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Drawing and interpreting histograms, frequency polygons, cumulative frequency diagrams and box plots for single-variable data. Questions usually ask you to use the fact that area represents frequency in a histogram, to estimate values from a diagram, or to compare two distributions in context.
Key facts
Frequency density = frequency ÷ class width
Histogram: area of a bar is proportional to frequency (frequency = \(k\) × area)
Cumulative frequency: plot at the upper class boundaries
Grouped data: median at \(\tfrac{n}{2}\), \(Q_1\) at \(\tfrac{n}{4}\), \(Q_3\) at \(\tfrac{3n}{4}\)
IQR \(= Q_3 - Q_1\) (the middle 50%)
10th to 90th interpercentile range \(= P_{90} - P_{10}\)
Histograms show grouped continuous data. The bars touch, and the area of each bar is proportional to the frequency.
Usually the vertical axis is frequency density = frequency ÷ class width, so that area = frequency.
Bar area = frequency. The tallest bar (2 to 3 kg) is not the class with the greatest frequency
Use class boundaries for widths: lengths recorded to the nearest cm in the class 10–14 have boundaries 9.5 and 14.5, so the width is 5.
If you are given one bar's frequency, find the scale first: frequency per unit of area (per cm² or per grid square) = that bar's frequency ÷ its area.
To estimate the frequency in part of a class, assume the values are spread evenly: 3 of the 5 units of width contain \(\tfrac{3}{5}\) of the frequency.
A frequency polygon joins the midpoints of the tops of the bars with straight lines.
Cumulative frequency
diagram
Plot cumulative frequency against the upper class boundary, starting at the lowest boundary with cumulative frequency 0, and join the points with a smooth curve or straight lines.
For \(n\) values, read across from \(\tfrac{n}{2}\) for the median, \(\tfrac{n}{4}\) for \(Q_1\), \(\tfrac{3n}{4}\) for \(Q_3\) and \(\tfrac{k}{100}n\) for the \(k\)th percentile.
Plot at upper class boundaries, then read across from n/4, n/2 and 3n/4 and down to the axis
To estimate how many values are above a given value, read up from it and subtract that cumulative frequency from \(n\)
Box plots
diagram
A box plot shows the minimum, \(Q_1\), median, \(Q_3\) and maximum. Outliers are plotted as crosses, and the whisker goes to the most extreme value that is not an outlier.
Box plot: the whisker stops at the most extreme value that is not an outlier
The box holds the middle 50% of the data; its length is the interquartile range, \(Q_3 - Q_1\)
Comparing distributions
Compare a measure of location (median or mean) and a measure of spread (IQR, range or standard deviation), quoting values and interpreting them in context.
Use the median and IQR when the data are skewed or contain outliers, as they are not affected by extreme values.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Calculate frequency densities and use areas
4 to 6 marksGrade D
Find each class width from the class boundaries.
Frequency density = frequency ÷ class width.
For part of a class, use (fraction of the class width) × frequency.
Example. The table shows the masses, \(m\) kg, of 60 parcels.
Mass, \(m\) kg
Frequency
\(0 \le m \lt 2\)
12
\(2 \le m \lt 3\)
18
\(3 \le m \lt 5\)
20
\(5 \le m \lt 10\)
10
(a) Calculate the frequency density of each class, and state which class has the tallest bar on a histogram. (b) A parcel is chosen at random. Estimate the probability that its mass is between 4 kg and 6 kg.Show the model answerHide the model answer
(a) Widths 2, 1, 2, 5, so frequency densities 6, 18, 10, 2 (M1 A1). The tallest bar is \(2 \le m \lt 3\) (B1), even though \(3 \le m \lt 5\) has the greatest frequency.
(b) 4 to 5 kg: \(\tfrac{1}{2} \times 20 = 10\) (M1); 5 to 6 kg: \(\tfrac{1}{5} \times 10 = 2\) (M1); probability \(= \dfrac{12}{60} = 0.2\)
(A1)
Compare two distributions from box plots
2 to 3 marksGrade C
Compare the medians (location), with values, in context.
Compare the IQRs or ranges (spread), with values, in context.
Mention outliers or skew only if they are relevant.
Example. A student's samples of daily mean windspeed (knots) from two stations give these box plot values.
Station
Minimum
\(Q_1\)
Median
\(Q_3\)
Maximum
A
3
7
9
12
18
B
6
10
12
13
17
Compare the windspeeds at the two stations.Show the model answerHide the model answer
The median at B (12 knots) is higher than at A (9 knots), so on average it was windier at B
(B1)
The IQR at A is \(12 - 7 = 5\) knots and at B is \(13 - 10 = 3\) knots, so the windspeeds at A were more variable
(B1)
Use a histogram when you are given one bar's frequency
3 to 4 marksGrade B
Find the area of the bar whose frequency you know.
Work out the frequency represented by one unit of area.
Multiply the area of each other bar (or part of a bar) by this.
Example. On a histogram of the times, \(t\) minutes, that people waited at a clinic, the bar for \(10 \le t \lt 15\) is 2 cm wide and 6 cm tall and represents 24 people. The bar for \(15 \le t \lt 25\) is 3.5 cm tall. (a) Find the number of people who waited between 15 and 25 minutes. (b) Estimate the number of people who waited between 18 and 25 minutes.
Show the model answerHide the model answer
(a) 5 minutes is 2 cm wide, so the second bar is 4 cm wide. First bar: \(2 \times 6 = 12\) cm² represents 24 people, so 1 cm² represents 2 people (M1). Second bar: \(4 \times 3.5 = 14\) cm², so 28 people
(A1)
(b) \(\tfrac{7}{10} \times 28\) (M1)\(= 19.6\), so about 20 people
(A1)
Don’t lose marks
Plotting frequency instead of frequency density when the class widths are unequal.
Plotting cumulative frequencies at class midpoints instead of upper class boundaries.
Using the stated class limits instead of the class boundaries to find widths.
Comparing only averages, or quoting numbers without saying what they mean in context.
More tips
Memory tricks
The tallest bar is not always the class with the greatest frequency: area, not height, shows frequency.
Scale questions: find the frequency per unit of area from one bar, then use it for every bar.
The last cumulative frequency always equals the total \(n\): a quick check of your table.
For comparisons think ‘one for the middle, one for the spread’.
Exam technique
‘Estimate’ tells you to assume values are spread evenly within a class; round sensibly if the answer is a number of people or items.
Check the vertical axis: if it is labelled frequency density, area = frequency; if not, find the scale from a bar you know.
When you draw a box plot, use the scale given, mark outliers with crosses and label which plot is which.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Calculate frequency densities for a histogramFrequency density = frequency ÷ class width, using the class boundaries.
Grade E
Read the median and quartiles from a box plotThe box shows \(Q_1\), the median and \(Q_3\); the whiskers show the extent of the data and crosses show outliers.
Grade D
Estimate quartiles from a cumulative frequency graphRead across from \(\tfrac{n}{4}\), \(\tfrac{n}{2}\) and \(\tfrac{3n}{4}\) on the cumulative frequency axis.
Grade D
Draw a histogram with unequal class widthsPlot frequency density on the vertical axis, with the bars touching at the class boundaries.
Grade C
Estimate frequencies from histogram bar areasUse area to find frequency, assuming values are spread evenly within a class for part of a bar.
Grade B
Find a histogram's scale from a known barWhen area is not equal to frequency, use one bar to find the frequency per unit of area.
Grade A
Compare distributions using location and spread in contextCompare a median (or mean) and an IQR (or standard deviation), with values and in the context of the data.
Grade A*
Link a histogram's shape to a probability modelRecognise that a roughly symmetrical, bell-shaped histogram suggests a normal model, and treat areas as probabilities.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy2 marks
The table summarises the journey times for a sample of journeys on a commuter train route.
[object Object]
Find the missing frequency density for the class 25 ≤ t < 30.[2]
The table shows the times, in minutes, taken by 80 people to complete a puzzle.
[object Object]
(a) A student draws a histogram of these data in which the height of each bar is equal to the frequency. Explain why the student’s histogram is misleading.[1]
(b) Calculate the frequency density for each class.[2]
(c) Estimate the number of people who took between 12 and 25 minutes.[2]
(d) Use linear interpolation to estimate the median time.[2]
Show the answer and mark scheme
(a)Answer: The classes have different widths; in a histogram the area of a bar, not its height, represents frequency, so wide classes (such as 30–50) look far too large.
B1 for the class widths are unequal, so heights should be frequency densities (area proportional to frequency); e.g. the 30–50 bar is 20 minutes wide and so looks much bigger than 6 people
Worked solution: The classes have unequal widths. In a histogram the area of each bar represents frequency, so heights must be frequency densities. Using frequency as height makes wide classes such as \(30 \le t \lt 50\) look far more important than they are.
(b)Answer: 1.2, 3.6, 4.8, 2, 0.3
M1 for frequency ÷ class width for at least two classes
Worked solution: The median is the 40th value. 30 people took less than 15 minutes, and the next class has 24 people, so the median \(\approx 15 + \frac{40 - 30}{24} \times 5 = 17.1\) minutes.
Question 3Hard7 marks
The table shows the journey times for a random sample of 139 journeys on a commuter train route. A histogram is drawn to represent these data. On the histogram, the bar for the class 20 ≤ t < 25 is 1 cm wide and 3 cm tall.
[object Object]
(a) Find the width and the height of the bar for the class 40 ≤ t < 50.[3]
(b) Estimate the probability that a journey chosen at random from this sample has \(t\) between 27 and 43.[3]
(c) State, giving a reason, whether a normal distribution is likely to be a suitable model for these journey times.[1]
Show the answer and mark scheme
(a)Answer: Width 2 cm, height 2.5 cm
B1 for width 2 cm (5 minutes is represented by 1 cm)
M1 for using area proportional to frequency, e.g. \(1 \times 3 = 3 \text{cm}^2\) represents 12, so \(1 \text{cm}^2\) represents 4 journeys
A1 for height 2.5 cm
Worked solution: Horizontal scale: 5 minutes is 1 cm, so the width is \(10 \times 0.2 = 2\) cm. Area: \(3 \text{cm}^2\) represents 12, so the bar for 20 journeys has area \(20 \div 4 = 5 \text{cm}^2\) and height \(5 \div 2 = 2.5\) cm.
M1 for a correct method for a part class, e.g. \(\dfrac{3}{5} \times 27\)
M1 for adding the frequencies of the whole classes and both part classes, and dividing by 139
A1 for awrt 0.663
Worked solution: \(5.4 \times 3 + 8.6 \times 5 + 5.4 \times 5 + 2.0 \times 3 = 92.2\), so the probability is \(\dfrac{92.2}{139} = 0.6633\).
(c)Answer: Not suitable: the distribution is not symmetrical (it has positive skew, with a longer tail to the right), whereas a normal distribution is symmetrical about its mean.
B1 for not suitable, with a reason based on the lack of symmetry (positive skew) of the data