Histograms with unequal class widths, cumulative frequency curves and scatter diagrams
are drawn by hand on both papers, and most of the marks are for where your pencil puts things — a point
half a square out, a bar at the frequency instead of the frequency density, a curve plotted at midpoints. The
arithmetic on this sheet is deliberately clean, because Paper 2 is non-calculator. Print it and work on paper.
In the print dialog choose A4 and Scale: 100% — not “Fit to page”.
Then work through it with a sharp pencil, a 30 cm ruler and an eraser,
exactly as you would in the exam. Check your answers on screen afterwards: the model answers below are
hidden when you print, so the sheet comes out blank to work on.
What you need before you start. A sharp HB pencil, a 30 cm ruler and an eraser. No compasses and no
protractor this time — and nothing on this sheet needs a calculator: every division has been chosen to come
out exactly, which is also how Cambridge sets the non-calculator paper.
The two marks this sheet exists to save. On a histogram, bar height is frequency density, not
frequency — frequency is the area of the bar. On a cumulative frequency curve, every point is
plotted at the upper class boundary, never the midpoint. Examiners’ reports name these two errors
every single year.
Section 1 · Histograms with unequal class widths (E9.7)
Frequency density = frequency ÷ class width. Height shows density; area shows
frequency. That one sentence is the whole topic.
1.1The table shows the masses of 75 parcels. Complete the class width and frequency density rows.[2]
Frequency density = frequency ÷ class width. Every division here is exact — if you get a messy decimal, the width is wrong.
mass m (kg)
0 < m ≤ 2
2 < m ≤ 3
3 < m ≤ 4
4 < m ≤ 6
6 < m ≤ 10
frequency
16
12
15
24
8
class width
2
freq. density
8
How the marks are awardedB2 for all four missing frequency densities correct (B1 for two or three). The widths themselves carry no marks, but a wrong width makes every density after it wrong — write them down anyway.
mass m (kg)
0 < m ≤ 2
2 < m ≤ 3
3 < m ≤ 4
4 < m ≤ 6
6 < m ≤ 10
frequency
16
12
15
24
8
class width
2
1
1
2
4
freq. density
8
12
15
12
2
Marking points
Widths: 2, 1, 1, 2, 4 — read them off the class limits, do not assume they are equal.
Sanity check: the largest frequency (24) does not have the largest density (15). That is the whole point of unequal widths.
Units of frequency density here are parcels per kg — per one unit of the variable.
1.2Draw a histogram for the data in 1.1 on the grid below.[3]
Plot heights equal to the frequency density column. Bars touch: each one runs from its lower class limit to its upper class limit.
How the marks are awardedB3 for all five bars with correct boundaries and correct heights (B2 for four bars, B1 for two or three). A histogram whose heights are the raw frequencies scores 0 of 3 — it is the single most common wrong answer, and the grid goes up to 18 precisely so that it is possible to draw.
Marking points
Bars at density 8, 12, 15, 12, 2, drawn with a ruler, no gaps between them.
Boundaries at 0, 2, 3, 4, 6 and 10 kg — the 6–10 bar is wide and low, even though 8 parcels is not the smallest frequency.
Check by area: first bar 2 × 8 = 16 parcels. Total area = 16 + 12 + 15 + 24 + 8 = 75 = the number of parcels. If your areas do not add to 75, a height is wrong.
Label the vertical axis frequency density — an unlabelled axis can cost the communication mark.
1.3The histogram shows the time 83 people spent on their phones one evening. Read the frequency density of each bar, then work out each frequency and check the total is 83.[3]
Reverse of 1.2: frequency = frequency density × class width. The minor gridlines are 0.2 apart on the density axis.
time t (min)
0 < t ≤ 20
20 < t ≤ 30
30 < t ≤ 35
35 < t ≤ 45
45 < t ≤ 60
freq. density
frequency
How the marks are awardedB1 for the densities read correctly off the grid. B2 for all five frequencies (B1 for three or four). The total is given as 83 so you can catch your own error — use it. In the exam this appears as “complete the frequency table” and the check row is up to you.
time t (min)
0 < t ≤ 20
20 < t ≤ 30
30 < t ≤ 35
35 < t ≤ 45
45 < t ≤ 60
freq. density
0.8
2.4
3.6
1.6
0.6
frequency
16
24
18
16
9
Marking points
Densities off the grid: 0.8, 2.4, 3.6, 1.6, 0.6 — each bar top sits exactly on a minor gridline.
16 + 24 + 18 + 16 + 9 = 83 ✓ — the check costs ten seconds and catches almost every slip.
The tallest bar (30–35 min) is not the most common class — the 20–30 class holds more people. Height is density, area is people.
Section 2 · Cumulative frequency (E9.6)
One data set carries the whole section: build the column, plot the curve, then read it.
2.1The heights of 60 tomato seedlings were measured. Complete the cumulative frequency column.[1]
Each entry is a running total: add the next frequency to the previous cumulative frequency. Never add the class widths.
height h (cm)
0 < h ≤ 10
10 < h ≤ 20
20 < h ≤ 30
30 < h ≤ 40
40 < h ≤ 50
50 < h ≤ 60
frequency
4
10
16
18
9
3
cum. freq.
4
14
How the marks are awardedB1 for all four missing values. The final value must equal 60, the number of seedlings — if it does not, find the slip now, before you plot anything.
height h (cm)
0 < h ≤ 10
10 < h ≤ 20
20 < h ≤ 30
30 < h ≤ 40
40 < h ≤ 50
50 < h ≤ 60
cum. freq.
4
14
30
48
57
60
Marking points
4, 14, 30, 48, 57, 60.
Each cumulative frequency answers “how many seedlings are this height or less”.
Last entry = 60 = total ✓. This check is free and the examiners’ reports beg for it.
2.2Plot the cumulative frequency curve for the data in 2.1 on the grid below.[3]
Plot each cumulative frequency at the upper end of its class — 4 at h = 10, 14 at h = 20, and so on — and start the curve at (0, 0).
How the marks are awardedP2 for all seven points plotted correctly, including (0, 0) (P1 for five or six). C1 for a single smooth increasing curve through them. Points plotted at the midpoints of the classes lose the plotting marks even if the curve through them is beautiful — the curve answers “how many are at most h”, and a class is only complete at its upper boundary.
Marking points
Points at (10, 4), (20, 14), (30, 30), (40, 48), (50, 57), (60, 60), plus the start (0, 0) because no seedling has height 0 or less.
Plotted at upper class boundaries, not midpoints — say it to yourself every time you place a point.
One smooth freehand curve, steepest in the middle where the frequencies are largest. It never goes down; if yours dips, a running total is wrong.
Small crosses or dots, not blobs — reading tolerance later is half a small square.
2.3Use your curve to estimate (a) the median, (b) the lower and upper quartiles, (c) the interquartile range, (d) the 90th percentile, (e) the number of seedlings taller than 45 cm.[5]
For 60 values read across from cf = 30 (median), 15 (LQ), 45 (UQ) and 54 (90th percentile). Draw the dashed reading lines — they carry credit.
How the marks are awardedB1 each for (a), (b) both values, (c) follow-through from your quartiles, (d), (e). Readings are accepted within half a small square of a correct curve, and dashed lines on the graph are the working — a bare number with nothing drawn is routinely refused the mark.
Marking points
(a) Median at cf = 30 → 30 cm exactly (the running total reaches 30 at the class boundary).
(b) LQ at cf = 15 → about 21 cm (accept 20–22). UQ at cf = 45 → about 38 cm (accept 37–39).
(c) IQR = UQ − LQ ≈ 38 − 21 = 17 cm (accept 15–19, and follow-through from your own quartiles).
(d) 90% of 60 = 54. Reading at cf = 54 → about 47 cm (accept 45–48).
(e) cf at 45 cm is about 52 or 53, so 60 − 52½ → about 7 or 8 seedlings (accept 6–9). Read below, then subtract — the curve counts “at most”, the question asks “more than”.
Section 3 · Scatter diagrams and the line of best fit (E9.5)
Plot, describe, draw one ruled line through the mean point, read from it — and know
where the line stops being trustworthy.
3.1A shop recorded the temperature and the number of cold drinks sold on ten days. The first six points are plotted. Plot the remaining four points.[2]
Small crosses, centred on the gridlines. The temperature axis starts at 10, not 0 — read it before you plot.
temperature (°C)
14
16
18
19
21
23
24
26
28
31
drinks sold
30
38
42
45
52
58
63
68
74
80
How the marks are awardedP2 for all four points within half a small square (P1 for two or three). Blobs the size of a pea lose plotting marks by themselves — the tolerance is half a square and a blob is bigger than that.
Marking points
The four points to add are (24, 63), (26, 68), (28, 74) and (31, 80) — shown on the model diagram in 3.3.
Check the axis scales first: 1 small square = 1 °C across, 2 drinks up. The broken temperature axis is a deliberate trap.
A miscounted square on one point loses that point only — plot all four, then re-check each against the table.
3.2Describe the correlation shown by the scatter diagram, and say what it means in this context.[1]
Two words for the type, one sentence for the meaning.
How the marks are awardedB1 for “strong positive” (or just “positive”) with the meaning in context. “The points go up” or “temperature and drinks are correlated” with no direction does not score.
Marking points
Strong positive correlation — the points lie close to a straight line that slopes upward.
In context: on warmer days, more cold drinks are sold.
Correlation is not causation, and Cambridge does occasionally ask: the diagram shows the two rise together, not that one causes the other.
3.3Work out the mean point and plot it. Draw a ruled line of best fit through the mean point, and use your line to estimate the number of drinks sold on a day when the temperature is 25 °C.[3]
Both means are exact: the temperatures sum to 220 and the drinks to 550. Then one ruled line through (22, 55) with the points balanced either side.
How the marks are awardedB1 for the mean point (22, 55) plotted. B1 for a single ruled line through the mean point passing through the cloud with balance — about as many points above the line as below it, all the way along. B1 for the reading at 25 °C with dashed lines drawn. A line forced through the origin, or a freehand line, loses the line mark however good the reading is.
Marking points
Mean temperature = 220 ÷ 10 = 22. Mean drinks = 550 ÷ 10 = 55. Plot (22, 55) and put your ruler on it.
One ruled line, drawn right across the data, roughly five points each side, through the mean point. Not joined dot-to-dot, not several attempts.
Reading at 25 °C → about 64 drinks (accept roughly 60–68 from a sensible line). Dashed lines up from 25 and across — that is the working and it carries the mark.
Any answer from your own correct line is accepted — the examiner checks your line, not a hidden official one.
3.4The shop owner uses the line to predict sales on a day when the temperature is 40 °C. Explain why this prediction is unreliable.[1]
One sentence. Name the range of the data.
How the marks are awardedB1 for saying 40 °C is outside the range of the data (14–31 °C), so there is no evidence the pattern continues. “The line might be wrong” or “it is too hot” scores 0 — the mark is for the word extrapolation or its meaning.
Marking points
40 °C is beyond the largest temperature recorded (31 °C). Using the line there is extrapolation: the data gives no evidence that the straight-line pattern continues.
A good answer names the data range: “the data only covers 14–31 °C, so predicting at 40 °C is outside the data and unreliable.”
Interpolation (reading inside the range, as at 25 °C) is fine; extrapolation is the one to flag.
How examiners award the marks on this sheet
Read this before you start, and again before you hand the paper in.
Histogram height is frequency density; area is frequency. Divide by the class width before you draw,
and check afterwards that the areas add to the stated total.
Cumulative frequency is plotted at the upper class boundary, never the midpoint. The curve answers
“how many are at most this value”, and a class is only complete at its upper end.
A cumulative frequency curve never goes down. If yours dips, a running total is wrong — and the
last cumulative frequency must equal the total.
Rule every straight line. Bars and the line of best fit. Freehand straight lines
lose accuracy marks even when the reading taken from them is right.
The line of best fit passes through the cloud with balance — about as many points above as below,
all the way along — and through the mean point when you have been asked to find it. Never force it through
the origin.
Show where you read a value from. Dashed lines to the axes on a cumulative frequency curve or a
scatter diagram carry marks of their own and cost ten seconds.
Plot with a small cross, not a blob. The tolerance is half a small square; a blob is bigger than the
tolerance.
Reading inside the data is estimation; reading beyond it is extrapolation — and extrapolation is
the standard one-mark “explain why this is unreliable” question.
Cambridge IGCSE Mathematics 0580 Extended · by-hand skills sheet 4 · 10 exercises ·
E9.7 histograms, E9.6 cumulative frequency, E9.5 scatter diagrams ·
Print at 100% on A4 and complete in pencil.