Sec 2 G2 Maths Bukit Timah | Unequal Histograms

Did you know that the tallest bar in a histogram might represent fewer observations than a shorter one? In an unequal-width histogram, a class five units wide with height four represents twenty observations, while a class ten units wide with height three represents thirty. The second bar is shorter—but it has greater area and more records. This is the idea that makes frequency density useful.

For Secondary 2 G2 Mathematics learners in Bukit Timah, histogram questions reward careful interpretation more than fancy drawing. If a pupil reads every vertical height as frequency without checking interval width, the answer can be wrong even with an otherwise accurate chart. A tutor should begin by asking what the axis means and how the width and height together represent the count.

This guide explains how to work from frequency tables to unequal-width histograms, how to recover counts from a drawn diagram and how to estimate means without mistaking class midpoints for raw data. The worked examples are original educational material, with quick-check practice and specific corrections for common mistakes.

The answer first: frequency is bar area

Quantity	Formula or meaning
Class width	Upper numerical boundary minus lower boundary.
Frequency density	Class frequency ÷ class width.
Bar height on density histogram	Frequency density, not raw count.
Class frequency	Width × density; the bar’s area in data units.
Whole sample size	Sum of all class frequencies or bar areas.
Grouped mean estimate	Sum(midpoint × frequency) ÷ sum(frequency).
Most observations	Compare class frequencies/areas.
Highest concentration	Compare frequency densities/heights.

Singapore Secondary 2 G2 Mathematics: the statistical-diagram context

G2 is a subject level under Full Subject-Based Banding. Secondary 2 Mathematics includes interpreting dot plots, histograms, grouped data and measures of central tendency, though schools differ in their lesson and assessment order. Unequal-width histograms require the frequency-density interpretation set out here; use the pupil’s own school materials to determine whether the present assignment uses equal intervals, unequal intervals, density or an extension beyond current coverage.

Curriculum context: a 2026 school’s G2 Mathematics outline and MOE G2 Mathematics syllabus. These examples illustrate the method rather than predicting a school assessment.

Twenty concepts behind accurate unequal histograms

1. Bar width carries information, not just decoration

A histogram represents observations grouped into numerical class intervals. A class from zero to five has width five; one from five to fifteen has width ten. Drawing both bars with identical widths would change the intervals shown along the horizontal axis. A histogram has a continuous numeric scale, so its horizontal dimensions need to be accurate before frequencies are considered.

How to teach: Draw classes [0,5), [5,15), [15,20) and mark every boundary on one numeric axis.

Watch for: Do not give all bars equal drawn widths simply because the class labels look similar.

2. The bar’s area represents frequency when widths differ

For unequal-width histograms, the rectangle’s area—not its height alone—is proportional to the frequency. A width-five bar with height four represents area twenty; a width-ten bar with height three represents area thirty. The shorter second bar contains more observations. This fact is the reason frequency density is needed and is the first idea worth teaching.

How to teach: Use rectangles of different widths on squared paper and count their square units.

Watch for: Never interpret the tallest unequal-width bar automatically as the greatest number of observations.

3. Frequency density is frequency divided by class width

For a class of frequency 20 and width five, density is 20÷5=4. Using density as the vertical-axis height makes bar area width×density = frequency. The equation links one known numerical relationship with the geometry of a rectangle; students who understand this are much less likely to memorise a fragile table rule.

How to teach: Write width, frequency and density in adjacent columns and verify area each time.

Watch for: Dividing width by frequency reverses the definition and gives incorrect heights.

4. Class width comes from interval boundaries

The interval 10≤x<15 has width 15−10=5, not fifteen or four. The convention tells us how values at a boundary are assigned, but the numerical width is the difference of the endpoints. Adjacent classes may have quite different widths, and the graph must preserve them on the same horizontal scale.

How to teach: Mark start and finish with arrows and subtract upper minus lower.

Watch for: Do not count integer labels inclusively as if a continuous class width were the number of whole numbers.

5. Half-open classes avoid double-counting boundaries

When 5≤x<15 follows 0≤x<5, a data value of five belongs to the second interval, while fifteen belongs to the next. The inequalities assign a boundary observation exactly once. This prevents overcounting in frequency tables. For measurements rounded to the nearest whole unit, continuous class boundaries may instead use half-unit values when the exercise specifies them.

How to teach: Check the membership of data values exactly at five, fifteen and twenty.

Watch for: Never put the same boundary observation in two neighbouring classes.

6. Equal widths make raw frequency bars simpler

If every histogram class has the same width, using frequency itself as height produces bar areas in a common constant proportion to frequency. This is why familiar school histograms often use frequency on the vertical axis. But once class widths vary, the vertical label should change to density or an equivalent frequency-per-width scale so comparisons remain fair.

How to teach: Compare equal-width class frequency heights with unequal-width density heights.

Watch for: A frequency-height convention from an equal-width histogram cannot simply be reused unchanged for unequal classes.

7. A taller density bar can hold fewer observations

Suppose class A has width five and density four: it contains twenty observations. Class B has width ten and density three: it contains thirty. A is taller, but B represents more observations. This is a powerful counterexample when a student judges a histogram by height rather than width times height.

How to teach: Calculate both rectangular areas and label them with frequencies.

Watch for: The highest density and the highest total frequency are different possible comparison questions.

8. The most concentrated interval has the greatest density

In a histogram, density represents observations per unit of horizontal measurement. The class with greatest density has the tallest bar and the highest concentration within its interval. For an unequal-width histogram, that class can be called the modal class in the density sense. Be careful not to confuse this with the class containing the greatest absolute frequency count.

How to teach: Ask separately for ‘highest density’ and ‘most observations’ and calculate both.

Watch for: Declaring that the tallest bar necessarily contains the highest count ignores class width.

9. Total sample size is the sum of bar areas

For classes with widths 5,10,5,20 and densities 2,3,4,1, the frequencies are 10,30,20,20. Their sum is 80, the total number of observations. Graphical area makes the count visible. A reliable final check compares this sum with any stated sample size and searches for missing or double-counted intervals.

How to teach: Multiply density by width for every bar and total the products.

Watch for: Adding just the bar heights 2+3+4+1=10 is not the sample size.

10. Frequency can be recovered when only density is supplied

If the horizontal class width is eight and the vertical density height is 2.5, frequency is 8×2.5=20. This reverse operation is often easier to see if the histogram’s rectangle is drawn and its area is labelled. The frequency should be a non-negative whole number for ordinary counts of people or objects, even though density itself can be fractional.

How to teach: Use rectangles on squared paper to build frequency from the plotted values.

Watch for: Reporting 2.5 people as the frequency confuses the vertical density with the count.

11. Density can be recovered from a known bar and used to calibrate scale

If one bar of width four contains twelve observations, its density is three. Its drawn height can therefore represent density three and help interpret other bars on the same density scale. This is useful when a graph’s vertical tick labels are omitted but one class’s frequency is known. The graphical scale must remain consistent across the bars.

How to teach: Calibrate with the known bar first, then infer another bar’s density and frequency.

Watch for: Do not invent arbitrary heights before identifying what the vertical scale represents.

12. A missing frequency can be found from the total

If the total sample is 80 and three known class counts are 10,30 and20, the fourth is 80−60=20. If that fourth class has width twenty, density is 20/20=1. This combines elementary algebra with a histogram. It also provides a useful check that the completed class table accounts for every observation.

How to teach: Make a frequency column that sums to the stated sample size.

Watch for: The missing value is a frequency first; its corresponding height must then be adjusted for width.

13. Midpoints support an estimated grouped mean

To estimate the mean from grouped data, multiply each class midpoint by that class’s frequency, sum the products, and divide by total frequency. For classes [0,5),[5,15),[15,20),[20,40) with frequencies 10,30,20,20, the midpoints are 2.5,10,17.5,30. Their weighted sum is 1275; estimated mean=1275/80=15.9375. It is an estimate because exact individual data have been grouped.

How to teach: Recover counts from bar areas before attempting any midpoint calculation.

Watch for: Do not use density as the multiplier in the grouped mean; the required weights are frequencies.

14. The class midpoint is not an observed data value

For class 20≤x<40, the midpoint is thirty, even if no observation was exactly thirty. Grouping has removed the individual numerical positions. The midpoint is a representative value for estimation and cannot reconstruct raw data exactly. When the question asks for an exact mean of raw measurements, the original ungrouped list is needed.

How to teach: Compare two different raw datasets that have the same class frequencies but different means.

Watch for: Do not report a grouped mean as the exact mean of all original observations.

15. Probability of a selected observation uses class frequency

If a sample of eighty observations has twenty in [15,20), the probability that one randomly selected observation is in that interval is 20/80=1/4, assuming each recorded observation is equally likely to be selected. The fact that the bar’s height is density four is irrelevant to the denominator. Recover area first, then compute probability.

How to teach: Show the event count and full sample count as separate integers.

Watch for: Using density 4/80 as the probability loses the class width.

16. Cumulative frequencies build a running count

Add class frequencies in order: 10, then 40, then 60, then 80. This tells how many recorded observations are below successive upper class boundaries, within the stated interval convention. It helps identify broad median regions and check the sample size, though the original data positions inside a class remain unknown.

How to teach: Write a cumulative column next to frequency and verify the last entry equals total.

Watch for: A cumulative frequency is not the same as a class’s individual bar height.

17. The graph must use a consistent horizontal numerical scale

If one centimetre of paper represents five units of data, an interval width twenty must occupy four times the space of a width-five class. Without a consistent horizontal axis, the bar areas cease to represent frequency. Scale selection therefore affects interpretation just as it does for linear graphs. A readable histogram labels the interval boundaries and the quantity being measured.

How to teach: Draw the same four-class table with a correct consistent scale before filling heights.

Watch for: Labels placed at equal distances despite unequal numerical intervals distort the graph.

18. Continuous classes usually have touching bars

For adjacent numerical intervals that cover consecutive values, histogram bars touch along their shared boundaries. Artificial gaps suggest unrelated categories, as in an ordinary bar chart. Any genuine class gap would need to be justified by the data and scale rather than inserted decoratively. The diagram is meant to preserve the structure of a numerical range.

How to teach: Compare a histogram of measurements with a bar chart of favourite foods.

Watch for: Do not treat category widths as numerical class widths.

19. A bar with zero frequency has zero area

If an interval has no observations, its frequency is zero and its density height is zero. The interval still occupies its correct width on the horizontal scale even though no rectangle rises above it. This matters when interpreting gaps in a numerical distribution. It also reminds students that an absent visible bar can represent a zero-frequency interval rather than missing information.

How to teach: Retain the interval boundaries in a frequency table even for a count of zero.

Watch for: Do not remove the interval and shift other bars left, which would alter the number-line scale.

20. Histogram interpretation must follow the exact question

One question may ask which interval contains the most observations, another which has the highest density, and another which contains at least twenty units. These are different tasks. Respond using area, height, or a boundary-conditioned sum as appropriate. The best practice is to write a sentence explaining what was counted and why before giving the numerical result.

How to teach: Underline ‘frequency’, ‘density’, ‘at least’ or ‘estimated mean’ in mixed questions.

Watch for: Using one automatic rule for every question hides whether the student understands the diagram.

Thirty original worked histogram examples

Worked example 1: Convert an unequal table to density

Question: Four classes have widths 5,10,5,20 and frequencies 10,30,20,20. Find densities.

Method: Divide frequency by width to obtain densities 2,3,4,1 respectively.

Check: Areas 5×2=10,10×3=30,5×4=20,20×1=20.

Wrong turn: Do not report heights as 10,30,20,20 for unequal widths.

Changed question: Change the last width to ten but keep its frequency twenty.

Worked example 2: The tallest is not the most numerous

Question: Class A width five density four; class B width ten density three. Which has more observations?

Method: A has 5×4=20. B has 10×3=30, so B has more.

Check: B is shorter but twice as wide.

Wrong turn: The vertical coordinate measures density, not frequency.

Changed question: What width would give a density-two bar frequency thirty?

Worked example 3: Find a missing bar frequency

Question: A bar represents width eight and density 2.5. Find its frequency.

Method: Frequency=8×2.5=20.

Check: Frequency is a whole observation count.

Wrong turn: Do not round density to two before multiplying.

Changed question: Try width six and density 1.5.

Worked example 4: Find density from frequency

Question: A class covers 10≤x<15 and has frequency twenty. Find density.

Method: Class width=5, density=20/5=4.

Check: Area 5×4=20.

Wrong turn: Do not use upper endpoint fifteen as width.

Changed question: Try class 15≤x<25 with same frequency.

Worked example 5: Class boundary inclusion

Question: Classes are 0≤x<5,5≤x<15,15≤x<20. Which class receives x=5 and x=15?

Method: Five belongs to second class; fifteen to third class, because lower boundaries are included and upper ones excluded.

Check: Neither point is counted twice.

Wrong turn: Do not put boundary values in adjacent classes simultaneously.

Changed question: Where does x=20 belong if next interval begins 20≤x<40?

Worked example 6: Calculate the whole sample

Question: Widths are 5,10,5,20 and density heights 2,3,4,1. Find total count.

Method: Frequencies 10,30,20,20 sum to 80.

Check: Areas sum 80.

Wrong turn: Adding heights gives ten, not 80.

Changed question: If the second density becomes four, find the new total.

Worked example 7: Find a missing class using total

Question: Total N=80; three class frequencies are 10,30,20. Find last frequency and density if its width is twenty.

Method: Missing f=80−10−30−20=20. Density=20/20=1.

Check: Total rebuilt is 80.

Wrong turn: Do not set density=20 directly.

Changed question: If width is ten, find new density.

Worked example 8: The grouped estimated mean

Question: Use classes [0,5),[5,15),[15,20),[20,40), counts 10,30,20,20. Estimate mean.

Method: Midpoints 2.5,10,17.5,30. Weighted sum 25+300+350+600=1275; divide 80 =>15.9375.

Check: Mean estimate lies between smallest zero and greatest forty.

Wrong turn: Use frequencies, not densities, as weights.

Changed question: Replace class [20,40) with [20,30) and discuss midpoint change.

Worked example 9: A grouped probability

Question: With total 80 and twenty observations in [15,20), find selection probability.

Method: 20/80=1/4.

Check: The event count comes from bar area width five×density four.

Wrong turn: Do not calculate 4/80 using density alone.

Changed question: Find probability for [5,15).

Worked example 10: Highest density versus highest frequency

Question: For counts 10,30,20,20 and densities 2,3,4,1, identify the class with largest density and the one with largest frequency.

Method: Largest density is [15,20) with four. Largest count is [5,15) with thirty.

Check: The answers differ because widths differ.

Wrong turn: Saying one is unambiguously ‘most common’ without clarifying the measure can mislead.

Changed question: Find the class with least frequency.

Worked example 11: A second unequal histogram

Question: Classes [0,10),[10,15),[15,25),[25,30) have frequencies 8,12,20,10. Find densities.

Method: Widths 10,5,10,5; densities 0.8,2.4,2,2.

Check: Areas recover 8+12+20+10=50.

Wrong turn: The tallest bar density 2.4 holds only twelve, while width-ten class density two holds twenty.

Changed question: Which class contains most observations?

Worked example 12: Estimated mean for second table

Question: Using the second four classes with counts 8,12,20,10, estimate the mean.

Method: Midpoints 5,12.5,20,27.5. Weighted sum 40+150+400+275=865; N=50, mean=17.3.

Check: All midpoints fall within respective classes.

Wrong turn: It is an estimate, not a raw-data exact mean.

Changed question: What happens if last class frequency doubles?

Worked example 13: Recover a bar height from area

Question: A histogram bar width ten has frequency thirty. What density height is needed?

Method: 30/10=3.

Check: Area equals ten times three = thirty.

Wrong turn: A height thirty would exaggerate frequency tenfold.

Changed question: Find height for width fifteen and frequency thirty.

Worked example 14: Compare same frequencies with different widths

Question: A width-five class and a width-ten class both have frequency twenty. Find density heights.

Method: Densities are four and two respectively.

Check: Each bar area is twenty.

Wrong turn: Equal counts need not produce equal bar heights.

Changed question: Which class is taller and why?

Worked example 15: Total from a graph without table

Question: Four drawn density bars have widths 4,6,10,5 and heights 2,3,1,4. Find total sample.

Method: Frequencies 8,18,10,20; total 56.

Check: Each product width×height counts observations.

Wrong turn: Do not add 2+3+1+4.

Changed question: Find probability for the fourth class.

Worked example 16: Probability from that same diagram

Question: In the sample of 56, fourth class has width 5 and density 4. Find selection probability.

Method: Fourth frequency=20, so P=20/56=5/14.

Check: Total 56 and event 20 are whole counts.

Wrong turn: Density four alone is not probability numerator.

Changed question: Calculate probability of first two classes combined.

Worked example 17: A known scale calibrates other bars

Question: A bar of width four has known frequency twelve and drawn height representing density three. Another bar has width six and density two. Find its frequency.

Method: Second frequency=6×2=12.

Check: Both bars represent twelve observations but their shapes differ.

Wrong turn: Do not treat the identical counts as equal heights.

Changed question: What if the second height represents density four?

Worked example 18: Class widths from decimals

Question: Class 2.5≤x<7.5 has frequency fifteen. Find width and density.

Method: Width=7.5−2.5=5, density=15/5=3.

Check: Bar area five×three=fifteen.

Wrong turn: Do not round endpoints before subtracting.

Changed question: Try 7.5≤x<17.5 with frequency twenty.

Worked example 19: A missing density

Question: Frequency of class [5,13) is twenty-four. Find density.

Method: Width eight, density=24/8=3.

Check: Eight×three=24.

Wrong turn: The bracket convention does not affect the numerical width.

Changed question: Change end boundary to eleven.

Worked example 20: What does density mean in words?

Question: A class spans ten units of time and contains thirty observations. Interpret density three.

Method: There are three recorded observations per one time-unit of class width on average as histogram density. It does not claim observations occur uniformly at each exact point.

Check: Integrating height across width gives thirty.

Wrong turn: Density is not three observations in total.

Changed question: Use a width-five class with same frequency.

Worked example 21: An interval with no records

Question: Class [10,20) has frequency zero; find density and bar appearance.

Method: Width ten and density 0/10=0; no raised rectangle, but interval remains on the horizontal scale.

Check: The frequency count zero is consistent with no area.

Wrong turn: Removing the class entirely could distort the axis.

Changed question: What happens to neighbouring bar boundaries?

Worked example 22: An unequal interval mode interpretation

Question: Which bar is tallest in the four-class table of densities 2,3,4,1?

Method: The [15,20) bar is tallest and has highest density; [5,15) has greatest frequency.

Check: Height and area measure distinct quantities.

Wrong turn: Do not call [15,20) the absolute most populous interval.

Changed question: State the answer for each version of the question.

Worked example 23: Find cumulative counts

Question: Counts are 10,30,20,20. Calculate cumulative frequencies.

Method: Cumulative totals 10,40,60,80.

Check: Final cumulative count equals N=80.

Wrong turn: Do not confuse cumulative 40 with frequency of just second interval.

Changed question: Where is the 60th observation located by class?

Worked example 24: Find which class contains 41st ranked observation

Question: For cumulative totals 10,40,60,80, which class contains 41st value?

Method: The third class [15,20) contains ranks 41 through 60.

Check: The second class ends at rank 40.

Wrong turn: A class boundary gives a rank region, not exact raw value.

Changed question: Which class contains 70th?

Worked example 25: Compare ratios from bar areas

Question: In first table, how many times as many observations are in [5,15) as in [0,5)?

Method: Counts are thirty and ten, so three times as many.

Check: Width differences accounted for via area.

Wrong turn: Do not compare height three to height two and claim ratio 3/2.

Changed question: Compare [5,15) and [15,20) frequencies.

Worked example 26: An incorrect raw-height histogram

Question: A student draws equal bar heights for frequency twenty across intervals width five and twenty. What is wrong?

Method: Unequal-width histogram should use densities 20/5=4 and 20/20=1 if frequency is represented by area. Equal heights would show areas in ratio one to four.

Check: Correct heights give equal bar area twenty.

Wrong turn: Copying a frequency-height convention is invalid with unequal widths.

Changed question: Redraw with one bar four times the height of the other.

Worked example 27: A measurement class with half-unit boundaries

Question: A whole-number measurement rounded to nearest unit is labelled 10–14 inclusive. What continuous boundaries might represent its values?

Method: Under nearest-unit rounding, boundaries can be 9.5≤x<14.5, width five.

Check: Values rounded as 10,11,12,13,14 all lie in that class.

Wrong turn: Do not assume these boundaries if the question provides a different measurement convention.

Changed question: Construct continuous boundaries for rounded labels 15–19.

Worked example 28: What the mean estimate cannot tell

Question: Two raw datasets lie within the same grouped class [0,10) and both have frequency two: values {1,2} and {8,9}. Do they have same actual mean?

Method: No. Means are 1.5 and 8.5, even though grouped midpoint estimate for each is five.

Check: The grouping discarded within-class positions.

Wrong turn: Do not treat all grouped midpoint calculations as exact.

Changed question: Explain why both would yield same histogram counts.

Worked example 29: Two density bars with equal areas

Question: Class A width six height three; class B width nine height two. Compare frequencies.

Method: A=6×3=18 and B=9×2=18. Their frequencies are equal.

Check: Both bar areas eighteen.

Wrong turn: A is taller but not more populous.

Changed question: Make a third different-width bar with frequency eighteen.

Worked example 30: An area-proportion reading

Question: In total 80, class [5,15) contains 30. What fraction and percentage of the sample is this?

Method: 30/80=3/8=37.5%.

Check: The bar area thirty supplies numerator.

Wrong turn: Using density three would give a wrong share.

Changed question: Calculate share of first and third classes together.

Ten mistakes and the first useful correction

1. Frequency and density mixed

Misconception: The graph’s vertical height is treated as number of students even though the bars have unequal widths. One class width ten, density three contains thirty students, not three. Ask the learner to say what the axis measures and multiply the height by interval width.

Repair: Change the interval width while keeping its frequency constant, then ask what happens to the height.

Independent evidence: A new rectangle should have correct area and height without a hint.

2. Widths assigned from labels instead of differences

Misconception: A 10–15 class is labelled width fifteen because fifteen is the upper boundary. Demonstrate that width is end minus start, five. Draw it on a correctly numbered horizontal axis.

Repair: Try class 2.5–7.5 and calculate width exactly.

Independent evidence: The student should handle fractional boundaries without rounding.

3. Tallest bar called greatest count

Misconception: The learner selects density-four class width five as having more records than density-three class width ten. Compute areas twenty and thirty, then explain highest density versus highest total frequency as two different questions.

Repair: Ask separately which class has highest density and which has most observations.

Independent evidence: The pupil should give two different correct answers where appropriate.

4. Bar boundaries overlap

Misconception: A measurement of 15 is counted in both 5≤x≤15 and 15≤x<20 classes. Use half-open interval notation, read endpoint conditions and allocate each observation once.

Repair: Build a small raw list including exact boundary values.

Independent evidence: No raw item should enter two classes.

5. The midpoint is mistaken for the exact data

Misconception: For class [20,40), the midpoint thirty is reported as a measured observation. Explain that grouping preserves how many records lie in the interval but loses their exact positions.

Repair: Construct two different raw lists that fit the same grouped counts.

Independent evidence: The learner describes estimated mean as estimated, not exact.

6. Density rather than frequency used as mean weights

Misconception: A grouped mean is calculated as a sum of midpoint×density instead of midpoint×frequency. Reconstruct frequencies from area first; they represent how many observations contribute to the average.

Repair: Use the four-class 80-observation table to compute correctly.

Independent evidence: The result should match a weighted-frequency estimate.

7. Histogram bars separated artificially

Misconception: Gaps are inserted between adjacent numerical classes, making the graph resemble unrelated categorical bars. Revisit the continuous x-scale and show why each class ends exactly where the next begins.

Repair: Draw a bar chart of colour categories beside a histogram of times.

Independent evidence: The pupil explains what distinguishes numerical intervals.

8. An inconsistent horizontal scale

Misconception: The intervals [0,5) and [5,15) are drawn the same width on paper. Relabel the horizontal axis with even numerical tick increments and draw the second twice as wide.

Repair: Use graph paper to trace a correctly scaled set of four intervals.

Independent evidence: Bar width ratios should match class width ratios.

9. Sample total obtained by adding heights

Misconception: Densities 2,3,4,1 are added to obtain ten, falsely taken as sample size. Multiply each height by its width first to get frequencies, then sum them to eighty.

Repair: Check areas and the last cumulative frequency against total.

Independent evidence: A fresh histogram is correctly totalled without any memorised answer.

10. The probability numerator comes from density alone

Misconception: A class with density four and width five has event frequency twenty, not four. Divide the event count twenty by the total sample eighty for the probability one-quarter.

Repair: Write a three-column table: density, frequency, selection probability.

Independent evidence: The final probability is based on counted observations.

Five small-group investigations with real mathematical purpose

One learner sees height, another sees area

Give three pupils the same unequal-width chart with class widths five and ten, densities four and three. One child may answer ‘four is bigger’ while another finds areas twenty and thirty. The third explains what frequency density means in words. Have everyone independently draw both bars before discussion. Then switch which question is asked: highest density, greatest frequency, or estimated share of the sample.

What to look for: Progress is shown when every pupil can distinguish height from area and recover the correct frequency for a changed width.

A child who knows the histogram formula but draws the axis incorrectly

A student calculates all densities accurately but places unequal interval boundaries at identical horizontal gaps. The resulting graph distorts frequencies. Draw the numeric x-axis from zero to forty at a consistent scale, mark 0,5,15,20,40 and then form each rectangle. Ask why the 20–40 interval has four times the width of 0–5.

What to look for: A changed class table should produce correct horizontal widths before any vertical heights are drawn.

A parent needs one simple graph-reading diagnostic

Show the four-class eighty-observation example and ask the child to identify which class has the tallest bar and which has most pupils. Then ask for total sample size. If the child can explain why density-four class contains twenty and density-three width-ten class contains thirty, the central misconception has been repaired. Finish by calculating probability of a randomly chosen observation being in the second class.

What to look for: The important evidence is an explanation of width×height, not memorisation of the numbers twenty and thirty.

Keep grouped-mean estimates separate from raw data

Give two pupils the same broad class [0,10) with frequency two, but one chooses raw values 1 and 2 while another chooses 8 and 9. Both yield identical histogram classes and midpoint estimate five, but very different actual means. This exercise makes the loss of information tangible. Then apply the same idea to several unequal intervals with midpoint weights.

What to look for: The learner should be able to say exactly why estimated grouped mean is not necessarily the raw-data mean.

Teaching a missing vertical scale

Show a histogram with one class of width four and known frequency twelve. Use this to infer its density three. On a correctly marked vertical scale, another bar’s height represents density two; its interval width six, so frequency twelve. This makes calibration a reasoning task rather than a visual guess. Reverse roles and ask students to calibrate another class independently.

What to look for: Success requires correct density, corresponding bar area and sample total after one known bar is changed.

Fifty quick retrieval questions

Quick check 1: Width of [10,15)

Answer: 5. Reason: 15−10.

Independent check: when a graph is involved, recover the rectangle area as class width multiplied by density. When the question asks for a mean, use the class’s frequency as the weight, not its density.

Quick check 2: Width of [5,15)

Answer: 10. Reason: 15−5.

Quick check 3: Width of [20,40)

Answer: 20. Reason: 40−20.

Quick check 4: Frequency20,width5: density

Answer: 4. Reason: 20/5.

Quick check 5: Frequency30,width10: density

Answer: 3. Reason: 30/10.

Quick check 6: Frequency10,width5: density

Answer: 2. Reason: 10/5.

Quick check 7: Frequency20,width20: density

Answer: 1. Reason: 20/20.

Quick check 8: Density4,width5: frequency

Answer: 20. Reason: Area.

Quick check 9: Density3,width10: frequency

Answer: 30. Reason: Area.

Quick check 10: Density1,width20: frequency

Answer: 20. Reason: Area.

Quick check 11: Counts 10,30,20,20: total

Answer: 80. Reason: Sum.

Quick check 12: Densities2,3,4,1: total frequency?

Answer: 80. Reason: Multiply widths.

Quick check 13: Tallest density of 2,3,4,1

Answer: 4. Reason: Third class.

Quick check 14: Most numerous f=10,30,20,20

Answer: Second class. Reason: Thirty.

Quick check 15: P(third class) if f20,N80

Answer: 1/4. Reason: 20/80.

Quick check 16: P(second class) if f30,N80

Answer: 3/8. Reason: 30/80.

Quick check 17: First class midpoint [0,5)

Answer: 2.5. Reason: Average bounds.

Quick check 18: Second class midpoint [5,15)

Answer: 10. Reason: Average bounds.

Quick check 19: Third midpoint [15,20)

Answer: 17.5. Reason: Average bounds.

Quick check 20: Fourth midpoint [20,40)

Answer: 30. Reason: Average bounds.

Quick check 21: Weighted sum for first table

Answer: 1275. Reason: 25+300+350+600.

Quick check 22: Estimated mean first table

Answer: 15.9375. Reason: 1275/80.

Quick check 23: First cumulative frequency

Answer: 10. Reason: First count.

Quick check 24: Second cumulative

Answer: 40. Reason: 10+30.

Quick check 25: Third cumulative

Answer: 60. Reason: 40+20.

Quick check 26: Final cumulative

Answer: 80. Reason: Whole sample.

Quick check 27: Class containing rank 41

Answer: [15,20). Reason: Cumulative 40 then 60.

Quick check 28: Class containing rank 70

Answer: [20,40). Reason: Ranks61–80.

Quick check 29: Bar width8,density2.5: f

Answer: 20. Reason: Product.

Quick check 30: Bar width6,density1.5: f

Answer: 9. Reason: Product.

Quick check 31: Bar width4,f12: density

Answer: 3. Reason: 12/4.

Quick check 32: Bar width10,f20: density

Answer: 2. Reason: 20/10.

Quick check 33: Second data sample N

Answer: 50. Reason: 8+12+20+10.

Quick check 34: Second table class0–10 density

Answer: 0.8. Reason: 8/10.

Quick check 35: Second class10–15 density

Answer: 2.4. Reason: 12/5.

Quick check 36: Second class15–25 density

Answer: 2. Reason: 20/10.

Quick check 37: Second class25–30 density

Answer: 2. Reason: 10/5.

Quick check 38: Second sample highest density class

Answer: [10,15). Reason: 2.4.

Quick check 39: Second sample most records class

Answer: [15,25). Reason: 20.

Quick check 40: Second weighted midpoint sum

Answer: 865. Reason: 40+150+400+275.

Quick check 41: Second estimated mean

Answer: 17.3. Reason: 865/50.

Quick check 42: Two equal frequencies width5 vs width10: heights

Answer: 4 and 2 for f20. Reason: f/width.

Quick check 43: Width6,height3: frequency

Answer: 18. Reason: 6×3.

Quick check 44: Width9,height2: frequency

Answer: 18. Reason: 9×2.

Quick check 45: Do bars touch for adjoining classes?

Answer: Yes. Reason: Continuous axis.

Quick check 46: Boundary x=5 in [0,5),[5,15)?

Answer: Second. Reason: Upper excluded first.

Quick check 47: Boundary x=15 in [5,15),[15,20)?

Answer: Third. Reason: Lower included.

Quick check 48: Raw values recoverable from wide classes?

Answer: No. Reason: Grouping loses detail.

Quick check 49: Mean from midpoints exact raw mean?

Answer: Generally no. Reason: Estimate.

Quick check 50: Frequency from vertical density alone?

Answer: No. Reason: Need class width.

Quick check 51: Highest density always greatest frequency?

Answer: No. Reason: Area matters.

Four weeks to move from graph reading to real understanding

Week 1: separate counts from density

Make simple frequency tables and calculate class widths. Compare equal-width and unequal-width graphs using physical paper rectangles, ensuring the student can explain why bar area represents count.

Progress check: A new two-bar comparison should be interpreted using width×height without tutoring hints.

Week 2: construct unequal-width histograms accurately

Mark real class boundaries on one consistent x-axis, calculate densities and draw adjoining bars with correct heights. Include an interval that has a fractional or zero density.

Progress check: A fresh frequency table produces the correct graph scale and bar areas independently.

Week 3: interpret sample size, proportions and missing values

Recover missing frequencies from known heights or sample totals. Calculate event probabilities using frequency over N. Compare tallest density bar with class containing most observations.

Progress check: The pupil can answer different wording correctly without relying on a one-size-fits-all bar-height shortcut.

Week 4: connect histograms to grouped mean

Use class midpoints with recovered frequencies to estimate the mean and explain why it is not exact. Mix boundary interpretation, density and midpoint questions in one short unlabelled diagnostic.

Progress check: Independent work shows the correct sample total, graphical dimensions, and a clearly labelled estimated mean.

How a 90-minute three-pupil tutorial can use histogram reasoning

The original eduKateSG Clementi Mathematics reference describes a small group of three in a ninety-minute lesson. An illustrative histogram session might use ten minutes to diagnose graph-reading errors, fifteen to derive density through rectangle area, twenty to build unequal-width bars, twenty-five for independent new tables, ten for grouped mean or probability connections and ten for reflective corrections. The schedule is an example of teaching design, not a statement of current vacancies.

Small-group teaching helps when each learner explains why a height-four class might have fewer observations than a height-three class. One pupil may misread the scale; another may calculate width incorrectly; a third may understand density but use it instead of frequency in an estimated mean. The tutor should choose a different repair for each and then check a new question independently.

Frequently asked questions

1. Why do unequal-width histograms use frequency density?

Because bar area must represent class frequency. If classes have different widths, identical numerical heights would give frequencies proportional to widths, not to the data counts. Frequency density is frequency divided by class width, ensuring that height multiplied by width recovers the correct number of observations.

2. What does a histogram bar’s height mean?

For equal-width histograms a frequency-scale height can represent frequency. For unequal-width histograms using a density scale, height represents frequency per unit of class width. Read the vertical-axis label before making any comparison.

3. What does a histogram bar’s area mean?

With density on the vertical axis and numerical class width on the horizontal axis, the area equals frequency (or is proportional to it if an arbitrary diagram scale is used). That is why a wide short bar may represent more observations than a narrow tall one.

4. How do I calculate class width?

Subtract the lower numerical boundary from the upper numerical boundary. The interval 10≤x<15 has width five. Boundary inclusion decides where exact endpoints are counted but does not change the numerical distance between boundaries.

5. Can the tallest bar contain fewer observations?

Yes. A width-five bar of density four contains twenty, while a width-ten bar of density three contains thirty. The first is taller, the second represents more records.

6. Which interval is the modal class when widths differ?

A density histogram’s highest bar gives the interval of greatest concentration per unit width, often called the modal class. If a question instead asks which class contains the most records, compare the areas or frequencies. Be explicit about what is being measured.

7. How can I find a class frequency if only its density is shown?

Multiply density by the class width. A width-eight class at density 2.5 contains twenty observations. The resulting frequency should be a non-negative whole number in ordinary count data.

8. How do I find the sample size from an unequal histogram?

Calculate each bar’s frequency from width×density and sum all the frequencies. The total is not the sum of heights. Check that any cumulative frequency table ends at the same total.

9. How can I estimate a mean from the histogram?

Recover class frequencies, find interval midpoints, multiply each midpoint by frequency, add the products and divide by total frequency. Label the result an estimate because the exact raw values within each interval are unknown.

10. Why should the bars touch?

Adjacent class intervals occupy contiguous numerical ranges on the x-axis. Histogram bars normally touch at those boundaries. A conventional categorical bar chart has different structural meaning and may separate categories.

11. How do I handle a value exactly at a class boundary?

Follow the given inequalities or stated measurement convention. With [0,5) and [5,15), five belongs to the second class. Each observation must enter exactly one interval.

12. Are unequal class-width histograms part of every G2 school assessment?

The Singapore Secondary 2 curriculum includes statistical diagrams and histogram interpretation; the depth of frequency-density calculations can vary by school and current subject-level scope. Follow the pupil’s classwork and teacher’s assessment notice before using extension examples.

The main takeaway

Did you know? In an unequal-width histogram, the picture only becomes reliable when the two dimensions of each rectangle tell the right story. Width describes the interval; height describes density; area recovers frequency. Once a learner can explain that relationship, missing-count questions, grouped mean estimates and probability readings become much easier to check.

The original Clementi three-pupil Mathematics tutorial reference remains unchanged. For class arrangements and suitability, contact eduKate Singapore directly. No particular result or subject-level change is guaranteed.

Read the other three Secondary 2 G2 Mathematics guides in this batch

This four-part collection connects algebraic fractions, unequal-width histograms, similar figures and hemisphere surface area. Choose the topic that matches a question from your current school work, explain its first rule, then test an independent changed problem.

Secondary 2 G2 Mathematics Tutor Bukit Timah | Unequal-Width Histograms and Frequency Density