The cumulative frequency curve · Dispersion & cumulative frequency

This is one of the most common Paper 2 questions. It has three jobs, and most candidates do the first well: the table, the curve and reading from the curve. The marks are lost in the second and third, so this lesson spends most of its time there.

Try it

Cumulative frequency curveFill in the running totals

marks	1–10	11–20	21–30	31–40	41–50	51–60	61–70	71–80
f	4	8	14	20	24	16	9	5
upper boundary	10.5	20.5	30.5	40.5	50.5	60.5	70.5	80.5
cum. f

Each cumulative frequency is the running total: this class's frequency added to all the ones before it. The last one must equal the total, 100.

Fill in the running totals (the cumulative frequencies), then draw the curve. Slide the percentage to read the median and quartiles, and switch to Count below a mark to see how many scored less than a mark.

Job 1: the cumulative frequency table

The cumulative frequency of a class is the number of values up to the end of that class: its frequency plus all the frequencies before it.

Marks	1–10	11–20	21–30	31–40	…
Frequency	4	8	14	20	…
Upper class boundary	10.5	20.5	30.5	40.5	…
Cumulative frequency	4	12	26	46	…

The last cumulative frequency must equal the total frequency. If it doesn’t, an addition has gone wrong.

For single scores rather than classes, the cumulative frequency of a score is the number of values less than or equal to it. With scores 1, 2, 3 having frequencies 4, 6, 5, the cumulative frequency of 2 is $4 + 6 = 10$ .

More: cumulative frequency of a score

Job 2: plotting and drawing

Horizontal axis: marks (or whatever the data is), using the upper class boundaries.
Vertical axis: cumulative frequency, from 0 to the total.
Plot each cumulative frequency at its upper class boundary: 4 at 10.5, 12 at 20.5, …. The running total is only complete at the end of the class.
Start the curve at 0, at the lower boundary of the first class (0.5 here): nobody scored less than that.
Join the points with a smooth S-shaped curve, drawn freehand.

Job 3: reading from the curve

Every reading works the same way. Find the right height on the cumulative frequency axis, go across to the curve, then down to the marks axis. (Or go up from a mark and across to read a count.) Draw the lines on your graph in pencil: they show the examiner your method.

With a total of $N$ :

To find	Go across from
median	$\frac12 N$
lower quartile $Q_1$	$\frac14 N$
upper quartile $Q_3$	$\frac34 N$
the $p$ th percentile	$\frac{p}{100} N$

Reading the curveAcross from ¼N, ½N and ¾N, then down

Counting: to find how many scored less than a mark, go up from that mark and across. To find how many scored more than it, take that reading away from $N$ .

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q10

Marks (%)	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Frequency	7	11	17	20	29	34	30	25	21	6

The table shows the distribution of marks obtained by students in an examination.

Construct a cumulative frequency table for the distribution.

Draw the cumulative frequency curve for the distribution.

Using the curve, find, correct to one decimal place, the: (i) median mark; (ii) lowest mark for distinction if $5\%$ of the students passed with distinction.

(a) The table
Upper class boundaries $9.5, 19.5, 29.5, \ldots, 99.5$ .
Cumulative frequencies: $7, 18, 35, 55, 84, 118, 148, 173, 194, 200$ . The last one is the total, 200 ✓.
Think first. What are the upper class boundaries? The running totals?
(b) The curve
Plot $(9.5, 7), (19.5, 18), \ldots, (99.5, 200)$ , starting from $(-0.5, 0)$ , and draw a smooth S-shaped curve through them.
(c)(i) The median
Half of 200 is 100. Go across from 100 to the curve, then down: about 54.5.
(As a check: 100 is between 84 at 49.5 and 118 at 59.5, and $49.5 + \frac{100 - 84}{34} \times 10 \approx 54.2$ . A reading from a hand-drawn curve will be close to this.)
Think first. Which cumulative frequency do you start from?
(c)(ii) Distinction
The top 5% are above it, so 95% are below it. $95\%$ of $200 = 190$ . Go across from 190: about 88.
Think first. If the top 5% got a distinction, what percentage are below the lowest distinction mark?

The median of grouped data without a graph

Treating the curve as straight between the two plotted points either side of $\frac12 N$ gives a formula. The median class is the class where the running total passes $\frac12 N$ . If $L$ is its lower class boundary, $F$ the cumulative frequency before it, $f$ its frequency and $c$ its width,

\text{median} = L + \frac{N \div 2 - F}{f} \times c

A straight piece of curveThe median is the same fraction of the way across as ½N is of the way up

More: the median of grouped data

More: percentiles and reading the curve

More: grouped frequency tables and histograms

Your turn

WAEC 2013 · Paper 2 · Q11

Marks (%)	1–10	11–20	21–30	31–40	41–50	51–60	61–70	71–80	81–90	91–100
Frequency	2	3	5	13	19	31	13	9	4	1

The frequency distribution table shows the marks obtained by 100 students in a Mathematics test.

(a)
Draw a cumulative frequency curve for the distribution.
Model answer
Plot each cumulative frequency against the upper class boundary ( $10.5, 20.5, \ldots, 100.5$ ), starting from $(0.5, 0)$ and ending at $(100.5, 100)$ , and join the points with a smooth S-shaped curve. Label both axes.

For (b): across from 60 the curve gives the 60th percentile, about 56; up from 34.5 it reads about 15, so about 85 of the 100 passed and the probability is about $0.85$ .
(b)
Use the graph to find the: (i) 60th percentile; (ii) probability that a student passed the test if the pass mark was fixed at $35\%$ .
Separate values with commas, e.g. 3, −2

Try it on a graph

The ogive with the 60th-percentile reading.

Worked solution (try it first)

(a)

Make the cumulative frequency table, with the upper class boundaries:

Marks	1–10	11–20	21–30	31–40	41–50	51–60	61–70	71–80	81–90	91–100
Upper boundary	10.5	20.5	30.5	40.5	50.5	60.5	70.5	80.5	90.5	100.5
Cumulative frequency	2	5	10	23	42	73	86	95	99	100

Plot each cumulative frequency at its upper boundary, starting from $(0.5, 0)$ , and join the points with a smooth S-shaped curve.

(b)(i)

The 60th percentile is at $\frac{60}{100} \times 100 = 60$ on the cumulative frequency axis.
Go across to the curve and down: about 56.
(As a check, 60 lies between 42 at 50.5 and 73 at 60.5: $50.5 + \frac{60 - 42}{31} \times 10 \approx 56.3$ .)

(ii)

A pass is 35 or more.
Go up from 34.5 (the boundary below 35) to the curve and across: about 15 students scored less than 35.
So about $100 - 15 = 85$ passed, and $P(\text{passed}) \approx \frac{85}{100}$
$= 0.85$ .
A reading close to this from your own curve is fine.

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