Relative frequency histograms

When you reach for it

A histogram groups numerical data into bins of equal width and draws one bar per bin, as tall as the number of values in it; Making a histogram and choosing the bin width is where the bins are chosen. A count answers “how many?”. That is the right question for one group and the wrong one for two groups of different sizes.

Teodora measured the 40 earthworms she dug up in her back garden, and Melchior measured 250 from a large orchard. (All the data in this lesson is made up.) Drawn as counts, the orchard has more worms of every length simply because it has more worms. Its bars tower over the garden’s, and setting them side by side says nothing about whose worms tend to be longer.

Relative frequency tables met the same problem with categories and solved it by replacing each count with its share of the total. A relative frequency histogram does the same for a histogram: the same bins, but each bar’s height is the share of the data in that bin, written as a decimal or a percent. Two groups of any sizes then stand on one footing.

The procedure

Equal bins, and a rule for the edges. Exactly as for a count histogram. A value landing on the edge between two bars has to belong to one of them, so state a rule and keep it. In this lesson a value on an edge goes in the bar to its right: the bin “8 to 10 cm” holds lengths x with 8 \le x < 10, and a worm exactly 10 cm long is counted in the 10 to 12 bar. The last bar has no bar to its right, so it keeps a value sitting on its top edge: a worm of exactly 16 cm would go in the 14 to 16 bar.

Divide each count by the total. A bin holding f of the n values gets the height

Multiply by 100 for a percent.

The heights add to 1. Every value sits in exactly one bin, so the counts f_1, \dots, f_k add to n, and

As percents, 100\%. Once the shares are rounded the total can come out slightly off, such as 0.99 or 101\%. That is rounding, not an error; the exact shares still add to 1.

The shape does not change. Every count is divided by the same number n, so every bar shrinks by the same factor. A bar twice as tall as another is still twice as tall, the tallest bar is still the tallest, and an empty bin stays empty. Put a share scale beside the count scale and the bars are the very same bars; only the numbers up the side are new.

So for one group on its own, the relative histogram shows nothing new about shape. Its use is setting two groups side by side.

A share of a range is a sum of bars. The bins do not overlap, so the share of the data in several bins is the sum of their heights: “at least 10 cm” is the 10 to 12, 12 to 14 and 14 to 16 bars together. Only whole bars can be added, because a bar does not show how its values spread inside it.

Back to counts by multiplying. From \displaystyle \text{share} = \frac{f}{n} the count is f = \text{share} \times n. That needs n, which the bars do not show, so a relative histogram should state its total.

Shape, center and spread are read as before, since the picture is the same. Shape: peaks, symmetry and tails (Shape of a distribution, Symmetric and skewed distributions). Center: the median lies in the bar where the running total of heights, added from the left, passes 0.5; if the total reaches 0.5 exactly at an edge, the median is close to that edge (center and spread). Spread: how far the bars reach and how wide their bulk is (Range and interquartile range). Then say what it means for the thing measured (Describing a distribution).

Worked examples

Boundaries

• Equal bin widths. A bin twice as wide collects more values just by being wide, so its bar overstates how crowded that stretch is. Histograms with unequal bins use a different scale, density, where a bar’s area rather than its height is the share; that idea leads to Density curves: area, mean and median.
• Numerical data. For categories the shares go in a bar graph, as in Relative frequency tables.
• Same bins, same scale. Two relative histograms with different bins or different vertical axes cannot be read bar against bar.
• The total is hidden. Shares alone cannot give counts, so state n. A small total also makes shares jumpy: in a group of 5, one value is 0.20 of the data.
• Nothing inside a bar. A share of a range can be read only when the range starts and ends on bin edges.

Where this goes next

The running totals of Example 2, graphed, make a Cumulative frequency graphs and percentiles, and reading one backwards for the length below which a chosen share of the data lies is Percentiles. With more data and narrower bins, a histogram scaled so its total area is 1 smooths into a density curve. Box plots compare groups of different sizes by another route (Comparing box plots), and Comparing two distributions brings the comparisons together.

Exercises

Building the bars

1. Xanthe dug 25 worms from a flower bed. The bins 4 to 6, 6 to 8, 8 to 10, 10 to 12 and 12 to 14 cm held 1, 4, 10, 6 and 4 worms. Find each height as a decimal and check the total.
2. Tancredi’s playlist has 80 songs: 8, 20, 28, 16 and 8 of them in the bins 2 to 3, 3 to 4, 4 to 5, 5 to 6 and 6 to 7 minutes. Find each height as a percent.
3. Twenty song lengths, in minutes: 2.5, 3.0, 3.2, 3.4, 3.5, 3.8, 4.0, 4.0, 4.1, 4.3, 4.5, 4.6, 4.9, 5.0, 5.2, 5.5, 5.8, 6.0, 6.4, 7.1. Use bins 1 minute wide from 2 to 8, a value on an edge going in the bar to its right. Find the counts and the heights as decimals.
4. A bucket holds 12 worms: 5 in the bin 6 to 8 cm, 5 in 8 to 10 cm and 2 in 10 to 12 cm. Find the heights as whole-number percents, add them, and explain the total.

Reading shares

5. In Teodora’s garden (Example 1), what share of the worms are at least 12 cm long? What share are at least 6 but less than 12 cm?
6. In Melchior’s orchard (Example 2), what share are at least 12 cm? What share are less than 10 cm?
7. On Silvan’s playlist, what percent of songs last less than 5 minutes? What percent last at least 6 minutes?
8. Using only the heights from Exercise 3, find the share of songs lasting at least 4 but less than 6 minutes. Then check it against the list.

Back to counts

9. Using the shares, how many of Melchior’s 250 worms are at least 12 cm long?
10. The five bars of a relative frequency histogram of 50 songs have heights 0.08, 0.30, 0.26, h and 0.10. Find h and the count in each bin.
11. A count histogram and a relative frequency histogram of the same worms stand side by side. One bar reads 21 on the count scale and 0.15 on the share scale. How many worms are there?
12. On a percent histogram of 90 songs, one bar reads 17\%, rounded to the nearest whole percent. How many songs are in that bin?

Comparing groups

13. Compare Silvan’s playlist with Tancredi’s (Exercise 2). Which has the larger share of songs shorter than 4 minutes? Which has more such songs?
14. A claim about Example 2: “Long worms are more common in the orchard. It has 40 worms of at least 12 cm and the garden has only 10.” Is the claim right?

Explaining

15. Explain why turning a count histogram into a relative frequency histogram keeps the tallest bar the tallest, and keeps a bar twice as tall as another twice as tall.
16. Explain why count histograms of groups of 40 and 250 cannot be compared bar by bar. If both groups had 40 values, would counts be enough?
17. Teodora regroups her worms into the bins 4 to 8, 8 to 10 and 10 to 16 cm. Find the heights, and explain why this histogram misleads even though they add to 1.