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Statistics · Histograms

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Interpreting histograms

Getting information back out of a histogram — a frequency is an area, so part of a class is the matching proportion of a bar's area. Why that rests on values being spread evenly within a class (and is therefore an estimate), and how to read off a total, a mean, the median class, a range that cuts across class boundaries, and a comparison of two groups.

Statistics · Histograms

Interpreting histograms

Getting information back out of a histogram — a frequency is an area, so part of a class is the matching proportion of a bar's area. Why that rests on values being spread evenly within a class (and is therefore an estimate), and how to read off a total, a mean, the median class, a range that cuts across class boundaries, and a comparison of two groups.

Why it works

Building a histogram turns frequencies into areas. Reading one turns areas back into frequencies. That is the whole of this page. The height of a bar is the frequency density, so

area of a bar  =  width×frequencywidth  =  frequency,\text{area of a bar} \;=\; \text{width} \times \frac{\text{frequency}}{\text{width}} \;=\; \text{frequency},

and every question below is secretly the same question: which area am I being asked for?

Here is the histogram we will read all the way through. It shows the times, in minutes, that 100100 people took to finish a task.10203040506070800.511.522.53Time (minutes)Frequency densityA whole bar is the easy case: 30≤t<5030 \le t < 50 is 2020 wide and 1.01.0 tall, so it holds 20×1.0=2020 \times 1.0 = 20 people. Do that for all five bars and you get 1515, 3030, 2020, 2020, 1515; adding those gives 100100. The total frequency is the total area — there is no other way to get the total off a histogram, because the heights on their own (1.5+3.0+2.0+1.0+0.5=81.5 + 3.0 + 2.0 + 1.0 + 0.5 = 8) count nobody.

Part of a bar is where the thinking is. How many people took between 2525 and 3030 minutes? The histogram does not record that. It was drawn from a grouped table, and grouping threw away where inside 20≤t<3020 \le t < 30 each person actually landed. All that survived is "2020 people, somewhere in there".

So you make the one assumption the picture itself suggests. The bar is a rectangle with a flat top: the same frequency density all the way across the class. Constant density means the values are spread evenly through the class. If they are, half the class width holds half the people, a quarter holds a quarter, and in general

frequency in part of a class  =  width of the partwidth of the class×frequency of the class.\text{frequency in part of a class} \;=\; \frac{\text{width of the part}}{\text{width of the class}} \times \text{frequency of the class}.

For 2525 to 3030 that is 510×20=10\frac{5}{10} \times 20 = 10 people — which is just the area of that slice of the bar, 5×2.0=105 \times 2.0 = 10. The same calculation said two ways.

Why the answer is only an estimate, made concrete. Suppose the truth was that all 2020 of those people finished between 2020 and 2222 minutes. The grouped table would still say "20≤t<3020 \le t < 30: 2020 people", the histogram would look identical, and yet the true number between 2525 and 3030 would be 00, not 1010. Nothing in the diagram can tell those two worlds apart. That is why every answer here is an estimate, and why "assuming the times are spread evenly throughout each class" is the sentence an examiner wants. It is not a polite hedge — it is the assumption doing all the work.

When the question's range doesn't line up with the classes. "How many took between 2525 and 4545 minutes?" That range cuts through 20≤t<3020 \le t < 30 and through 30≤t<5030 \le t < 50. Chop it at every class boundary inside it and do each piece on its own:
  • 2525 to 3030 is 55 of the 1010 wide class: 510×20=10\frac{5}{10} \times 20 = 10.
  • 3030 to 4545 is 1515 of the 2020 wide class: 1520×20=15\frac{15}{20} \times 20 = 15.
Estimate =10+15=25= 10 + 15 = 25 people. The tempting wrong move is to count whole bars — "2525 to 4545 touches the 2020–3030 bar and the 3030–5050 bar, so 20+20=4020 + 20 = 40" — which quietly counts everyone from 2020 to 2525 and everyone from 4545 to 5050, none of whom belong.

An estimate for the mean. You never see the individual times, so each class is represented by its midpoint — and that is the even-spread assumption again, because evenly spread values average out at the middle. Recover each frequency from its area, multiply by the midpoint, add, and divide by the total:

mean  ≈  ∑(midpoint×frequency)∑frequency.\text{mean} \;\approx\; \frac{\sum (\text{midpoint} \times \text{frequency})}{\sum \text{frequency}}.

Both of those must be frequencies, not densities: what you multiply the midpoints by, and what you divide by. Dividing by 55 because there are five bars is the classic wreck — a mean is per person, not per class.

The median. The median is the middle value once everything is in order, so build up running totals from the left until you pass half the total. Half of 100100 is 5050; the running totals are 1515, 4545, 6565, so the 5050th person sits in 20≤t<3020 \le t < 30 — that is the median class. Notice it is not the tallest bar: 10≤t<2010 \le t < 20 is taller, but height alone says nothing about where the middle of the data lies. If you want a value rather than a class, run the even-spread idea backwards: you need 50−45=550 - 45 = 5 more people out of the 2020 in that class, so travel 520\frac{5}{20} of the way across a class of width 1010, giving 20+2.5=22.520 + 2.5 = 22.5 minutes.

Comparing two groups. Two histograms drawn for groups of different sizes cannot be compared by raw counts: 2020 out of 5050 is a bigger share (40%40\%) than 3030 out of 100100 (30%30\%), even though 20<3020 < 30. Turn each count into a proportion of its own total first. And before you compare bar heights by eye, check the two frequency-density scales actually match — two histograms can look alike and mean very different things.