Frequency Density and Unequal-Width Histograms

Why unequal widths need frequency density

A histogram displays continuous numerical data grouped into class intervals. The horizontal axis shows the measured variable, such as time in minutes. Each bar extends from the lower boundary to the upper boundary of its class, so neighbouring classes have touching bars.

With unequal class widths, using frequency as the height would give wider classes disproportionately large bars. Instead, the area of each bar represents its frequency. A wide bar containing six observations must have the same area as a narrow bar containing six observations, so it must be shorter.

The height needed to make this work is called frequency density: the frequency per unit of class width. It describes how concentrated the observations are within the interval, rather than simply how many observations the interval contains.

Because a rectangle’s area is its width multiplied by its height:

$$\text{frequency}=\text{class width}\times\text{frequency density}$$

Rearranging gives:

$$\text{frequency density}=\frac{\text{frequency}}{\text{class width}}$$

Calculate widths and densities before drawing

Consider these grouped times spent watching a television channel on one day. The interval $100 \leq t < 200$ includes 100 minutes but excludes 200 minutes, which belongs in the next class.

Find each width by subtracting the lower boundary from the upper boundary. For example, the fourth class has width $500-300=200$ minutes, so its density is $16\div200=0.08$ observations per minute.

The first and last classes both contain six observations. Their densities are different because those observations are spread across intervals of different widths: $6\div100=0.06$, but $6\div300=0.02$.

Turn the table into a histogram

Choose a horizontal scale that accommodates all the class boundaries, from 0 to 800 minutes. The spacing must reflect numerical distance: the interval from 500 to 800 needs three times the width of the interval from 0 to 100.

Label the vertical axis Frequency density and choose a scale that accommodates the largest density, 0.36. Draw each rectangle across its complete class interval, with its height equal to the calculated density.

The last bar is three times as wide as the first but only one-third as high. Their areas are therefore equal, correctly representing six people in each class.

To complete a partly drawn histogram, use exactly the same method: calculate the missing class’s width and density, then draw the rectangle between its boundaries using the existing axis scales.

Recover frequencies from bar areas

To read a frequency, multiply the bar’s width by its density. The bar from 300 to 500 minutes has frequency $200\times0.08=16$, not 0.08.

Adding all the bar areas gives the total number of observations:

$$6+36+24+16+6=88$$

You can also estimate how many observations lie in part of a class. For example, to estimate how many people watched for more than 400 minutes, take the part from 400 to 500 and the whole final class:

$$(500-400)\times0.08+(800-500)\times0.02=8+6=14$$

This is about $\frac{14}{88}\times100\%=15.9\%$ of the group. It is an estimate because taking part of a bar assumes that observations are evenly distributed within that class. The grouped data do not tell us their exact positions.

Estimate the middle of a distribution

The median divides the histogram’s total area into equal halves. With 88 observations, locate the point with a frequency of $88\div2=44$ to its left.

The first two bars contain $6+36=42$ observations, so the median lies in the class from 200 to 300 minutes. We need an additional frequency of $44-42=2$ within that bar.

Its density is 0.24, so the required horizontal distance into the bar is $2\div0.24\approx8.3$ minutes. The estimated median is therefore $200+8.3=208.3$ minutes. Again, this assumes an even distribution within the class.

Compare distributions using evidence

When comparing histograms, check the class intervals and axis scales before comparing their shapes. Matching intervals and density scales allow direct visual comparison, but frequencies still come from areas. If the groups have different totals, use proportions rather than raw frequencies to compare how common a range of values is.

For example, suppose a second group of 88 people has frequencies of 18, 36, 18, 12 and 4 in the same five viewing-time classes. Its first two classes contain $18+36=54$ people, so its median lies between 100 and 200 minutes. The original group’s median lies between 200 and 300 minutes. This supports the conclusion that the second group typically spent less time watching the channel.

The second group also has fewer people watching for at least 300 minutes: $12+4=16$, compared with $16+6=22$ in the original group. Both groups have observations in the same lowest and highest classes, however, so these grouped data do not establish their exact ranges.