Histograms and Percentiles

Representing measurements with a histogram

Geographical measurements such as rainfall, river depth and sediment size can be grouped into class intervals. A frequency table records how many observations fall within each interval. A histogram turns this table into a picture of the distribution: where observations are concentrated and how widely they are spread.

The horizontal axis shows the measured variable, including its units. Each bar extends from the lower to the upper boundary of its class. Adjacent classes have touching bars because they represent a continuous numerical scale, rather than separate categories.

For example, $10\leq x<20$ includes values from 10 up to, but not including, 20. The next class can begin at 20 without counting that value twice.

Making bar areas represent frequency

A histogram represents frequency through bar area. This matters especially when class widths differ: a wider class must not appear to contain more observations merely because its bar is wider.

The height needed is called frequency density:

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

Since a rectangle’s area is width multiplied by height:

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

Consider a group of television viewers. Six people watched for between 0 and 100 minutes, while another six watched for between 500 and 800 minutes. Their class widths are 100 and 300 minutes, so their frequency densities are $6\div100=0.06$ and $6\div300=0.02$. The second bar is three times as wide but one-third as high. Both have an area representing six people.

With equal class widths, frequency can be used on the vertical axis because all bar areas have the same proportional relationship to frequency. Using frequency density also works and makes the area calculation consistent across classes.

Drawing and recognising distributions

To draw a histogram, first calculate each class width and, if using a frequency-density axis, each density. Choose evenly spaced numerical scales that accommodate the full range and greatest height. Label both axes, then draw each bar across its actual class boundaries at the appropriate height.

For example, recorded worm lengths were grouped as follows:

To plot these data, mark the boundaries 15, 20, 30, 40 and 60 on a length axis. Draw touching bars with heights 1.2, 1.4, 2.6 and 0.1 on a frequency-density axis. The last bar is twice as wide as the preceding one, but much lower: its area represents only two worms.

The completed histogram reveals the shape of the distribution. A normal distribution is symmetrical and bell-shaped: the tallest bars are near the centre, with heights generally decreasing towards both ends. Most observations cluster around the centre, with relatively few very low or very high values. A histogram of sampled measurements may only approximately resemble this shape.

A skewed distribution is not symmetrical. In a positively skewed distribution, most observations are towards the lower end and a long tail extends towards higher values, to the right. In a negatively skewed distribution, most observations are towards the higher end and the long tail extends towards lower values, to the left.

When constructing either shape from a table, plot the given frequencies faithfully. Do not adjust the bars to make a smoother bell or a more obvious tail. With unequal widths, a tall bar indicates a high concentration of observations per unit of the measured variable, not necessarily the largest class frequency.

Percentiles divide the observations, not the horizontal scale

A percentile identifies a position within an ordered distribution. The 50th percentile is the median: it divides the observations into lower and upper halves. The 90th percentile has approximately 90% of observations below it and 10% above it.

For geographical measurements, a percentile keeps the units of the original variable. For example, the 90th percentile of a set of rainfall measurements is a rainfall amount in millimetres. It is not 90% of the greatest rainfall recorded. Percentiles help identify relatively low or high measurements within a dataset without relying only on its extremes.

On a histogram, frequencies are represented by areas. To find the $p$th percentile, find the point on the horizontal axis with $p\%$ of the total bar area to its left.

Estimating a percentile from a histogram

A running club recorded 400 m times for 36 members aged under 30. The histogram has four classes: 50–54 seconds, 54–58 seconds, 58–62 seconds and 62–70 seconds. Their frequency densities are 1, 4, 3 and 0.5 members per second.

Multiplying each width by its density gives frequencies of 4, 16, 12 and 4. Their total is 36. The cumulative frequencies are therefore 4, 20, 32 and 36.

To estimate the 50th percentile, find the value with half the observations below it: $36\div2=18$. The first class accounts for four observations; the first two account for 20. The required value therefore lies in the 54–58 second class.

After allowing for the first four observations, we need an area representing $18-4=14$ observations inside this class. Its height is 4, so the required horizontal distance is $14\div4=3.5$ seconds. Adding this distance to the lower boundary gives $54+3.5=57.5$ seconds. The estimated median time is therefore 57.5 seconds.

The same method applies to other percentiles. First calculate the target cumulative frequency:

$$\text{target cumulative frequency}=\frac{p}{100}\times N$$

Here, $p$ is the required percentile and $N$ is the total frequency. Locate the class containing this target, subtract the frequency before that class, then divide the remaining frequency by the class’s density and add the lower boundary.

For instance, the 75th percentile requires three-quarters of the histogram area to its left, rather than half. It will lie farther to the right than the median. The 25th percentile requires one-quarter of the area and will lie farther to the left. The spaces between these values need not be equal, especially in a skewed distribution.

These calculations give estimates because grouped data do not reveal each individual measurement. Taking a proportional part of a bar assumes observations are evenly spread within that class.

Construct a histogram

• Horizontal axis: measured variable and units; bars span actual class boundaries.
• Adjacent classes have touching bars.
• With unequal class widths, use frequency density on the vertical axis.

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

Recognise the shape

• Normal: approximately symmetrical and bell-shaped.
• Positive skew: long tail to the right, towards higher values.
• Negative skew: long tail to the left, towards lower values.

Estimate a percentile

The $p$th percentile has approximately $p\%$ of observations below it; the 50th percentile is the median.

1. Recover class frequencies from bar areas and total them to find $N$.
2. Calculate the target cumulative frequency: $pN\div100$.
3. Locate its class using cumulative frequencies.
4. Subtract the frequency before that class.
5. Divide the remaining frequency by the class’s density; add the lower class boundary.

The answer has the measured variable’s units. It is an estimate assuming an even spread within the class.