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Cambridge 0580 IGCSE  ·  Topic E9.6

E9.6 Cumulative Frequency Diagrams

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01 — Explanation

What does a cumulative frequency diagram show?

Cumulative frequency is a running total of frequencies, built up to each class's upper boundary. Plotted against the upper boundaries, it forms a smooth S-shaped curve (an ogive) that lets the median, quartiles and percentiles be read off directly, without needing the raw data.

02 — Explanation

How do I decide what to plot?

Plot each point at (upper class boundary,cumulative frequency)(\text{upper class boundary}, \text{cumulative frequency}), plus a starting point at the lower boundary of the first class with cumulative frequency 00.

Example: times of 8080 runners.

Time (tts)ffUpper boundaryCumulative ff
40<t≤5040<t\le5066505066
50<t≤6050<t\le60161660602222
60<t≤7060<t\le70343470705656
70<t≤8070<t\le80181880807474
80<t≤9080<t\le906690908080

Points to plot: (40,0),(50,6),(60,22),(70,56),(80,74),(90,80)(40,0),(50,6),(60,22),(70,56),(80,74),(90,80), joined with a smooth curve — never straight segments, since an ogive summarises a continuous accumulation.

Examiner Tip

Plot at the upper class boundary, never the mid-interval value — that habit is carried over incorrectly from estimating a mean, where mid-points are used instead.

03 — Explanation

How do I read the median and quartiles from the curve?

With N=80N=80 total: read across from y=12Ny=\frac12N, 14N\frac14N and 34N\frac34N to the curve, then down to the xx-axis.

Median at y=40≈65 s,Q1 at y=20≈59 s,Q3 at y=60≈71.5 s\text{Median at } y=40 \approx 65\text{ s}, \qquad Q_1 \text{ at } y=20 \approx 59\text{ s}, \qquad Q_3 \text{ at } y=60 \approx 71.5\text{ s}
IQR=Q3−Q1=71.5−59=12.5 s\text{IQR} = Q_3-Q_1 = 71.5-59 = 12.5\text{ s}

04 — Explanation

How do I estimate a percentile other than the median or quartiles?

Any percentile is read off using exactly the same "across then down" method, just starting from a different fraction of NN on the vertical axis. The ppth percentile is read from height y=p100×Ny=\dfrac{p}{100}\times N.

Example: using the same 8080 runners, estimate the 9090th percentile.

y=90100×80=72y = \frac{90}{100}\times80 = 72

Reading across from y=72y=72 to the curve and down to the xx-axis gives approximately 7878 seconds — meaning about 90%90\% of runners completed the race in under 7878 seconds.

05 — Explanation

How do I find a frequency above or below a given value?

Reading up from x=75x=75 to the curve gives about 6666 runners taking 7575 seconds or less. Since the curve only ever shows "up to" a value, the number taking longer than 7575 seconds needs a subtraction from the total:

80−66=14 runners80 - 66 = 14 \text{ runners}

The same subtraction idea extends to finding a range: the number of runners taking between two times is found by reading off the cumulative frequency at each boundary and subtracting one from the other, rather than trying to read a "band" directly off the curve.

Examiner Tip

Every reading taken off an ogive is an estimate, since the smooth curve only approximates the true accumulation between plotted points — answers read this way should never be presented as exact values, even when they come out as whole numbers.