Maths · 10 Statistics

10.8 Cumulative frequency diagrams

Before you start

Reading a grouped frequency table whose classes are written as inequalities, such as $10 < t \leq 20$ (10.1 and 10.3).
The median, quartiles and interquartile range of a list of values (10.4).
Choosing a scale, plotting points and reading values off a graph.
Finding a percentage of an amount, and writing one number as a percentage of another.

Syllabus objectives covered (2)

Draw and interpret cumulative frequency tables and diagrams.
Estimate and interpret the median, percentiles, quartiles and interquartile range from cumulative frequency diagrams.

Part 1 of 6

Cumulative frequency is a running total

Start with an ordinary grouped frequency table. These are the times, in minutes, that $80$ people took to finish a puzzle.
$\begin{matrix} Time, t minutes & Frequency \\ 0 < t \leq 10 & 6 \\ 10 < t \leq 20 & 14 \\ 20 < t \leq 30 & 24 \\ 30 < t \leq 40 & 20 \\ 40 < t \leq 50 & 11 \\ 50 < t \leq 60 & 5 \end{matrix}$
The cumulative frequency of a class is the frequency of that class added to the frequencies of all the classes before it. You get it by keeping a running total down the frequency column.
Add each new frequency to the total so far. The first class gives $6$ . The second adds $14$ to make $20$ , and so on.
$6, 6 + 14 = 20, 20 + 24 = 44, 44 + 20 = 64, 64 + 11 = 75, 75 + 5 = 80$
Each cumulative frequency answers one question: how many values are less than or equal to the upper end of that class? The $44$ says that $44$ people took $30$ minutes or less. That is why a cumulative frequency table uses headings such as $t \leq 30$ .
$\begin{matrix} Time, t minutes & t \leq 10 & t \leq 20 & t \leq 30 & t \leq 40 & t \leq 50 & t \leq 60 \\ Cumulative frequency & 6 & 20 & 44 & 64 & 75 & 80 \end{matrix}$
Check the last entry. The final cumulative frequency must equal the total number of values, here $80$ . If it does not, there is an adding error somewhere in the column.
You can also go backwards. The frequency of one class is the difference between two neighbouring cumulative frequencies: $64 - 44 = 20$ people took more than $30$ minutes but no more than $40$ minutes.

Part 2 of 6

Drawing the diagram: upper bounds, crosses, a smooth curve

Put the variable, here time, on the horizontal axis. Put cumulative frequency on the vertical axis, with a scale that reaches the total, $80$ .
Plot each cumulative frequency above the upper bound of its class. For the puzzle that gives $(10, 6)$ , $(20, 20)$ , $(30, 44)$ , $(40, 64)$ , $(50, 75)$ and $(60, 80)$ .
Why the upper bound and not the midpoint? The $20$ means that $20$ people took $20$ minutes or less. You only know that all $14$ people in the class $10 < t \leq 20$ have finished when you reach $t = 20$ , the end of the class. At the midpoint, $t = 15$ , some of them may still be working.
Add a starting point. Nobody took $0$ minutes or less, so plot $(0, 0)$ . The start point is always the lower bound of the first class, with cumulative frequency $0$ . If the first class were $5 < t \leq 10$ , the curve would start at $(5, 0)$ , not at the origin.
Mark each point clearly with a small cross ( $\times$ ). Then join the crosses with a smooth curve that passes through every one of them. The syllabus asks for a smooth curve, so do not join the points with ruled straight lines, and do not draw a line that misses some of the points.
graph diagram — not drawn yet
Cumulative frequency diagram with horizontal axis 'Time (t minutes)' from 0 to 60 in steps of 10 and vertical axis 'Cumulative frequency' from 0 to 80 in steps of 10, with small crosses at (0, 0), (10, 6), (20, 20), (30, 44), (40, 64), (50, 75), (60, 80) joined by one smooth S-shaped curve through every cross.
Each cross sits above the upper bound of its class, the curve starts at (0, 0), and it is steepest between 20 and 30 minutes, where the most people are.
The curve never goes down as you move to the right, because a running total cannot get smaller. When most of the data are bunched in the middle, the curve has an S shape: flat at the ends, steep in the middle.
The steepness tells you where the values are crowded. The steepest part here is between $20$ and $30$ minutes, which is the class with the largest frequency, $24$ . A flat part means few values in that stretch.

Part 3 of 6

Reading the curve: how many, more than, between

To find how many values are less than or equal to a given value, start on the horizontal axis. Go up to the curve, then across to the vertical axis, and read the number there.
How many people took $35$ minutes or less? Go up from $35$ to the curve and across. The reading is about $55$ . It must lie between $44$ (at $30$ minutes) and $64$ (at $40$ minutes), which is a quick check on your reading.
For more than, subtract the reading from the total. The curve always counts from the bottom, so it gives ' $35$ or less'. People who took more than $35$ minutes:
$80 - 55 = 25$
For between, take two readings and subtract. At $25$ minutes the curve gives about $32$ , so about $55 - 32 = 23$ people took between $25$ and $35$ minutes.
Draw the reading lines on the graph with a ruler every time: up and across, or across and down. They show your method, and they stop you reading off the wrong axis.
Every answer you read from the curve is an estimate. The grouped table does not say exactly where the values lie inside each class; the curve is a sensible guess at how they are spread. That is why a question asks you to 'find an estimate', and why two careful people can get readings that differ slightly.

Part 4 of 6

The median and the quartiles: read across from the side

The median is the middle value. On a cumulative frequency diagram, find half of the total frequency on the vertical axis. Go across to the curve, then down to the horizontal axis, and read the value there.
For the puzzle, $n = 80$ , so read across at $\frac{1}{2} \times 80 = 40$ . Across from $40$ and down gives a median of about $28$ minutes.
This is the reverse of the last section. A number of people starts on the vertical axis; a time is read on the horizontal axis. Before you read, say which axis your answer belongs on.
For a list of values you used the $\frac{n + 1}{2}$ th value (10.4). For a cumulative frequency diagram, use $\frac{n}{2}$ . The curve treats the data as spread smoothly along the scale, and for a large data set the two positions are too close to tell apart on a graph: $40$ and $40.5$ here.
The lower quartile (LQ) is read across at a quarter of the total, and the upper quartile (UQ) at three quarters of the total.
$LQ: \frac{1}{4} \times 80 = 20, median: \frac{1}{2} \times 80 = 40, UQ: \frac{3}{4} \times 80 = 60$
Across from $20$ , the line meets the curve at the plotted point $(20, 20)$ , so the LQ is $20$ minutes. Across from $60$ , the line meets the curve between the points $(30, 44)$ and $(40, 64)$ , and the UQ is about $38$ minutes.
graph diagram — not drawn yet
The puzzle-time cumulative frequency curve through (0, 0), (10, 6), (20, 20), (30, 44), (40, 64), (50, 75), (60, 80), with three dashed horizontal lines from the vertical axis at 20, 40 and 60 to the curve and dashed vertical lines down to the time axis at 20, 28 and 38, labelled LQ, median and UQ.
Each reading goes across from the cumulative frequency axis to the curve, then down to the time axis; the IQR is the gap between 20 and 38 on the time axis.
The interquartile range (IQR) is the upper quartile minus the lower quartile. Subtract the two times, not the cumulative frequencies $60$ and $20$ .
$IQR = UQ - LQ \approx 38 - 20 = 18 minutes$
Check that the results make sense: the median must lie between the two quartiles, and all three must lie inside the range of the data. Here $20 < 28 < 38$ , all between $0$ and $60$ .

Part 5 of 6

Percentiles

A percentile splits the data by percentage. The $p$ th percentile is the value that has $p %$ of the data at or below it.
To find it, read across at $p %$ of the total frequency, then down.
$p th percentile: read across at \frac{p}{100} \times n$
For the $90$ th percentile of the puzzle times, read across at $\frac{90}{100} \times 80 = 72$ . Across and down gives about $47$ minutes. So about $90 %$ of the people finished within $47$ minutes, and the slowest $10 %$ took longer than that.
The quartiles and the median are percentiles with their own names: the LQ is the $25$ th percentile, the median is the $50$ th and the UQ is the $75$ th.
Watch the wording. 'The slowest $10 %$ ', 'the heaviest $10 %$ ' and 'the top $10 %$ of marks' all point to the $90$ th percentile, because $90 %$ of the values lie below it. Think about which end of the scale the words mean: with times, 'the fastest $15 %$ ' are the smallest values, so they point to the $15$ th percentile.
The question can also run the other way: what percentage of the people took $35$ minutes or less? Read up from $35$ to get about $55$ , then write $55$ as a percentage of the total.
$\frac{55}{80} \times 100 = 68.75 % \approx 69 %$

Part 6 of 6

What the median and the IQR tell you

The median is an average: a typical value. 'The median time is about $28$ minutes' means that about half of the people took $28$ minutes or less, and about half took longer.
The interquartile range measures spread. The middle half of the data lie between the LQ and the UQ, so the IQR is the width of the stretch that holds the middle half. A small IQR means the values are close together, or consistent. A large IQR means they are spread out.
The IQR leaves out the lowest quarter and the highest quarter of the data. One very slow person can change the range a lot, because the range uses the most extreme value, but that person hardly moves the IQR.
To compare two sets of data, write two sentences. One compares the medians and says what that means for a typical value. The other compares the IQRs and says what that means for consistency. Each sentence names which group is higher and uses the context.
For example: 'Group B has the lower median, so on average Group B finished the puzzle faster. Group A has the smaller IQR, so Group A's times were more consistent.' A sentence such as 'B is $4$ less' without saying what it means does not compare the data.

Worked examples

Example 1. The heights, $h$ cm, of $120$ plants are shown in the table. $\begin{matrix} Height, h cm & Frequency \\ 10 < h \leq 15 & 8 \\ 15 < h \leq 20 & 22 \\ 20 < h \leq 25 & 30 \\ 25 < h \leq 30 & 42 \\ 30 < h \leq 35 & 14 \\ 35 < h \leq 40 & 4 \end{matrix}$ (a) Complete a cumulative frequency table. (b) Write down the points you would plot for a cumulative frequency diagram, and describe how to draw it.

Show the working

Keep a running total down the frequency column.
$8, 8 + 22 = 30, 30 + 30 = 60, 60 + 42 = 102, 102 + 14 = 116, 116 + 4 = 120$
Check: the last cumulative frequency is $120$ , the number of plants. Write the table with 'less than or equal to' headings.
$\begin{matrix} Height, h cm & h \leq 15 & h \leq 20 & h \leq 25 & h \leq 30 & h \leq 35 & h \leq 40 \\ Cumulative frequency & 8 & 30 & 60 & 102 & 116 & 120 \end{matrix}$
Plot each cumulative frequency above the upper bound of its class: $(15, 8)$ , $(20, 30)$ , $(25, 60)$ , $(30, 102)$ , $(35, 116)$ , $(40, 120)$ .
Now the starting point. The first class is $10 < h \leq 15$ , so no plant is $10$ cm or less. The curve starts at $(10, 0)$ , not at the origin. The trap here is a first class that does not begin at $0$ .
Choose axes: height from $10$ to $40$ cm along the bottom, cumulative frequency from $0$ to $120$ up the side. Mark all seven points with small crosses and join them with one smooth curve through every cross.
graph diagram — not drawn yet
Cumulative frequency diagram with horizontal axis 'Height (h cm)' from 10 to 40 in steps of 5 and vertical axis 'Cumulative frequency' from 0 to 120 in steps of 20, with crosses at (10, 0), (15, 8), (20, 30), (25, 60), (30, 102), (35, 116), (40, 120) joined by a smooth curve.
The curve starts at (10, 0), the lower bound of the first class, and is steepest between 25 and 30 cm, the class with the largest frequency.

Answer: (a) Cumulative frequencies $8, 30, 60, 102, 116, 120$ . (b) Plot $(10, 0)$ , $(15, 8)$ , $(20, 30)$ , $(25, 60)$ , $(30, 102)$ , $(35, 116)$ , $(40, 120)$ as crosses and join them with a smooth curve.

Example 2. Use the cumulative frequency diagram for the $120$ plants in the previous worked example to find an estimate of (a) the median height, (b) the interquartile range, (c) the number of plants taller than $30$ cm.

Show the working

Work out where to read across. With $n = 120$ : the median is at $\frac{1}{2} \times 120 = 60$ , the LQ at $\frac{1}{4} \times 120 = 30$ and the UQ at $\frac{3}{4} \times 120 = 90$ .
(a) Across from $60$ , the line meets the curve at the plotted point $(25, 60)$ . Down to the axis: the median is $25$ cm.
(b) Across from $30$ , the line meets the curve at the point $(20, 30)$ , so the LQ is $20$ cm.
Across from $90$ : this lies between the cumulative frequencies $60$ (at $25$ cm) and $102$ (at $30$ cm), so the UQ is between $25$ and $30$ cm. It is $30$ of the $42$ steps up from $60$ , roughly seven tenths of the way, so expect a reading roughly seven tenths of the way from $25$ to $30$ cm. Across and down from the curve gives about $28.5$ cm.
Subtract the heights, not the cumulative frequencies.
$IQR \approx 28.5 - 20 = 8.5 cm$
(c) 'Taller than $30$ cm' is a 'more than' question. The curve gives the number $30$ cm or less: $102$ , at a plotted point. Subtract from the total.
$120 - 102 = 18$

Answer: (a) $25$ cm. (b) About $8.5$ cm (UQ about $28.5$ cm, LQ $20$ cm). (c) $18$ plants.

Example 3. A courier weighs $200$ parcels. The cumulative frequency diagram of their masses, $m$ kg, is a smooth curve through the points $(0, 0)$ , $(2, 12)$ , $(4, 46)$ , $(6, 118)$ , $(8, 180)$ and $(10, 200)$ . (a) The heaviest $10 %$ of the parcels get a 'heavy' label. Find an estimate of the mass above which a parcel gets the label. (b) Parcels heavier than $5$ kg cost extra to send. Find an estimate of the number of parcels that cost extra. (c) Find the percentage of the parcels with a mass of $6$ kg or less.

Show the working

(a) The heaviest $10 %$ lie above the $90$ th percentile, because $90 %$ of the parcels are below it. Read across at $90 %$ of the total.
$\frac{90}{100} \times 200 = 180$
graph diagram — not drawn yet
Cumulative frequency diagram with horizontal axis 'Mass (m kg)' from 0 to 10 and vertical axis 'Cumulative frequency' from 0 to 200, a smooth curve through crosses at (0, 0), (2, 12), (4, 46), (6, 118), (8, 180), (10, 200), with a dashed line across from 180 to the curve and down to 8, and a dashed line up from 5 to the curve and across to about 80.
A percentile starts on the vertical axis and ends on the mass axis; a 'heavier than 5 kg' count starts on the mass axis and ends on the vertical axis, then is taken from 200.
Across from $180$ , the line meets the curve at the plotted point $(8, 180)$ . So the label goes on parcels heavier than about $8$ kg.
(b) Now you start from a mass, so go up from $5$ kg to the curve and across. The reading lies between $46$ (at $4$ kg) and $118$ (at $6$ kg), and the curve gives about $80$ . That is the number of parcels of $5$ kg or less.
'Heavier than' needs the subtraction from the total.
$200 - 80 = 120$
(c) At $6$ kg the cumulative frequency is $118$ , a plotted point, so $118$ parcels have a mass of $6$ kg or less. Write that as a percentage of $200$ .
$\frac{118}{200} \times 100 = 59 %$

Answer: (a) About $8$ kg. (b) About $120$ parcels. (c) $59 %$ .

Example 4. Ravi records the waiting times, in minutes, of $100$ patients at each of two clinics. He draws a cumulative frequency diagram for each clinic and reads these values from them. $\begin{matrix} Lower quartile & Median & Upper quartile \\ Clinic X & 12 & 18 & 31 \\ Clinic Y & 15 & 20 & 24 \end{matrix}$ (a) Work out the interquartile range for each clinic. (b) Compare the waiting times at the two clinics. (c) Mr Osei must catch a bus soon after his appointment, so he wants a wait he can predict. Comment on which clinic suits him better.

Show the working

(a) The IQR is UQ minus LQ for each clinic.
$X: 31 - 12 = 19 minutes, Y: 24 - 15 = 9 minutes$
(b) Compare the medians first and say what they mean. Clinic X has the lower median, $18$ minutes against $20$ , so on average patients at Clinic X wait less time.
Then compare the IQRs and say what they mean. Clinic Y has the smaller IQR, $9$ minutes against $19$ , so the waiting times at Clinic Y are more consistent: the middle half of its patients all wait between $15$ and $24$ minutes.
(c) A Comment needs an opinion and a reason from the data. Mr Osei cares about predictability, which is spread, so the IQR decides it. Clinic Y has the smaller IQR, so he should choose Clinic Y.
The upper quartiles back this up: a quarter of the patients at Clinic X waited more than $31$ minutes, but at Clinic Y only a quarter waited more than $24$ minutes. The typical wait at Y is only $2$ minutes longer.

Answer: (a) X: $19$ minutes; Y: $9$ minutes. (b) Clinic X has the lower median, so its patients wait less time on average; Clinic Y has the smaller IQR, so its waiting times are more consistent. (c) Clinic Y: its smaller IQR means his wait is more predictable, and its median is only $2$ minutes longer.

Where people go wrong

Mistake: Plotting each cumulative frequency above the midpoint of its class, as you would for a frequency polygon.

Instead: Plot above the upper bound of the class: $(20, 20)$ , not $(15, 20)$ .

A cumulative frequency counts the values less than or equal to the upper bound. Only at the end of the class do you know that all of that class has been counted.

Mistake: Starting the curve at the origin when the first class is, say, $10 < h \leq 15$ , or leaving out the starting point.

Instead: Start at the lower bound of the first class with cumulative frequency $0$ : here $(10, 0)$ .

No value is less than the lower bound of the first class, so the count there is $0$ . The curve has nothing to say about values below $10$ .

Mistake: Finding the median by halving the horizontal axis, for example saying the median time is $30$ minutes because the axis goes from $0$ to $60$ .

Instead: Halve the total frequency on the vertical axis, go across to the curve, then down to read the median.

The median is the value of the middle person. Halfway along the scale is only the middle of the axis; it says nothing about where the middle person is.

Mistake: Answering a 'more than' question with the reading itself, for example saying $55$ people took more than $35$ minutes.

Instead: Subtract the reading from the total: $80 - 55 = 25$ took more than $35$ minutes.

The curve always counts from the bottom, so a reading gives 'less than or equal to'. 'More than' is everyone else.

Mistake: Working out the IQR by subtracting the cumulative frequencies, such as $60 - 20 = 40$ .

Instead: Read the two quartiles on the horizontal axis and subtract those: $38 - 20 = 18$ minutes.

The IQR is a spread of values (minutes, cm, kg). The numbers $60$ and $20$ are just the positions where you read across, and their difference is always half of the total.

Check yourself

The frequencies of five classes, in order, are $5, 9, 14, 8, 4$ . Write down the cumulative frequencies.
Answer
$5, 14, 28, 36, 40$ . The last one, $40$ , equals the total frequency.
In a cumulative frequency table, the class $20 < t \leq 30$ has a cumulative frequency of $44$ . Write down the point you plot for it, and say what the $44$ means.
Answer
$(30, 44)$ , above the upper bound. It means $44$ values are $30$ or less.
The first class in a grouped table is $50 < h \leq 60$ . Where does the cumulative frequency curve start?
Answer
At $(50, 0)$ : the lower bound of the first class, with cumulative frequency $0$ .
A curve for $80$ people gives a reading of $55$ at $35$ minutes. How many people took more than $35$ minutes, and what percentage of the people is that?
Answer
$80 - 55 = 25$ people, and $\frac{25}{80} \times 100 = 31.25 %$ .
A cumulative frequency diagram shows $240$ values. At which cumulative frequencies do you read across for the lower quartile, the median and the upper quartile?
Answer
LQ at $60$ , median at $120$ , UQ at $180$ (a quarter, a half and three quarters of $240$ ).
A cumulative frequency diagram shows $150$ values. At which cumulative frequency do you read across for the $70$ th percentile?
Answer
$\frac{70}{100} \times 150 = 105$ .
On a curve for $120$ bags, the LQ is $34$ kg and the UQ is $52$ kg, read across at $30$ and $90$ . Ali says the interquartile range is $60$ . Is he right?
Answer
No. The IQR is $52 - 34 = 18$ kg. Ali subtracted the cumulative frequencies, $90 - 30$ , instead of the masses.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The masses, $m$ kg, of $100$ suitcases checked in for a flight are shown in the table. $\begin{matrix} Mass, m kg & Frequency \\ 0 < m \leq 5 & 4 \\ 5 < m \leq 10 & 12 \\ 10 < m \leq 15 & 26 \\ 15 < m \leq 20 & 33 \\ 20 < m \leq 25 & 18 \\ 25 < m \leq 30 & 7 \end{matrix}$

(a) Complete the cumulative frequency table. $\begin{matrix} Mass, m kg & m \leq 5 & m \leq 10 & m \leq 15 & m \leq 20 & m \leq 25 & m \leq 30 \\ Cumulative frequency \end{matrix}$ 2 marks

Mark scheme

B2 4, 16, 42, 75, 93, 100. B1 for four or five correct values

(b) On a grid with $0 \leq m \leq 30$ on the horizontal axis and cumulative frequency from $0$ to $100$ on the vertical axis, draw a cumulative frequency diagram for these masses. 3 marks

Mark scheme

B2 all six points plotted correctly at the upper bounds: (5, 4), (10, 16), (15, 42), (20, 75), (25, 93), (30, 100), FT their table. B1 for four or five correct points
B1 a smooth increasing curve through all their points, starting at (0, 0)

(c) Use your diagram to find an estimate of the median mass. 1 mark

Mark scheme

B1 16 to 16.5 kg (read across at 50)

(d) Use your diagram to find an estimate of the interquartile range. 2 marks

Mark scheme

M1 UQ in the range 19.8 to 20.2 and LQ in the range 11.5 to 12.2 seen, or readings at cumulative frequencies 75 and 25 shown on the diagram
A1 7.8 to 8.5 kg

(e) Suitcases with a mass of more than $23$ kg pay an extra charge. Use your diagram to find an estimate of the number of suitcases that pay the extra charge. 2 marks

Mark scheme

M1 a reading of 85 to 88 at m = 23, or 100 − their reading seen
A1 12 to 15

Question 2. The cumulative frequency table shows the finishing times, $t$ minutes, of $160$ runners in a $10$ km race. No runner took $35$ minutes or less. $\begin{matrix} Time, t minutes & t \leq 40 & t \leq 45 & t \leq 50 & t \leq 55 & t \leq 60 & t \leq 65 & t \leq 70 \\ Cumulative frequency & 8 & 24 & 56 & 100 & 128 & 148 & 160 \end{matrix}$

(a) Find an estimate of the $80$ th percentile of the finishing times. 2 marks No calculator

Mark scheme

M1 $0.8 \times 160$ oe, or 128 seen
A1 60 minutes

(b) Calculate the percentage of the runners who took more than $55$ minutes. 2 marks No calculator

Mark scheme

M1 $\frac{160 - 100}{160} \times 100$ oe, or 60 runners seen
A1 37.5%

(c) Without drawing the cumulative frequency diagram, explain why the median time is between $50$ and $55$ minutes. 1 mark No calculator

Mark scheme

B1 half of 160 is 80, and 80 lies between the cumulative frequencies 56 (at 50 minutes) and 100 (at 55 minutes) oe

Question 3. Lucia tests $200$ batteries of brand P and $200$ batteries of brand Q. She draws a cumulative frequency diagram of the lifetimes, in hours, for each brand, and reads these values from her diagrams. $\begin{matrix} Lower quartile & Median & Upper quartile \\ Brand P & 41 & 46 & 50 \\ Brand Q & 38 & 49 & 57 \end{matrix}$

(a) Calculate the interquartile range of the lifetimes for each brand. 2 marks No calculator

Mark scheme

B1 brand P: 9 hours
B1 brand Q: 19 hours

(b) Compare the lifetimes of the two brands. 2 marks No calculator

Mark scheme

B1 brand Q has the higher median, so brand Q batteries last longer on average oe
B1 brand P has the smaller IQR, so brand P lifetimes are more consistent / less spread out oe

(c) Write down an estimate of the number of brand P batteries that lasted more than $50$ hours. 1 mark No calculator

Mark scheme

B1 50 (50 hours is the upper quartile, so a quarter of the 200 lasted longer)

How this is examined

Turning a frequency table into a cumulative frequency table, then drawing the curve: a cumulative frequency table to complete, followed by drawing the diagram on a grid Running totals that end at the total frequency. Points plotted at the upper bound of each class, not the midpoint, marked as small crosses. A starting point at the lower bound of the first class with cumulative frequency 0. One smooth increasing curve through every plotted point.
Reading the median, the quartiles and the interquartile range from the curve: a run of short parts: 'Use your diagram to find an estimate of the median', then of the lower quartile, the upper quartile or the interquartile range Reading across at $\frac{1}{2} n$ for the median, $\frac{1}{4} n$ for the LQ and $\frac{3}{4} n$ for the UQ, then down to the value axis. The IQR is UQ − LQ worked out with the two values from the horizontal axis. Draw the reading lines on the diagram so your method is visible.
Counting how many values are more than (or between) given values: a two-mark part: a threshold such as a price band, a pass mark or a weight limit, and 'find an estimate of the number of …' A reading up from the threshold on the horizontal axis and across to the cumulative frequency. For 'more than', the total minus that reading; for 'between', the difference of two readings. The reading alone answers 'at most', not 'more than'.
Percentiles given in words, such as 'the top 10%' or 'the fastest 15%': a two-mark part in context, worded with a percentage such as 'the top 10%' instead of the word percentile Turning the words into a percentile (top 10% means the 90th percentile), then $\frac{p}{100} \times n$ seen as the cumulative frequency to read across at, and the value read from the curve.
Comparing two groups with the median and the interquartile range: a two-mark Compare or Comment after you have found the median and IQR of two groups One sentence on the medians (which group has the higher typical value, in the context) and one on the IQRs (which group is more consistent or more spread out). Name the group and say what the difference means; listing the numbers is not a comparison.

Words to know

Cumulative frequency: A running total of the frequencies. The cumulative frequency at a value is the number of data values less than or equal to it.
Cumulative frequency diagram: A graph of cumulative frequency (vertical axis) against the variable (horizontal axis), with points plotted at the upper bounds of the classes and joined by a smooth curve.
Upper bound of a class: The largest value a class can contain, such as $20$ in $10 < t \leq 20$ . Cumulative frequencies are plotted above it.
Median: The middle value. On a cumulative frequency diagram, read across at half of the total frequency.
Lower quartile (LQ): The value with a quarter of the data at or below it; the $25$ th percentile. Read across at $\frac{1}{4} n$ .
Upper quartile (UQ): The value with three quarters of the data at or below it; the $75$ th percentile. Read across at $\frac{3}{4} n$ .
Interquartile range (IQR): UQ − LQ. The width of the stretch that holds the middle half of the data; a measure of spread that extreme values hardly affect.
Percentile: The $p$ th percentile is the value with $p %$ of the data at or below it. Read across at $\frac{p}{100} \times n$ .

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