Read a histogram with unequal class widths by area
A histogram with unequal class widths uses area to represent frequency. Its rectangle heights show frequency density: count divided by interval width. To compare how many observations lie in two intervals, multiply each height by its own width. This guide builds an original example and separates what the graph proves from what grouped data leave unknown.
A histogram with unequal class widths needs a density axis
First ask what the horizontal axis measures. Our example records how long an activity takes, so each rectangle covers a numerical interval of durations. A bar chart comparing named categories, such as three materials, does a different job. Category slots can have equal visual widths without representing equal numerical intervals. Here, width carries information and cannot be chosen merely to make the page tidy.
Read the vertical label before interpreting any height. In the convention used throughout this guide, height is the absolute frequency density, measured in observations per minute. A height of two means two observations per minute of interval width on average across that class. It does not mean two observations in the entire rectangle, nor does it tell you the individual values inside it.
Adjacent intervals share a boundary without overlapping. We use 0 ≤ t < 10, then 10 ≤ t < 30, then 30 ≤ t ≤ 35. A duration of exactly ten minutes belongs only to the second class. The final upper boundary is included by explicit agreement. In another task, follow its stated convention. Our math homework guide covers the broader habit of identifying the requested quantity before calculating.
Keep count, width and density in separate columns
Imagine thirty adult participants completing a short craft activity. The following durations and counts are invented for practice, not observations from a real workshop. Eight durations lie in the first interval, twelve in the second and ten in the third. The table includes the calculations that we will justify. You can reproduce everything on paper; the photographs are conceptual illustrations and contain no additional measurements.
| Duration t (min) | Count | Width (min) | Density (per min) |
|---|---|---|---|
| 0 ≤ t < 10 | 8 | 10 | 0.8 |
| 10 ≤ t < 30 | 12 | 20 | 0.6 |
| 30 ≤ t ≤ 35 | 10 | 5 | 2 |
Find each width by subtracting the lower boundary from the upper boundary: 10 − 0 = 10 min, 30 − 10 = 20 min, and 35 − 30 = 5 min. Do not call the second width thirty just because thirty is its upper boundary. Nor should you count the written endpoint numbers as though these were lists of whole-minute values. The variable is duration, and an interval from ten to thirty spans twenty minutes.
Check the count column separately: 8 + 12 + 10 = 30 observations. Its total is useful later, but the widths do not count people. This separation makes an error easier to locate: a wrong width changes density even when the copied count is correct.
Divide by width, then check by multiplying back
Calculate density as count ÷ width. The three heights are 8/10 = 0.8, 12/20 = 0.6 and 10/5 = 2 observations per minute. The inverse check is count = density × width. Multiplying back gives 0.8 × 10 = 8, 0.6 × 20 = 12 and 2 × 5 = 10. The same calculation checks both the table and the intended meaning of each rectangle.
Use a linear horizontal scale: twenty minutes must occupy twice the physical width of ten minutes. Do not draw three equally wide rectangles merely because the table has three rows. Start every rectangle at the baseline, use a consistent vertical scale, and label the axis as density. The BBC Bitesize histogram reference explains this frequency-density convention. Our data and diagram are original.
The numerical area, in minute × observations-per-minute units, gives observations. Physical square centimetres on your paper depend on the chosen drawing scales; they are not automatically one observation each. Changing the page scale changes physical area without changing the data. Keep axis units attached when explaining the multiplication.
The tallest class and the largest count are different answers
The last rectangle is tallest because its density is two, compared with 0.8 and 0.6. Yet it contains ten observations, fewer than the twelve in the middle class. The wider middle interval has more observations in total despite having the lowest average concentration per minute. Answer “which interval contains the most observations?” using areas or the count column, not the tallest shape.
For “which interval has the greatest density?”, the last class is the answer. These are two legitimate questions about the same data. A complete sentence names the quantity: “The 10-to-under-30-minute interval contains twelve observations”; “The 30-to-35-minute interval has the greatest density.” Avoid the unqualified phrase “the biggest group” when its meaning is unclear.
Course conventions for “modal class” can vary with presentation. If the question uses that term with unequal widths, inspect its definition rather than silently treating largest count and greatest density as interchangeable. You can state both explicitly. You also cannot recover an exact most-common individual duration from this grouped table. Many different lists fit the same three counts.
Try the next step with your own homework.
Bring your question and attempt into Lirno. Ask for a hint, discuss the reasoning, then try the next step yourself.
Recover counts only after checking the vertical scale
A supplied histogram may use relative frequency density instead. Divide each count by the total of thirty, then by its width. The first height becomes (8/30)/10 = 8/300 per minute. Its area is 8/30, the fraction of observations in that class. All three areas then sum to one, whereas the absolute-density areas in our graph sum to thirty. Neither convention permits ignoring width.
If the vertical axis is unlabeled and no class frequency or total is provided, rectangle proportions alone may let you compare relative amounts, but they do not establish an absolute count. You need a scale anchor. For instance, a clearly defined ten-minute class known to contain eight observations must have absolute density 0.8 per minute. That known rectangle can calibrate a consistent drawing scale.
For an equal-width histogram, heights are proportional to counts, so comparing heights ranks the counts correctly. That convenient special case does not transfer to unequal widths. The University of Oldenburg’s density lesson discusses why class width matters. Read your own graph’s label rather than assuming every histogram has the same numerical normalization.
A partial interval needs an assumption, not an invented count
How many participants took less than twenty minutes? The first class supplies eight. The second class spans ten to thirty minutes, and its twelve observations are not individually located. Between zero and all twelve of those could lie below twenty. Therefore the exact total below twenty can range from eight to twenty using only the grouped information. The rectangle does not reveal twelve hidden timestamps.
If a task explicitly asks for an estimate assuming uniform spread within each class, twenty cuts the middle interval in half. Assign half of twelve to that portion, giving an estimated 8 + 6 = 14. Call it an estimate under that assumption. It is not an exact count established by the table. A flat rectangle is a display of average density, not proof that observations were evenly spaced.
You can answer whole-class questions exactly from these counts: at least ten minutes accounts for 12 + 10 = 22 observations. Precise values within a class, an exact median and an exact mean generally need more information. Our word-problem guide helps connect a calculation to its actual question; here the crucial extra decision is whether the requested cutoff matches a class boundary.
Repair the first mismatch between the table and the picture
Suppose someone plots the raw counts 8, 12 and 10 as heights while retaining the unequal widths 10, 20 and 5. The resulting numerical areas are 80, 240 and 50. They do not preserve the count ratio 8:12:10. The middle rectangle is disproportionately large. Repair its height rule, rather than changing the correct data until the picture looks familiar.
Check in a useful order: boundary subtraction, count copied from the same row, density division, horizontal spacing, and vertical scale. Then multiply width by height for every rectangle and compare with the original count. Summing heights is not a total-frequency check when widths differ. In our graph, 0.8 + 0.6 + 2 = 3.4 is neither thirty people nor a meaningful combined count.
If the same mistake returns, record the earliest faulty decision and one fresh test in a focused mistake-review note. A useful cue is “recover the count from area before comparing classes.” It directs attention to the operation that failed instead of merely recording that the final answer was wrong.
Test yourself with three new intervals
Close the worked example. A second invented activity groups durations into 0 ≤ t < 5 with six observations, 5 ≤ t < 15 with eight, and 15 ≤ t ≤ 30 with six. Find the widths and heights. Which class contains the most observations, and which has the greatest density? Explain why the two classes with six observations need different heights.
The widths are 5, 10 and 15 minutes. Densities are 6/5 = 1.2, 8/10 = 0.8 and 6/15 = 0.4 per minute. The middle class contains the most observations, eight; the first has the greatest density, 1.2. The two six-observation rectangles have equal numerical areas, but the last is three times as wide and one third as high as the first. Multiplying each width by its height recovers 6, 8 and 6, totaling twenty.
Now decide whether the exact number below ten minutes is known. Six are in the first class; anywhere from zero to eight of the middle class could be below ten. Only a uniform-within-class estimate would assign four there and report ten altogether. Finish by stating your conclusion and its conditions in words. That final distinction tests whether you can read evidence, rather than only operate a density formula.
Questions about this method
Can I add the bar heights to get the total?
Not with unequal widths. Multiply each density by its own width, then add those counts. Relative-density areas instead sum to the proportion represented.
Does a flat rectangle mean equal spacing of the observations?
No. It displays one average density for a class. An exact within-class count needs the original observations; proportional estimates require an explicit assumption.
Can the most populated interval be shorter on the graph?
Yes. Height describes density. A wider interval can have a greater area and count while its height is lower. Name the quantity being compared.