- To describe a distribution from a histogram, always comment on four things, in context: shape, centre, spread and unusual features (outliers, gaps, clusters).
- A histogram groups a quantitative variable into classes and draws a bar for each class. When class widths differ, the height must be frequency density (frequency ÷ class width), so that area shows frequency.
- In our case, 40 students’ commute times are unimodal and skewed to the right, centred at a median of 21.5 minutes, with a range of 88 minutes, a gap between 80 and 90 minutes and one very long commute of 92 minutes.
- Skewness tells you which centre to report: the mean (26.95) is pulled into the long right tail, so the median is the better “typical” value here.
- Exam point: “describe the distribution” answers lose marks for missing context (units and variable name) or for describing only shape.
The case: how long do students take to get to college?
A sixth-form college is reviewing its start time. Before it changes the timetable, it wants a clear picture of how long students spend travelling in each morning. A statistics class surveys a random sample of 40 students and records each one’s door-to-door commute time in minutes. Here are the 40 times, sorted from shortest to longest:
4, 6, 7, 8, 9, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 18, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 30, 32, 34, 36, 38, 41, 44, 47, 52, 58, 63, 71, 92.
A list of 40 numbers is hard to read. The principal wants to know: what is a typical commute? How much do commutes vary? Are there a few students with very long journeys who might need special arrangements? The class decides to draw a histogram and then describe what it shows, the way an examiner would expect. (The figures are illustrative.)
The concept: histograms and the language of distributions
The distribution of a variable tells us what values it takes and how often it takes them. For a quantitative variable with many different values, the standard picture is a histogram. We split the number line into non-overlapping classes (also called bins or intervals), count how many observations fall into each class, and draw touching bars whose sizes show those counts. The bars touch because the variable is continuous: there is no gap between “10 to under 20” and “20 to under 30”.
A histogram is different from a bar chart. A bar chart shows a categorical variable, the order of the bars is often arbitrary, and the bars are separated. In a histogram the horizontal axis is a number line, the order is fixed, and the bars touch.
Once the histogram is drawn, you describe it with four headings. Many teachers remember them with the phrase “SOCS” or “shape, outliers, centre, spread”.
Shape. Is there one peak (unimodal), two (bimodal) or none (uniform)? Is it roughly symmetric, or does one tail stretch further than the other? If the long tail is on the right (towards larger values), the distribution is skewed to the right, or positively skewed. If the long tail is on the left, it is skewed to the left, or negatively skewed.
Centre. Where is the middle of the data? Report the median or the mean, and say which.
Spread. How variable are the values? Report the range, the interquartile range (IQR) or the standard deviation.
Unusual features. Look for outliers (individual values far from the rest), gaps (classes with no data inside the range) and clusters (separate groups of values).
When every class has the same width, the bar height can simply be the frequency. When the class widths differ, a wide class would look too important if we used frequency as height. The fix is to plot frequency density:
Here frequency is the number of observations in the class, and class width is the upper boundary minus the lower boundary (for “20 to under 40”, the width is 20). The second form says that the area of each bar equals the frequency in that class. That is the key property of a correctly drawn histogram: area, not height, represents how many observations there are. With equal widths, height and area tell the same story, which is why many histograms simply use frequency on the vertical axis.
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Why this method fits the case
Commute time is a continuous quantitative variable, measured in minutes, with almost as many different values as there are students. That is exactly the situation a histogram is built for. Grouping into classes hides the tiny differences (is 17 minutes really different from 18?) and reveals the overall pattern: where most students are, how far the tail stretches, and whether anyone stands apart.
A histogram is also honest about shape in a way a single number is not. If the principal were told only “the mean commute is 27 minutes”, she might picture most students taking about half an hour. The histogram shows that most actually take 10 to 30 minutes and a few take much longer[2].
When is a histogram not the best tool? With very small data sets (say fewer than 15 values), the picture depends heavily on the choice of classes; a dotplot or stemplot keeps every value and is clearer. If you want to compare several groups side by side, box plots are more compact. If you need the exact values, a stemplot keeps them. And for a categorical variable, such as mode of transport, use a bar chart instead. Finally, a histogram can hide or invent features if the class width is badly chosen. Too few classes smooth away a gap; too many classes create jagged peaks that are just noise. It is good practice to try more than one class width before you describe the shape.
Setting it up
We choose classes of width 10 minutes, starting at 0. Each class includes its lower boundary but not its upper one, so a commute of exactly 20 minutes goes in “20 to under 30”.
| Commute time \( t \) (minutes) | Frequency | Cumulative frequency |
|---|---|---|
| \( 0 \le t < 10 \) | 5 | 5 |
| \( 10 \le t < 20 \) | 13 | 18 |
| \( 20 \le t < 30 \) | 9 | 27 |
| \( 30 \le t < 40 \) | 5 | 32 |
| \( 40 \le t < 50 \) | 3 | 35 |
| \( 50 \le t < 60 \) | 2 | 37 |
| \( 60 \le t < 70 \) | 1 | 38 |
| \( 70 \le t < 80 \) | 1 | 39 |
| \( 80 \le t < 90 \) | 0 | 39 |
| \( 90 \le t < 100 \) | 1 | 40 |
From the raw data: the sum of the 40 times is 1,078 minutes, so the mean is \( 1078/40 = 26.95 \) minutes. The median is the average of the 20th and 21st values, \( (21+22)/2 = 21.5 \) minutes. The minimum is 4 and the maximum is 92, so the range is 88 minutes. The lower quartile (median of the bottom 20 values) is 13.5 and the upper quartile (median of the top 20) is 35, so the IQR is \( 35 – 13.5 = 21.5 \) minutes. The standard deviation is about 19.3 minutes.
Solving it
Example 1 (describe the distribution). Using the histogram in Figure 1 and the summary statistics above, describe the distribution of commute times.
A typical student commutes for about 20 minutes, but a long right tail means a few students travel for an hour or more.
Example 2 (unequal class widths). The college regroups the same data into wider classes at the top: 0–10 (5 students), 10–20 (13), 20–30 (9), 30–50 (8) and 50–100 (5). Find the frequency density for each class, and explain what goes wrong if frequency is used as the height.
With unequal classes, plot frequency density so that the area of each bar, not its height, shows the number of students.
Example 3 (how skew affects mean and median). Remove the 92-minute commute and recompute the mean and median. What does the change tell you?
The mean is sensitive to extreme values in the tail; the median is resistant, which is why it is the better centre for a skewed distribution.
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What the answer tells us
For the college, the message is clear. A typical student commutes for about 20 minutes, and half of all students arrive within 14 to 35 minutes. Moving the start time by 10 minutes would make little difference to most students. But the long right tail matters too: 8 of the 40 students (20%) travel for 40 minutes or more, and one travels for over an hour and a half. Those students may need a different solution, such as flexible first lessons, rather than a change for everyone.
Reasonableness check. Does right skew make sense? Yes. Commute times cannot go below zero, and in practice few students live right next to the college, so there is a “floor” on the left. There is no hard ceiling on the right: a student who lives far away or takes two buses can take a very long time. Variables with a floor and no ceiling, such as income, waiting times and house prices, are very often skewed to the right.
The positions of the mean and median confirm the shape. In a right-skewed distribution the mean is usually larger than the median, because the few large values in the tail pull the mean towards them. Here the mean is 26.95 and the median is 21.5. In a roughly symmetric distribution they would be close together.
Sensitivity to class width. If we redraw the histogram with 5-minute classes, the frequencies become 1, 4, 6, 7, 5, 4, 3, 2, 2, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0. The peak moves to 15 to 20 minutes and the picture is more jagged, but the overall story is the same: one peak near 15 to 20 minutes, a long right tail and an isolated value above 90. When the main features survive a change of class width, you can trust them[1].
Is 92 really an outlier? A histogram can only flag values that look unusual. A formal check uses the 1.5 × IQR rule: the upper fence is \( 35 + 1.5 \times 21.5 = 67.25 \) minutes, so both 71 and 92 would be flagged. That rule, and how to show it on a box plot, belongs with the five-number summary.
Our Statistics Made Simple textbook covers histograms, stemplots, dotplots and frequency density step by step, and the Statistics Made Simple Practice Questions book has hundreds of “describe the distribution” questions with model answers.
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Practice questions
Try each question before opening the worked answer. The last one is written in exam style with a mark scheme.
EasyQ1. A histogram of the ages at which people in a town retire has a peak at 60 to 65 and a long tail stretching to the left down to age 45. Describe the shape, and say whether the mean or the median is likely to be larger. Show worked answer
The few early retirements pull the mean down, so the mean is likely to be smaller than the median; the median is larger.
EasyQ2. In the commute data, what percentage of students take under 20 minutes? Show worked answer
Percentage \( = 18/40 \times 100 = 45\% \).
MediumQ3. Heights of seedlings (cm) are grouped as 0–5 (12 seedlings), 5–10 (18), 10–20 (20) and 20–40 (10). Calculate the frequency density for each class and identify the modal class. Show worked answer
Frequency densities: \( 12/5 = 2.4 \), \( 18/5 = 3.6 \), \( 20/10 = 2.0 \), \( 10/20 = 0.5 \) seedlings per cm.
The modal class is the one with the greatest frequency density: 5–10 cm (even though 10–20 has the largest frequency).
Check: total \( = 12+18+20+10 = 60 \) seedlings.
MediumQ4. Phone battery life (hours) for 40 phones: 2–4 (6 phones), 4–6 (14), 6–8 (16), 8–10 (4). Estimate the mean using class midpoints, and explain why it is only an estimate. Show worked answer
\( \sum f x = 6\times3 + 14\times5 + 16\times7 + 4\times9 = 18+70+112+36 = 236 \).
Estimated mean \( = 236/40 = 5.9 \) hours.
It is an estimate because we assume every value in a class sits at the midpoint; the raw data are not known.
ExamQ5. Waiting times (minutes) at a clinic for 50 patients: 10–20 (4), 20–30 (18), 30–40 (14), 40–50 (8), 50–60 (4), 60–90 (2). (a) Calculate the frequency density of each class. (b) Estimate the median. (c) Describe the distribution in context. [7 marks] Show worked answer
(b) Median is the 25th value; cumulative frequencies 4, 22, 36, so it lies in 30–40. [1] Estimate \( 30 + \frac{25-22}{14}\times10 \approx 32.1 \) minutes. [1]
(c) The distribution of waiting times is unimodal with its peak in the 20–30 minute class, [1] and skewed to the right, with a tail stretching up to 90 minutes. [1] The median wait is about 32 minutes, and most patients (32 of 50) wait between 20 and 40 minutes. [1]
Frequently asked questions
How many classes should a histogram have?
There is no single correct number. For school-sized data sets, 5 to 12 classes usually work well. Choose round class widths (2, 5, 10, 20) so the boundaries are easy to read, and try at least two widths to make sure the shape you describe is real, not an artefact of the grouping.
What is the difference between a histogram and a bar chart?
A histogram shows a quantitative variable on a number line, so the bars touch and their order is fixed. A bar chart shows a categorical variable, the bars are separated and their order can be changed. With unequal class widths, a histogram uses area to show frequency; a bar chart always uses height.
Can a distribution have no outliers but still be skewed?
Yes. Skew is about the overall shape: one tail being longer than the other. Outliers are individual values far from the rest. A distribution can have a gently stretched right tail with no single value extreme enough to count as an outlier.
Which measure of spread should I use for a skewed distribution?
Pair the measure of spread with the measure of centre. For a skewed distribution or one with outliers, report the median with the IQR, which are both resistant to extreme values. For a roughly symmetric distribution without outliers, report the mean with the standard deviation.
References
- NIST/SEMATECH. e-Handbook of Statistical Methods, Exploratory Data Analysis: histogram. https://www.itl.nist.gov/div898/handbook/
- Illowsky, B. and Dean, S. (2023) Introductory Statistics 2e. Houston, TX: OpenStax. https://openstax.org/details/books/introductory-statistics-2e
- Starnes, D. S. and Tabor, J. The Practice of Statistics. New York: W. H. Freeman.
- Cambridge International. Cambridge International AS & A Level Mathematics 9709 syllabus 2026–2027. https://www.cambridgeinternational.org/programmes-and-qualifications/cambridge-international-as-and-a-level-mathematics-9709/
- College Board. AP Statistics Course and Exam Description. New York: College Board.
