A histogram represents grouped numerical data through adjoining class intervals. Its important feature is that the area of each bar represents frequency. When class widths differ, height alone cannot represent frequency fairly. Frequency density adjusts the height so that a wide interval does not appear more common simply because it covers more values.
For a histogram whose vertical axis is frequency density, frequency density equals frequency divided by class width. Equivalently, frequency equals density multiplied by class width. Understanding these relationships is more reliable than memorising a formula without checking the axes. This S1 guide uses original examples to explain construction, interpretation and checking. Confirm the exact syllabus and examination requirements for your course before selecting practice materials.
Begin with class intervals and boundaries
Read how the grouped data are defined. An interval such as 10 ≤ t < 15 has width 5 because its boundaries are 10 and 15. Do not count the possible printed integer labels as if they were the width. The interval describes a continuous span, even when the original measurements were recorded to a particular precision.
If a question gives rounded observations in inclusive-looking groups, determine the appropriate boundaries from the recording convention. For instance, measurements recorded to the nearest whole unit in groups 10–14 and 15–19 may use continuous boundaries 9.5–14.5 and 14.5–19.5. That interpretation depends on the context; do not add or subtract 0.5 automatically from every interval. Follow the information actually provided.
Understand why area represents frequency
Think of a bar as a rectangle. Its area is horizontal width multiplied by vertical height. If height is frequency density, multiplying by the class width gives frequency. A class with a larger frequency therefore has a larger bar area, regardless of how wide the class is. This keeps unequal intervals comparable within the same histogram.
Suppose two classes each contain 12 observations. One spans 4 units and the other spans 8 units. Their densities are 12/4 = 3 and 12/8 = 1.5. The second bar is half as tall but twice as wide, so both have area 12. Looking only at height would incorrectly suggest that the first class contains twice as many observations.
Calculate density from a grouped table
Consider fictional journey times in minutes. The interval 0 ≤ t < 5 contains 10 journeys; 5 ≤ t < 15 contains 30; and 15 ≤ t < 30 contains 15. The widths are 5, 10 and 15. Divide each frequency by its width to obtain densities 2, 3 and 1 respectively.
Draw the bars over the correct numerical intervals and use those densities as heights. The bar areas are 5 × 2 = 10, 10 × 3 = 30 and 15 × 1 = 15, recovering the frequencies. The total represented frequency is 55. These original values allow you to check the complete construction rather than relying on the apparent shape of the bars.
Label the axes according to their meaning
The horizontal axis represents the measured variable and should include its unit where relevant. The vertical axis represents frequency density in this construction. If journey time is measured in minutes, the density can be interpreted as observations per minute of class width. The precise labelling expected in a task should follow the question and teaching guidance.
Do not label the vertical axis “frequency” when you have plotted frequency divided by width. A correct calculation with an incorrect label communicates the wrong relationship. Also use a continuous numerical scale along the horizontal axis. Different class widths must occupy proportionally different horizontal distances; drawing every interval as an equally wide category box defeats the purpose of the histogram.
Recover frequency from a bar
If a bar covers 20 ≤ x < 28 and has frequency density 4.5, the width is 8 and the frequency is 8 × 4.5 = 36. The density need not be an integer even though a count frequency normally is. What matters is that the interpreted bar area gives the relevant number of observations.
Read the scale carefully before multiplying. If one vertical grid square represents 0.5 density units, a bar nine squares tall has density 4.5, not 9. The O Level graph-reading guide provides a related approach to checking axis intervals. Accurate scale reading is a prerequisite to an otherwise simple frequency calculation.
Compare bars using both height and width
A taller bar indicates a greater concentration of observations per unit interval, not automatically a larger frequency. For example, a width-2 class with density 9 contains 18 observations. A width-6 class with density 4 contains 24 observations. The shorter second bar represents more observations because its width is sufficiently larger.
When explaining a comparison, name the quantity you mean. “The first class has higher density” and “the second class has higher frequency” can both be true in this example. Avoid saying only that one class is “bigger”. Statistical language should distinguish concentration from count so the reader can tell what the graph supports.
Find a missing height
Suppose an interval has width 12 and frequency 42. Its density is 42/12 = 3.5. Plot the bar to that height on the stated density scale. Check by multiplying 12 × 3.5 to recover 42. This reverse operation is a direct way to detect a mistaken division or an incorrectly measured width.
If the axis uses a scaled measure rather than ordinary frequency density, read the question's calibration carefully. Some tasks use known bar information to establish how drawn area corresponds to frequency. Do not impose a density value on an unlabelled scale without using the supplied information. First identify the relationship between the diagram and the counts, then apply it consistently.
Estimate part of a class carefully
Grouped data do not show the exact positions of observations within an interval. Estimating the frequency in part of a class commonly involves assuming an even distribution within that class. State or recognise that assumption when it matters. An estimate is not an exact count recovered from information that the table does not contain.
Using the earlier journey-time table, estimate the number with 5 ≤ t < 11. This is 6 of the 10 minutes in the class 5 ≤ t < 15, which contains 30 journeys. Under a uniform-within-class assumption, the estimate is 6/10 × 30 = 18. The table alone does not prove that exactly 18 journeys lie in that narrower interval.
Combine intervals without mixing quantities
To find the total frequency across several complete classes, add their frequencies or calculate each bar area and then add. Do not add densities unless the task explicitly requires a quantity for which that operation makes sense. Density is a rate per unit width; its sum across unrelated widths is not the total count.
For the journey-time example, the frequency below 15 minutes is 10 + 30 = 40. Adding densities 2 + 3 = 5 would not answer that question. Write the quantity you are calculating before choosing an operation. The S1 clear-working approach can help you separate the interpretation step from the arithmetic.
Distinguish a histogram from a categorical bar chart
A histogram's horizontal positions represent numerical intervals, and the widths matter mathematically. A categorical bar chart displays categories whose widths do not carry the same measurement meaning. In a histogram, adjoining intervals normally meet because the horizontal scale represents their continuous boundaries. A genuine empty interval can have zero height.
Do not decide the graph type only from whether the bars visually touch. Read the variable and axes. A badly drawn chart can resemble another format while representing something different. Ask what the horizontal distance means and whether bar area or height carries frequency. Those questions provide a more reliable distinction than memorising a rule about gaps alone.
Use equal-width cases to check understanding
When every class has the same width, density is proportional to frequency by a common factor. This explains why relative heights can still reflect relative frequencies in that special case. It does not remove the need to label the axis accurately or understand what the question expects you to plot.
For width-5 classes with frequencies 10, 20 and 30, the densities are 2, 4 and 6. The ratios between heights match the frequency ratios. If one class later changes to width 10 while retaining frequency 30, its density becomes 3. That change shows why an equal-width shortcut cannot be carried into an unequal-width histogram without adjustment.
Build a short checking routine
Check boundaries, widths, density calculations, axis labels and bar areas. Make sure the total recovered frequency matches the table where all classes are shown. Inspect whether your horizontal scale represents unequal widths proportionally. If a comparison seems surprising, calculate the two areas before concluding that the drawing is wrong.
Practise with fresh tables containing unequal widths and at least one non-integer density. Then try reading frequencies from a completed histogram. The AS Level S1 course, course catalogue and admissions FAQ provide routes to ask about current subject support. Use the AS and A2 study plan to connect this focused work with broader revision.
Frequently asked questions
Why do we divide frequency by class width?
The bar area must represent frequency. Since area equals width multiplied by height, dividing frequency by width gives the required height. This adjustment allows classes of different widths to be compared without confusing width with a larger count.
Is the tallest histogram bar always the most frequent class?
Not when class widths differ. Height shows frequency density, while area represents frequency. A wider, shorter bar can contain more observations. Compare each width multiplied by its height when the question asks about counts.
Can frequency density be a decimal?
Yes. Density is frequency divided by class width, so it need not be an integer. Check that multiplying the density by the width recovers the frequency and that you have read the vertical scale correctly.
Do I always adjust class boundaries by 0.5?
No. Boundaries depend on the stated intervals and recording convention. An interval already written as 10 ≤ x < 15 has boundaries 10 and 15. Use the context to interpret rounded inclusive-looking groups instead of applying 0.5 automatically.
Is a frequency estimated inside one class exact?
Usually the grouped table does not reveal exact positions within that class. A proportional estimate assumes a distribution within the interval, often uniform. Recognise that assumption and distinguish the estimate from a known count across complete classes.
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