Secondary Mathematics: Histograms, Frequency Density and Grouped Data

Secondary Mathematics · Worked Repair Guide 30

A histogram is not a bar chart with the gaps removed. Its horizontal axis is numerical and continuous, its class intervals have widths, and when widths differ the height must be adjusted so that area, not height alone, represents frequency.

This guide develops one central habit: read each rectangle as width × density = frequency. A tall narrow bar can represent fewer observations than a shorter wide bar. The picture only makes sense when the class widths and vertical-axis meaning are read together.

All data below are invented for teaching. Use Averages, Spread and Data Interpretation for basic frequency summaries and Quartiles, Cumulative Frequency, Box Plots and Standard Deviation for neighbouring grouped-data work.

1. Grouped data compress exact observations into intervals

Suppose twenty invented times are grouped into 0≤t<10, 10≤t<20 and 20≤t<30 minutes.

Once only the grouped table remains, we know how many observations lie in each interval but not their exact individual values.

Entry check: if five values lie in 10≤t<20, we cannot conclude that all five equal 15. The midpoint 15 may later be used as an estimate, not as recovered raw data.

2. Class width is upper boundary minus lower boundary

For 10≤t<20, class width=20−10=10.

For 20≤t<35, class width=15.

Class width is measured on the horizontal variable’s scale. It is not the frequency and not the drawn width in centimetres on the page.

If the horizontal axis is rescaled, the physical picture changes size while the numerical class width remains determined by the axis values.

3. Histograms use adjacent numerical intervals

A bar chart can display categories such as bus, train and walk. A histogram displays numerical intervals such as 0–10, 10–20 and 20–30.

The class rectangles touch because the scale is continuous across adjacent boundaries.

The order of histogram classes is determined by numerical position, not by an arbitrary category order.

Do not infer that every touching-bar chart is a valid histogram; check the axis and area meaning.

4. Equal class widths can hide the density idea

If every class has width 10, plotting frequency as height produces rectangles whose areas are all 10×frequency. The same width factor applies to every class, so comparing heights also compares frequencies.

This convenience disappears when class widths differ.

That is why learners sometimes believe histogram height always equals frequency: they first meet equal-width classes where the distinction is invisible.

Frequency density makes the area principle explicit.

5. Frequency density equals frequency divided by class width

Frequency density=frequency/class width.

If frequency is 18 in a class of width 6, density=18/6=3.

The rectangle area is width×density=6×3=18, reproducing frequency.

If the vertical axis is explicitly labelled frequency density, this relationship is the primary reading rule.

6. Unequal widths require different heights for equal frequencies

Class A has width 5 and frequency 20. Density=4.

Class B has width 10 and frequency 20. Density=2.

Both rectangles represent the same frequency, but A is twice as tall because it is half as wide.

Therefore the tallest histogram rectangle does not necessarily contain the largest frequency.

7. Build a histogram table before drawing

Interval	Width	Frequency	Density
0≤x<5	5	10	2
5≤x<15	10	30	3
15≤x<20	5	20	4
20≤x<40	20	20	1

The final class has frequency 20, equal to the third class, but density only 1 because its width is four times as large.

The total frequency is 80. The sum of rectangle areas on the density scale is also 80.

8. Recover frequency from a histogram rectangle

If a class runs from 12 to18 and its density height is 2.5, class width=6.

Frequency=6×2.5=15.

If the density scale is read incorrectly as frequency, the answer would be 2.5 observations, which may be impossible in a simple count context and is structurally wrong.

Always inspect the vertical-axis label before reading a rectangle.

9. Recover a missing density from a total frequency

Suppose total frequency is 100. Three known classes contain 18,27 and35 observations. A fourth class therefore contains 20.

If its width is 8, density=20/8=2.5.

The problem combines a frequency-total calculation with the histogram area rule.

Do not subtract density values from total frequency; those quantities are not directly comparable.

10. Continuous class boundaries remove artificial gaps

Suppose lengths are recorded to the nearest centimetre and grouped as 10–14 cm, 15–19 cm, 20–24 cm.

If the recorded integers represent rounded continuous measurements, true-value boundaries can be written 9.5≤L<14.5, 14.5≤L<19.5 and 19.5≤L<24.5.

The boundary 14.5 is shared; there is no real measurement gap from 14 to15.

This links histogram construction to the Batch 08 guide on Approximation, Significant Figures, Bounds and Error Intervals.

11. Midpoints estimate grouped-data means

When raw observations are unavailable, a common estimate replaces every value in a class by its midpoint.

For interval 10≤x<20, midpoint=15. If frequency is 8, estimated contribution to the total is 8×15=120.

Estimated mean=Σ(f×midpoint)/Σf.

It is an estimate because actual within-class values need not equal the midpoint.

12. Work a complete estimated mean

Interval	Frequency	Midpoint	f×midpoint
0≤x<10	4	5	20
10≤x<20	10	15	150
20≤x<30	6	25	150

Total estimated value=320 and total frequency=20, so estimated mean=16.

The true raw-data mean could differ while remaining compatible with the grouped frequencies.

Report it as an estimated mean when the midpoint method was required.

13. The modal class is not automatically the tallest histogram bar

The modal class is the class with greatest frequency.

If class widths are unequal, the greatest frequency may not have the greatest density.

For width 20, frequency 30 gives density 1.5. A width 5 class with frequency 20 gives density 4 and appears taller.

Read area or reconstruct frequency before identifying the modal class from an unequal-width histogram.

14. Frequency density is not probability density unless normalised appropriately

In a school histogram labelled frequency density, rectangle area equals frequency.

A probability-density graph uses area to represent probability, with total area 1.

The visual idea is related, but the vertical scale has a different normalisation.

Do not assume every density graph’s total area equals the number of observations; read the definition supplied by the problem.

15. Grouped data hides within-class variation

Two raw data sets can produce the same grouped frequency table while having different exact means or standard deviations.

For one class 10≤x<20 with two values, {10.1,19.9} and {14.9,15.1} both contribute frequency 2 to that class.

Their internal spread differs greatly.

Grouping compresses information, which is why midpoint-based statistics are estimates.

16. Compare histograms only after checking their scales

Two histograms can use different class widths or different vertical-axis scales.

A bar twice as tall in one graph does not necessarily represent twice the frequency of a bar in another graph.

Check class boundaries, density scale and total sample size before making cross-graph claims.

Visual comparison without scale control can create a false conclusion even when both graphs are individually correct.

17. Estimate a count inside part of a class only with an assumption

Suppose 20 observations lie in 10≤x<20. From the grouped table alone, the number below15 could be anywhere from 0 to20 depending on the unseen values.

If a question explicitly assumes uniform distribution within the class, an estimate below15 would be half of20=10.

The estimate comes from the assumption, not from information contained in the grouped count alone.

State whether interpolation or uniformity is being used.

18. Capstone: reconstruct an unequal-width histogram table

An invented histogram has classes 0–5, 5–15, 15–25 and25–40 with density heights 2,3,1.5 and2.

Frequencies are 5×2=10, 10×3=30, 10×1.5=15 and15×2=30.

Total frequency=85.

The tallest bar is the second class, which has frequency30. The fourth bar is shorter but also has frequency30 because it is wider.

This one example shows why area, not height alone, controls interpretation.

19. Independent practice

1. Find the width of class 12≤x<19.
2. A class has width5 and frequency20. Find density.
3. A class has width8 and density2.5. Find frequency.
4. Explain why a histogram with unequal class widths cannot use frequency as bar height directly.
5. Class A width4 frequency12; class B width8 frequency16. Find both densities.
6. Which class in Question5 is taller?
7. Which class in Question5 has greater frequency?
8. For intervals 0–10,10–20,20–30 with frequencies4,10,6, find midpoints.
9. Using Question8, estimate the mean.
10. A class 20≤x<35 has frequency30. Find density.
11. A histogram class spans 40 to50 with density1.8. Find frequency.
12. Frequencies are18,27,35 and unknown; total100. Find the missing frequency.
13. If the missing class in Question12 has width8, find its density.
14. Recorded integer lengths 10–14 cm represent values rounded to nearest cm. State continuous boundaries.
15. Explain why grouped data cannot normally recover the exact mean.
16. A class width20 frequency30 and another width5 frequency20. Which has larger density?
17. Which class in Question16 has larger frequency?
18. An unequal-width histogram rectangle has width6 and area15. Find its density height.
19. Under a stated uniform-within-class assumption, 24 observations lie in 10≤x<20. Estimate how many are below15.
20. Explain why the tallest rectangle need not be the modal class.

20. Worked answers

1. 7.

2. 4. 20/5.

3. 20. 8×2.5.

4. Because frequency must be represented by rectangle area; unequal widths require height=frequency/width.

5. A density3; B density2.

6. A.

7. B. Frequency16 versus12.

8. 5,15,25.

9. 16. (4×5+10×15+6×25)/20=320/20.

10. 2. 30/15.

11. 18. 10×1.8.

12. 20.

13. 2.5.

14. 9.5≤L<14.5.

15. Exact positions within each class are no longer known; midpoint replacement is an approximation.

16. The width5 class. Densities1.5 and4.

17. The width20 class. Frequencies30 and20.

18. 2.5. Area/width=15/6.

19. 12. Half the class width under the stated uniform assumption.

20. Height represents density; a wider, shorter bar can have greater area and therefore greater frequency.

21. Diagnose histogram errors by checking area first

Common failures include reading density as frequency, ignoring unequal class widths, using midpoints as though they were observed values, leaving artificial gaps between continuous rounded classes, or identifying the tallest bar as the modal class without recovering its frequency.

A useful repair note says “area=frequency”, “density=frequency÷width”, “grouped mean is estimated”, or “continuous boundaries come from rounding intervals”.

Then change only one class width while preserving its frequency. If the learner understands the histogram, the rectangle height should change automatically.

22. Continue through the BTT learning routes

Return to the BTT Mathematics Hub or BTT Mathematical Lab. Use Averages, Spread and Data Interpretation for raw summaries and Quartiles, Cumulative Frequency, Box Plots and Standard Deviation for grouped cumulative analysis.

Within Batch08, continue to Probability Trees, Conditional Events and Without-Replacement Reasoning, Distance–Time, Speed–Time and Motion Graphs, or Approximation, Significant Figures, Bounds and Error Intervals.

23. Sources and scope

The grouped tables, histogram values and practice questions are original teaching material. Frequency density is used in the standard school sense where rectangle area represents frequency.

For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match unequal-width histograms, interpolation and grouped-statistics depth to the learner’s actual subject level and school programme.