40-to 60 classes. Both contain 4 observations, but their widths are 10 and 20. Their correct heights are 0.4 and 0.2; the narrower bar is twice as tall while the areas remain equal. A taller histogram bar can reflect greater concentration per minute rather than a greater number of observations.

The original Commons reference uses different bin counts in three separate panels, not our five classes. Its vertical scales also differ, so comparing raw pixel heights across panels would be misleading. Use it to discuss how grouping hides or exposes detail; use our supplied class boundaries and density calculations to construct the requested unequal-width chart.

The actual Corbettmaths Drawing Histograms preview labels the horizontal axis Hours and the vertical axis Frequency Density. Observe the distinction between frequency and density before plotting a rectangle. Its hours-per-week example is different from our completion times measured in minutes. Original: corbettmaths, Drawing Histograms - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Plot and interpret cumulative counts

Cumulative frequency counts observations at or below each upper boundary for these stated classes. Add the class frequencies in order: 20, 30, 34, 36, 40. In the source example, the point(40, 36)means 36 observations are at or below 40 minutes. It does not mean that the 20-to 40 class contains 36 observations.

Start at the lower boundary with zero count. The complete boundary/count pairs are(0, 0), (5, 20), (10, 30), (20, 34), (40, 36), (60, 40). The original schematic below plots these six points; the horizontal values are minutes and the vertical values are cumulative numbers of observations. Empty spaces do not represent missing measured data.

40 |                        * 
38 |                          
36 |                *         
34 |        *                 
32 |                          
30 |    *                     
28 |                          
26 |                          
24 |                          
22 |                          
20 |  *                       
18 |                          
16 |                          
14 |                          
12 |                          
10 |                          
 8 |                          
 6 |                          
 4 |                          
 2 |                          
 0 |*                         
   +--------------------------
    0   10  20  30  40  50  60

Text alternative: the cumulative count rises from 0 at time 0 to 20 at 5, to 30 at 10, to 34 at 20, to 36 at 40, and to 40 at 60. On graph paper, join neighbouring boundary points with straight segments for the interpolation model used here. A smoothly drawn curve is another common visual estimate; it should not be treated as extra measured observations between boundaries.

Recover a class frequency by subtracting consecutive cumulative values. The 20-to 40 class has 36−34=2 observations. The 40-to 60 class has 40−36=4. Every difference is non-negative, and the final cumulative count 40 equals the total frequency. A downward segment or final count different from 40 would signal a recording or construction error.

Using the conventional grouped-graph positions N/4, N/2, 3 N/4 gives cumulative levels 10, 20, 30. Straight-line interpolation estimates Q 1=2.5 minutes, median=5 minutes, Q 3=10 minutes, so IQR=7.5 minutes. These are grouped-data estimates under the stated graph convention, not the exact quartiles or exact raw-data median, which cannot generally be recovered from class totals alone.

For example, at cumulative level 10, you are halfway through the first class’s 20 observations, so interpolate halfway across 0 to 5. The median and upper-quartile levels meet listed boundary points in this example. Even then, the raw middle observations within neighbouring classes remain unknown; do not replace a conventional graph estimate with a claim of exact individual values.

At 15 minutes, the straight segment from(10, 30)to(20, 34)estimates 32 observations at or below 15. Consequently 8 are estimated above 15. The count 36 at or below 40 is exact from the supplied class totals; the count 32 inside a class relies on an assumed within-class distribution. State which type of result you are reporting.

Estimate quartiles and audit grouped conclusions

A grouped-data calculation must distinguish exact counts from estimates of unknown observations. A midpoint estimate of the mean substitutes one representative value for every observation in a class. It can be useful, but it cannot recover the exact original total when only grouped counts are known.

For our five classes, the midpoints are 2.5, 7.5, 15, 30, 50 minutes. Multiplying by frequencies 20, 10, 4, 2, 4 gives 50, 75, 60, 60, 200. Their sum 445 divided by 40 gives estimated mean 11.125 minutes. This is an estimated mean, not an actual recorded mean or a median.

The quartile interpolation rule for a target cumulative level r inside a class is $x \approx L + \frac{r - C}{f} w .$ Here L is its lower boundary, C the count before that class, f its class frequency, and w its width. This distributes observations uniformly within that class for the straight-line model; the assumption and inequality boundaries matter.

Near transfer with unequal widths. The following fictional grouped data also contain 40 observations, but they use different intervals and frequencies. Calculate densities before deciding which bar is tallest.

Time class in minutes	Width	Frequency	Density per minute	Cumulative count
0 < x ≤ 10	10	8	0.8	8
10 < x ≤ 30	20	12	0.6	20
30 < x ≤ 40	10	20	2	40

The histogram heights are 0.8, 0.6, 2. The middle class contains 12 observations versus 8 in the first, but it is twice as wide, so its bar is shorter. Areas 10×0.8=8, 20×0.6=12 and 10×2=20 recover every frequency and total 40.

The cumulative boundary points are(0, 0), (10, 8), (30, 20), (40, 40). Under the same N/4, N/2, 3 N/4 straight-line convention, the median estimate is 30. Q 1 is 10+(10−8)/12×20=40/3, about 13.33 minutes; Q 3 is 30+(30−20)/20×10=35 minutes. IQR=35−40/3=65/3, about 21.67 minutes. Retain exact fractions until final display rounding.

The transfer midpoint mean uses 5×8+20×12+35×20=980, so its estimate is 24.5 minutes. Compare that with the original 11.125 estimate only with the stated sampling and grouping limitations. Group summaries alone do not establish why two real populations differ or whether a difference is causal.

First-error repair. If someone uses frequency as height, divide by the actual width and recheck all areas. If someone reads cumulative 36 as a single-class frequency, subtract the preceding 34. If someone calls an interpolated quartile exact, identify the unknown within-class values and state the graph convention. Correct the earliest interpretation error rather than adjusting an isolated final answer.

Retain both complete grouped tables, a labelled unequal-width histogram construction, the six-point cumulative plot, recovered class counts, quartile estimates and a midpoint-mean calculation. Explain which results follow exactly from supplied totals and which require an approximation. Practical graph-drawing accuracy or later independent work should be recorded as completed only when actually observed.

Interactive Tasks

Quiz: Test Your Knowledge

Class width 5 and frequency 20: density? (4) (!20) (!100) (!0.25)

Width 20 and frequency 2: density? (0.1) (!2) (!40) (!10)

Equal frequency 4 in widths 10 and 20: which height is larger? (The width 10 bar) (!The width 20 bar) (!They must have equal height) (!Neither can be calculated)

Cumulative count 36 at upper boundary 40 means? (36 observations at or below 40) (!36 in the 20 to 40 class) (!Density 36) (!All 40 observations are below 36)

Consecutive cumulative values 34 and 36: class frequency? (2) (!36) (!34) (!70)

Forty observations: conventional graph median level? (20) (!40) (!10) (!30)

Original straight-line grouped Q 1 estimate? (2.5 minutes) (!5 minutes) (!10 minutes) (!11.125 minutes)

Original grouped midpoint mean estimate? (11.125 minutes) (!5 minutes) (!7.5 minutes) (!445 minutes)

Transfer width 20 and frequency 12: density? (0.6) (!12) (!240) (!1.2)

Transfer straight-line Q 3 estimate? (35 minutes) (!30 minutes) (!20 minutes) (!40 minutes)

Memory Game

Frequency density	Class frequency divided by its width
Histogram area	Represents the class frequency on consistent axes
Cumulative frequency	Count accumulated up to a stated boundary
Class midpoint	Representative value halfway between two class boundaries
Interpolation	Estimating an input or count between known boundary points
Interquartile range	Upper quartile minus lower quartile

Drag and Drop

Unequal class widths	Calculate density before drawing bar heights
Recover class count	Subtract consecutive cumulative values
Estimate grouped mean	Weight each class midpoint by its frequency
Estimate a quartile	Locate its cumulative level and state an interpolation convention
Check total frequency	Match the final cumulative count and sum of histogram areas

Crossword Puzzle

HISTOGRAM	Numerical class diagram whose areas represent counts
DENSITY	Frequency divided by class width
CUMULATIVE	Describes a running total
QUARTILE	One of the quarter-position values
MIDPOINT	Value halfway across a class interval
ESTIMATE	Result inferred when the exact underlying observations are unknown

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Identify the class boundaries: Explain where observations 5 and 10 belong.
Calculate the source density: Use 20 observations across width 5.
Read the cumulative 40 point: Explain 36 without calling it a class frequency.
Recover one interval count: Subtract 34 from 36.

Standard

Construct the five-bar histogram: Use true widths, correct density heights and labelled units.
Plot all six cumulative points: Use the table and check the final total 40.
Estimate the original quartiles: State N/4, N/2, 3 N/4 and straight-line interpolation.
Calculate the midpoint mean: Show all weighted products and mark it as an estimate.

Advanced

Solve the full transfer table: Find densities, quartiles and midpoint mean for the changed intervals.
Compare height with area: Explain why the middle transfer bar is shorter despite containing more observations.
Critique an exact-quartile claim: Explain what the grouped totals do not reveal about raw observations.
Repair a misleading chart or cumulative reading: Identify the first interpretation error and verify every corrected count.

Learning Check

Original grouped table and histogram: Preserve interval convention, widths and area.
Complete cumulative representation: Read boundary totals and recover classes.
Grouped quartile estimates: State graph convention and approximation.
Original midpoint estimate: Keep weights and denominator.
Full changed-data transfer: Calculate all representations independently.
First errors and honest evidence: Separate exact counts from inferred values.

Worked checks and assessment guidance

Frequencies 20, 10, 4, 2, 4 and widths 5, 5, 10, 20, 20 give densities 4, 2, 0.4, 0.1, 0.2. Each height-times-width recovers its frequency and their sum 40. Original classes are lower-open and upper-closed.
Points 0, 0; 5, 20; 10, 30; 20, 34; 40, 36; 60, 40 match the original data. The 20 to 40 class has 36−34=2. At or below 40 there are 36 observations, not 36 in that one class.
For N 40, use cumulative levels 10, 20, 30. Straight segments give Q 1=2.5, median 5, Q 3=10 and IQR 7.5 minutes. These are grouped-graph estimates, not exact raw-data order statistics. At 15 the estimated count 32 depends on interpolation; at 40 the count 36 comes directly from complete classes.
Midpoints 2.5, 7.5, 15, 30, 50 weighted by 20, 10, 4, 2, 4 give products 50, 75, 60, 60, 200, total 445. Divide by 40 to estimate mean 11.125 minutes. Unknown within-class observations prevent calling it exact.
Frequencies 8, 12, 20 over widths 10, 20, 10 give densities 0.8, 0.6, 2 and cumulative 8, 20, 40. Straight-line quartiles are 40/3, 30, 35 with IQR 65/3 minutes. Midpoint total 980 over 40 gives estimated mean 24.5.
Frequency-height confusion is repaired by dividing by width; class counts come from cumulative differences. An interpolation rule assumes a within-class pattern and cannot recover exact raw values. Keep labelled tables/plots and specification scope; do not claim unobserved drawing accuracy or uncompleted later work.

Learning evidence

Retain two complete grouped tables, correct density-height and area checks, a labelled six-point cumulative plot, exact cumulative differences, explicitly approximate quartiles and means, and first-error repairs.

OER and Wikipedia

Department for Education:GCSE Mathematics subject content and assessment objectives. Source area:Key stage 4 > Subject content; GCSE Mathematics subject content > Statistics. Year10 allocation is editorial within the GCSE pathway;Higher-only extensions are explicitly labelled. Original examples,teaching and feedback;Commons images retain their actual attribution,and videos identify original publishers. Reviewed12November2026.

Wikipedia bincounts.svg, Zckub, CC0. Unchanged original; caption and learning activities are original.

2D Cartesian Coordinates.svg, H Padleckas, Public domain. Unchanged original; caption and learning activities are original.