5.1 - Cumulative frequency and quartiles

Cumulative frequency

If $30$ people record their waiting times, how can you estimate the fraction who wait between $15$ and $25$ minutes without knowing each time? A running total turns that question into the difference between two graph readings.

Suppose $4$ waits are below $10$ minutes and a further $12$ are between $10$ and $20$. Then $4+12=16$ waits are below $20$, not $12$.

Label both axes and use linear numerical scales. The graph cannot decrease: increasing the boundary cannot remove observations already counted. Join plotted points with an increasing curve when requested. Joining them with straight line segments gives the linear interpolation model used in this lesson: it assumes values are spread evenly within each class.

The dashed reading at $15$ minutes will help answer the opening question.

[DIAGRAM: asset_name: Boundaries and running totals; asset_slug: data-cumulative-boundaries; caption: Plot cumulative frequencies at upper boundaries and begin at zero at the lowest boundary; description: Increasing piecewise linear cumulative frequency graph of waiting time with labelled points 0 comma 0, 10 comma 4, 20 comma 16, 30 comma 24, 50 comma 30. Dashed lines read across and down from the point 15 comma 10 to show an estimated count of 10 below 15 minutes. Both axes use numbered linear scales.]

Now keep the same running-total method but change how the measurements are recorded. Whole-number class labels for rounded measurements need continuous boundaries.

If a table instead says $10\le x<20$, those are already its boundaries; do not subtract $0.5$. Nor should you introduce half-unit boundaries for an exact discrete count unless a continuous grouped model is actually being used.

To read a count below a given value, go vertically from the value to the graph, then horizontally to the cumulative-frequency axis. In Example 1, linear interpolation gives

Here $F(t)$ denotes the estimated count below $t$, rather than a proportion. Hence the estimated fraction between $15$ and $25$ minutes is $(20-10)/30=1/3$. Above $25$ minutes, subtract from the total: $30-20=10$ waits. Boundary inclusion does not change these continuous-model estimates; exact discrete counts require their stated inequalities.

Convert the next rounded table before choosing the plotting positions.

Medians, quartiles and percentiles

A percentile answers the reverse question: what value has a specified percentage of the observations below it? Instead of starting on the horizontal axis, start with a cumulative count.

For raw data, sort first. The median is the middle observation if the count is odd, or the mean of the two middle observations if it is even. To find raw-data quartiles here, take the median of the lower half and of the upper half; exclude the overall middle observation when the count is odd. For example, in $2,4,5,7,9,11,13,16,20$, the median is $9$, while the halves give $Q_1=(4+5)/2=4.5$ and $Q_3=(13+16)/2=14.5$. Follow a convention explicitly stated in a question if one is supplied.

The next example changes the information available: only grouped counts are known. Use cumulative levels $n/4,n/2,3n/4$, or $pn/100$ for a percentile. Read across to the graph and down to the value axis. These are distribution estimates, not a rule telling you to round raw-data positions.

For straight-line interpolation within a class, the fraction of the class frequency needed is also the fraction of the class width needed.