Cumulative Frequency Graphs

AS · S1 · 13 min

A cumulative frequency graph answers the question "how many values are less than this?" for every value at once. That makes it the tool for estimating the median, the quartiles and any percentile of a large grouped data set, and for finding how many values lie above, below or between given values. The syllabus asks you to draw and interpret these graphs, and to use them "to estimate medians, quartiles, percentiles, the proportion of a distribution above (or below) a given value, or between two values". Expect a full question: build the table, draw the graph, read off estimates, then often draw a box plot or compare with another data set.

Cumulative frequency

Definition

The cumulative frequency at a value $x$ is the number of data values less than (or equal to) $x$ . For grouped data it is found at each upper class boundary by adding up the frequencies of all classes up to and including that class.

A cumulative frequency table adds a running total to the frequency table. The key idea is where each running total belongs: when you have counted all the values in the class $20 \le t < 30$ , you know how many are less than $30$ . So each cumulative frequency is plotted at the upper class boundary, never at the mid-point.

Drawing the graph

Method

Find the upper class boundary of each class (see Histograms for boundaries of rounded, discrete and age data).
Add up the frequencies to get the cumulative frequency at each upper boundary. The last one must equal the total $n$ .
Draw axes: the variable (with units) across, cumulative frequency up, from $0$ to $n$ .
Plot the point (lower boundary of the first class, $0$ ): no values are below the start of the first class.
Plot (upper boundary, cumulative frequency) for each class.
Join the points with a smooth curve or with straight line segments. Do not extend the graph beyond the last point.

Straight segments correspond to assuming the values are spread evenly within each class, the same assumption used for histograms. A smooth curve usually gives readings very close to this. Either is accepted unless the question says otherwise. The worked answers below use straight segments (linear interpolation), so they can be checked exactly; readings from your own smooth curve may differ slightly and are accepted within a tolerance.

Reading the graph

For large grouped data sets with $n$ values, read across from these cumulative frequencies.

Key result

Quantity	Read across at cumulative frequency
Median	$\tfrac{1}{2} n$
Lower quartile $Q_1$	$\tfrac{1}{4} n$
Upper quartile $Q_3$	$\tfrac{3}{4} n$
$k$ th percentile	$\tfrac{k}{100} n$

Number below a value $x$ : read up from $x$ to the curve, then across.
Number above $x$ : $n$ minus the number below.
Number between $a$ and $b$ : (number below $b$ ) $-$ (number below $a$ ).

For grouped data you use $\tfrac{n}{2}$ , not the $\tfrac{n+1}{2}$ used for raw data: the graph is a continuous model of the data, and the half-way point of the total is what you want. Every value read from the graph is an estimate.

Here are the journey times of $120$ people.

Time $t$ (min)	$0 \le t < 10$	$10 \le t < 20$	$20 \le t < 30$	$30 \le t < 40$	$40 \le t < 60$	$60 \le t < 80$
Frequency	$6$	$18$	$36$	$32$	$20$	$8$

Time less than (min)	$0$	$10$	$20$	$30$	$40$	$60$	$80$
Cumulative frequency	$0$	$6$	$24$	$60$	$92$	$112$	$120$

Points at upper class boundaries joined by straight segments. Dashed lines read across at 30, 60 and 90 (a quarter, half and three quarters of 120).

Calculating instead of reading: linear interpolation

When a question gives a cumulative frequency table but asks for an estimate "by calculation", or when you want to check a graph reading, use linear interpolation: assume the values in a class are spread evenly.

Method

To estimate the value at cumulative frequency $c$ :

Find the class whose cumulative frequencies straddle $c$ : say it runs from boundary $L$ to boundary $U$ , with cumulative frequency $F$ before it and frequency $f$ in it.
The fraction of the way through the class is $\dfrac{c - F}{f}$ .
Estimate $= L + \dfrac{c - F}{f} \times (U - L)$ .

For the journey times, $Q_1$ is at $c = 30$ . That lies in $20 \le t < 30$ , with $F = 24$ and $f = 36$ :

Q_1 \approx 20 + \frac{30 - 24}{36} \times 10 = 21.7\ \text{minutes}.

Worked examples

Median and quartiles from the graph (routine)

Use the journey-time graph above to estimate the median and the interquartile range.

Solution

$n = 120$ . Read across at $60$ for the median, $30$ for $Q_1$ and $90$ for $Q_3$ .

Median: cumulative frequency $60$ is reached exactly at $t = 30$ . Median $\approx 30$ minutes.
$Q_1$ : $20 + \tfrac{30 - 24}{36} \times 10 = 21.7$ minutes.
$Q_3$ : $90$ lies in $30 \le t < 40$ , $F = 60$ , $f = 32$ : $30 + \tfrac{90 - 60}{32} \times 10 = 39.4$ minutes.

IQR $\approx 39.4 - 21.7 = 17.7$ minutes. (Readings from a smooth curve between about $17$ and $19$ minutes would be accepted.)

Above, between and percentiles

For the same $120$ journey times, estimate:

(a) the number of people whose journey took more than $50$ minutes; (b) the $90$ th percentile; (c) the percentage of people whose journey took between $15$ and $35$ minutes.

Solution

(a) At $t = 50$ : halfway through $40 \le t < 60$ , so cumulative frequency $\approx 92 + \tfrac{10}{20} \times 20 = 102$ . More than $50$ minutes: $120 - 102 = 18$ people.

(b) $90\%$ of $120$ is $108$ . This lies in $40 \le t < 60$ : $40 + \tfrac{108 - 92}{20} \times 20 = 56$ minutes.

(c) At $t = 15$ : $6 + \tfrac{5}{10} \times 18 = 15$ . At $t = 35$ : $60 + \tfrac{5}{10} \times 32 = 76$ . Between: $76 - 15 = 61$ people, which is $\tfrac{61}{120} \times 100 = 50.8\%$ , about $51\%$ .

Discrete data and grade boundaries (exam style)

The marks of $200$ students in an exam (whole numbers from $0$ to $100$ ) are grouped.

Mark	$0$ – $19$	$20$ – $39$	$40$ – $59$	$60$ – $79$	$80$ – $100$
Frequency	$14$	$36$	$70$	$56$	$24$

(a) State the coordinates of the points you would plot for a cumulative frequency graph. (b) The top $20\%$ of students get grade A. Estimate the lowest mark for grade A. (c) $85\%$ of students pass. Estimate the pass mark.

Solution

(a) Marks are discrete, so the upper boundaries are $19.5, 39.5, 59.5, 79.5, 100.5$ , and the graph starts at the lower boundary $-0.5$ . Points: $(-0.5, 0)$ , $(19.5, 14)$ , $(39.5, 50)$ , $(59.5, 120)$ , $(79.5, 176)$ , $(100.5, 200)$ .

(b) The top $20\%$ is $40$ students, so the boundary is where cumulative frequency $= 160$ . This lies in $59.5$ to $79.5$ : $59.5 + \tfrac{160 - 120}{56} \times 20 = 73.8$ . Students scoring $74$ or more get grade A.

(c) $15\%$ fail: $30$ students, so read at cumulative frequency $30$ , in $19.5$ to $39.5$ : $19.5 + \tfrac{30 - 14}{36} \times 20 = 28.4$ . The pass mark is about $29$ (students scoring $29$ or more pass).

In context, round grade boundaries to whole marks and say which way you rounded.

From the graph to a box plot, and a comparison

For the $120$ journey times, the shortest journey was $2$ minutes and the longest $78$ minutes. A second group of $120$ people had median $34$ minutes, $Q_1 = 28$ , $Q_3 = 41$ , shortest $12$ and longest $70$ minutes.

(a) Describe the box plot for the first group. (b) Compare the two groups' journey times.

Solution

(a) From the earlier results: minimum $2$ , $Q_1 \approx 21.7$ , median $\approx 30$ , $Q_3 \approx 39.4$ , maximum $78$ . Box from $21.7$ to $39.4$ with the median at $30$ ; whiskers to $2$ and $78$ .

(b) The second group's journeys took longer on average (median $34$ against $30$ minutes). The second group's times were less variable (IQR $41 - 28 = 13$ against about $17.7$ minutes; range $58$ against $76$ minutes).

Unknown frequencies from an estimated median (exam-hard)

The heights, $h$ cm, of $80$ seedlings are grouped as $0 \le h < 10$ (frequency $12$ ), $10 \le h < 20$ (frequency $p$ ), $20 \le h < 30$ (frequency $q$ ) and $30 \le h < 50$ (frequency $16$ ). Using linear interpolation, the median is estimated to be $22\ \text{cm}$ . Find $p$ and $q$ .

Solution

Total: $12 + p + q + 16 = 80$ , so $p + q = 52$ .

The median ( $40$ th value, at cumulative frequency $40$ ) is $22$ , which lies in $20 \le h < 30$ . Before this class the cumulative frequency is $12 + p$ . By interpolation:

20 + \frac{40 - (12 + p)}{q} \times 10 = 22 \quad\Rightarrow\quad \frac{28 - p}{q} = 0.2 \quad\Rightarrow\quad 28 - p = 0.2q.

Substitute $p = 52 - q$ : $28 - 52 + q = 0.2q$ , so $0.8q = 24$ , $q = 30$ and $p = 22$ .

Check: cumulative frequency at $20$ is $34$ ; we need $6$ more out of $30$ , a fifth of the class width, giving $22$ .

Watch out

Plotting at mid-points or lower boundaries. Cumulative frequency is "how many are less than", so it belongs at the upper boundary. Plotting at mid-points shifts the whole graph left by half a class.

Watch out

Forgetting the starting point. The graph must start at (lowest boundary, $0$ ). Without it the first class cannot be read.

Watch out

Reading "more than" as "less than". The graph gives numbers below a value. For "more than", subtract from $n$ . For percentages, divide by $n$ and multiply by $100$ .

Exam tip

Show the cumulative frequency table; the plotted points earn marks independently of the curve.
Draw your reading lines on the graph (across, then down). Examiners look for them, and they protect method marks.
Use $\tfrac{n}{2}$ , $\tfrac{n}{4}$ and $\tfrac{3n}{4}$ for grouped data on a cumulative frequency graph.
Give answers as estimates, to a sensible accuracy (usually what you can read from the graph, such as to the nearest $0.5$ or $1$ unit).
If a question asks for the "least mark to get a grade" or a similar practical value, give a whole number and make sure it makes sense in context.

Summary

Cumulative frequency at an upper class boundary $=$ total of all frequencies up to that class.
Plot (upper boundary, cumulative frequency), starting from (lowest boundary, $0$ ). Join with a smooth curve or straight lines.
Median at $\tfrac{n}{2}$ , quartiles at $\tfrac{n}{4}$ and $\tfrac{3n}{4}$ , $k$ th percentile at $\tfrac{k}{100}n$ .
Below $x$ : read up and across. Above $x$ : $n$ minus that. Between: subtract.
Linear interpolation: $L + \tfrac{c - F}{f}(U - L)$ .
All values are estimates, because individual values within classes are unknown.

Practice questions

Question

The masses, $m$ grams, of $100$ apples are grouped as $100 \le m < 120$ ( $10$ ), $120 \le m < 130$ ( $22$ ), $130 \le m < 140$ ( $35$ ), $140 \le m < 150$ ( $21$ ), $150 \le m < 170$ ( $12$ ). Construct a cumulative frequency table, and estimate the median and interquartile range.
For the apples in question 1, estimate the number of apples lighter than $125\ \text{g}$ and the number heavier than $155\ \text{g}$ .
For the journey times in this note, estimate the time exceeded by the slowest $15\%$ of people.
The numbers of text messages sent by $80$ people in a day are grouped as $0$ – $9$ , $10$ – $19$ , $20$ – $29$ , $30$ – $59$ with frequencies $10, 24, 30, 16$ . Write down the coordinates of the points to plot for a cumulative frequency graph.
Explain why the median estimated from a cumulative frequency graph is only an estimate.
The times taken by two groups of $50$ people to complete a task are grouped (minutes):

Time	$0 \le t < 5$	$5 \le t < 10$	$10 \le t < 15$	$15 \le t < 20$	$20 \le t < 30$
Group A	$4$	$10$	$16$	$12$	$8$
Group B	$2$	$6$	$14$	$18$	$10$

Estimate the median and interquartile range for each group, and compare the groups. 7. The lengths, $x$ cm, of $60$ objects are grouped as $0 \le x < 20$ ( $8$ ), $20 \le x < 30$ ( $a$ ), $30 \le x < 40$ ( $b$ ), $40 \le x < 60$ ( $10$ ). The median, estimated by linear interpolation, is $32\ \text{cm}$ . Find $a$ and $b$ . 8. A newspaper claims that "more than $40\%$ of the people in the journey-time survey in this note took more than $35$ minutes". Use the data to decide whether the claim is justified.

Answers

Cumulative frequencies at $120, 130, 140, 150, 170$ : $10, 32, 67, 88, 100$ (starting from $0$ at $100$ ). Median at $50$ : $130 + \tfrac{50 - 32}{35} \times 10 = 135.1\ \text{g}$ . $Q_1$ at $25$ : $120 + \tfrac{25 - 10}{22} \times 10 = 126.8\ \text{g}$ . $Q_3$ at $75$ : $140 + \tfrac{75 - 67}{21} \times 10 = 143.8\ \text{g}$ . IQR $\approx 143.8 - 126.8 = 17.0\ \text{g}$ .
At $125$ : $10 + \tfrac{5}{10} \times 22 = 21$ apples lighter. At $155$ : $88 + \tfrac{5}{20} \times 12 = 91$ , so $100 - 91 = 9$ apples heavier.
The slowest $15\%$ means the $85$ th percentile: cumulative frequency $0.85 \times 120 = 102$ . From the earlier example, cumulative frequency $102$ is at $50$ minutes. The slowest $15\%$ took more than about $50$ minutes.
Discrete counts, so boundaries at $-0.5, 9.5, 19.5, 29.5, 59.5$ . Points: $(-0.5, 0)$ , $(9.5, 10)$ , $(19.5, 34)$ , $(29.5, 64)$ , $(59.5, 80)$ .
The data are grouped, so the individual values are unknown. The graph assumes values are spread evenly (or smoothly) through each class, which is unlikely to be exactly true.
Group A: cumulative frequencies $4, 14, 30, 42, 50$ . Median at $25$ : $10 + \tfrac{25 - 14}{16} \times 5 = 13.4$ . $Q_1$ at $12.5$ : $5 + \tfrac{12.5 - 4}{10} \times 5 = 9.25$ . $Q_3$ at $37.5$ : $15 + \tfrac{37.5 - 30}{12} \times 5 = 18.1$ . IQR $\approx 8.9$ minutes. Group B: cumulative frequencies $2, 8, 22, 40, 50$ . Median: $15 + \tfrac{25 - 22}{18} \times 5 = 15.8$ . $Q_1$ : $10 + \tfrac{12.5 - 8}{14} \times 5 = 11.6$ . $Q_3$ : $15 + \tfrac{37.5 - 22}{18} \times 5 = 19.3$ . IQR $\approx 7.7$ minutes. Group A were quicker on average (median about $13.4$ against $15.8$ minutes); Group B's times were less spread out (IQR about $7.7$ against $8.9$ minutes).
$8 + a + b + 10 = 60$ , so $a + b = 42$ . Median at $30$ in $30 \le x < 40$ : $30 + \tfrac{30 - (8 + a)}{b} \times 10 = 32$ , so $\tfrac{22 - a}{b} = 0.2$ , i.e. $22 - a = 0.2b$ . With $a = 42 - b$ : $-20 + b = 0.2b$ , so $b = 25$ and $a = 17$ .
At $35$ minutes: cumulative frequency $\approx 60 + \tfrac{5}{10} \times 32 = 76$ . More than $35$ minutes: $120 - 76 = 44$ people, which is $\tfrac{44}{120} = 36.7\%$ . The claim is not justified by the data (the estimate is under $40\%$ ).