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How do you display grouped continuous data with frequency polygons, histograms using frequency density, and cumulative frequency curves?

Construct and interpret frequency polygons, histograms with equal and unequal class widths using frequency density, and cumulative frequency curves, and read the median and quartiles from a cumulative frequency curve.

A CCEA GCSE Statistics answer on representing grouped continuous data: frequency polygons, histograms with frequency density and unequal class widths, cumulative frequency tables and curves, and reading the median and quartiles.

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  1. What this dot point is asking
  2. Frequency polygons
  3. Histograms and frequency density
  4. Cumulative frequency
  5. Why this matters

What this dot point is asking

Grouped continuous data needs its own displays. CCEA expects you to draw and read frequency polygons, histograms (including with unequal class widths using frequency density), and cumulative frequency curves, and to estimate the median and quartiles from a cumulative frequency curve. The histogram with unequal widths and the cumulative frequency curve are among the most demanding representation skills in the course, and both appear regularly at Higher tier.

Frequency polygons

A frequency polygon shows the shape of grouped data by plotting the frequency against the midpoint of each class and joining the points with straight lines.

It is especially useful for comparing two distributions on the same axes, because two polygons overlaid show which group is higher or more spread out without the clutter of two sets of bars. The points are plotted at class midpoints, not boundaries, and the lines are not extended beyond the data.

Histograms and frequency density

A histogram looks like a bar chart but is for continuous grouped data, so the bars touch. The crucial idea is that area, not height, represents frequency.

To find a missing frequency from a histogram, reverse the process: read the frequency density (height), multiply by the class width, and you have the frequency.

Cumulative frequency

Cumulative frequency is a running total of the frequencies, used to estimate the median and quartiles for grouped data.

To read the curve, use the cumulative frequency (vertical) axis: the median is at n2\tfrac{n}{2}, the lower quartile Q1Q_1 at n4\tfrac{n}{4}, and the upper quartile Q3Q_3 at 3n4\tfrac{3n}{4}. Read across to the curve and down to the value axis. The interquartile range is Q3Q1Q_3 - Q_1, the spread of the middle half of the data, and these are all estimates because the exact values within each class are unknown.

Why this matters

These displays are the gateway to the spread topic: the cumulative frequency curve feeds straight into quartiles, the interquartile range and box plots, while the histogram links to frequency density and the shape of a distribution, which underlies the normal distribution at Higher tier. Reading a missing frequency from a histogram, and reading quartiles from a curve, are reliable sources of marks once the area-equals-frequency idea is secure. Misreading a histogram as a bar chart is one of the most common and costly errors in the whole subject.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA-style4 marksA histogram class is 10<t3010 < t \le 30 with frequency 24, and another is 30<t4030 < t \le 40 with frequency 18. Work out the frequency density of each class.
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Frequency density =frequencyclass width= \dfrac{\text{frequency}}{\text{class width}}.

First class: width =3010=20= 30 - 10 = 20, so density =2420=1.2= \dfrac{24}{20} = 1.2. Two marks (one for the width, one for the density).

Second class: width =4030=10= 40 - 30 = 10, so density =1810=1.8= \dfrac{18}{10} = 1.8. Two marks. On a histogram it is the area of each bar that equals the frequency, so unequal widths must use frequency density on the vertical axis, never raw frequency.

CCEA-style4 marksA cumulative frequency curve is drawn for 80 runners' times. Describe how to estimate the median and the interquartile range from it.
Show worked answer →

Median: go to n2=802=40\dfrac{n}{2} = \dfrac{80}{2} = 40 on the cumulative frequency axis, read across to the curve and down to the time axis. One mark for 4040, one for the method.

Quartiles: read across at n4=20\dfrac{n}{4} = 20 for the lower quartile Q1Q_1 and at 3n4=60\dfrac{3n}{4} = 60 for the upper quartile Q3Q_3, each time down to the time axis. One mark.

Interquartile range =Q3Q1= Q_3 - Q_1. One mark. Cumulative frequency is plotted against the upper class boundary, and these readings give estimates because the original values within each class are unknown.

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