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E-Math notes · Chapter 17

Data Handling and Analysis: O-Level E-Math Notes

Statistics turns lists of numbers into answers. In this chapter you collect and tabulate data, draw and read the standard diagrams, spot diagrams that mislead, find averages and spread (including quartiles and the standard deviation), use the cumulative frequency curve, and compare two data sets with the mean and the standard deviation.

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Figure 1 · Cumulative frequency curve, drawn smooth. Read across from 10, 20 and 30 to the curve, then down to the axis.
Cumulative frequency curveSmooth cumulative frequency curve0510152025303540Cumulative frequency1015202530354045505560Time in minutesmedian ≈ 33.8Q₁ ≈ 26.7Q₃ ≈ 40.0
12 outcomes · 30 worked examples · every step shown ✦
Maths in real life

Statistics is how we turn lists of numbers into decisions. Three real situations come first, so you can see what the chapter is for. Each is solved in full, with a Your Turn that has its own data. Come back to them when the ideas are new.

Collecting and tabulating

Tally first, then total.

Bar, line and pie

Read the scale and the key before the picture.

By the end of this chapter
  • Collect, classify and tabulate data, and read tables, bar graphs, pictograms, line graphs, pie charts, dot diagrams, histograms and stem-and-leaf diagrams.
  • Choose a diagram, draw inferences from it, and explain why a diagram can mislead.
  • Find the mean, median and mode, including the mean of grouped data, and say when each is the right average.
  • Find quartiles and percentiles, draw and read a cumulative frequency curve, and read a box-and-whisker plot.
  • Find the range, the interquartile range and the standard deviation, for grouped and ungrouped data.
  • Compare two data sets using the mean and the standard deviation.
Supports S1.8 · S1.6 · S1.2.g · S1.2 · S1.10 · S1.11 ...

Maths in real life: where statistics is used

Statistics is how we turn lists of numbers into decisions. Three real situations come first, so you can see what the chapter is for. Each is solved in full, with a Your Turn that has its own data. Come back to them when the ideas are new.

Waiting for a bus

When data are grouped, a histogram shows the shape, the modal class shows the most common wait, and the mid-values give an estimate of the mean.

Example 1.1

Waiting times at a bus stop

A bus company records how long, in minutes, 50 passengers waited at a stop. The results are grouped in classes of width 5 minutes: 0≤t<50\le t<5: 6 passengers; 5≤t<105\le t<10: 14 passengers; 10≤t<1510\le t<15: 18 passengers; 15≤t<2015\le t<20: 9 passengers; 20≤t<2520\le t<25: 3 passengers.

  • (i)Which class is the modal class?
  • (ii)Estimate the mean waiting time.
  • (iii)What percentage of the passengers waited at least fifteen minutes?
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare
(i)
largest frequency  =18\displaystyle \text{largest frequency}\;=18
the modal class is the class with the highest frequency (the tallest bar)
modal class: 10≤t<15\displaystyle \text{modal class: }10\le t<15
(ii)
mid-value  =lower+upper2\displaystyle \text{mid-value}\;=\dfrac{\text{lower}+\text{upper}}{2}
one value stands for the whole class
classx0≤x<52.55≤x<107.510≤x<1512.515≤x<2017.520≤x<2522.5\displaystyle \begin{array}{c|c}\text{class} & x\\ \hline 0\le x<5 & 2.5\\ 5\le x<10 & 7.5\\ 10\le x<15 & 12.5\\ 15\le x<20 & 17.5\\ 20\le x<25 & 22.5\end{array}
the mid-value of each class
xffx2.56157.51410512.51822517.59157.522.5367.5\displaystyle \begin{array}{c|cc}x & f & fx\\ \hline 2.5 & 6 & 15\\ 7.5 & 14 & 105\\ 12.5 & 18 & 225\\ 17.5 & 9 & 157.5\\ 22.5 & 3 & 67.5\end{array}
add the ff row and a fxfx row (each mid-value times its frequency)
∑f  =50, ∑fx=570\displaystyle \sum f\;=50,\allowbreak\ \qquad \sum fx=570
estimated mean  =57050\displaystyle \text{estimated mean}\;=\dfrac{570}{50}
an estimate, because we used mid-values
=11.4\displaystyle =11.4
(iii)
9+3  =12\displaystyle 9+3\;=12
the last two classes start at fifteen minutes or more
1250×100  =24\displaystyle \dfrac{12}{50}\times100\;=24
fraction of the passengers, times one hundred
Waiting times at a bus stopHistogram of waiting times in classes of five minutes02468101214161820Number of passengers0510152025Waiting time in minutes
Waiting times, in classes of 5 minutes. Equal class widths, so the bar heights are the frequencies.
Final answer
(i)10≤t<1510\le t<15 minutes
(ii)Estimated mean =11.4=11.4 minutes
(iii)24%24\% of the passengers

Two classes, one test

The mean tells you the typical mark. The standard deviation tells you how spread out the marks are. You need both to compare two groups fairly.

Example 1.2

Two classes, one test

Two classes of 8 students sit the same test. Class A scores: 58, 59, 60, 60, 61, 61, 62, 63. Class B scores: 42, 50, 57, 60, 63, 67, 72, 73.

  • (i)Find the mean mark of Class A.
  • (ii)Find the mean mark of Class B.
  • (iii)Find the standard deviation of Class A.
  • (iv)Find the standard deviation of Class B.
  • (v)Compare the two classes.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplifyround, estimate or compare
(i)
∑x  =58+59+60+60+61+61+62+63=484\displaystyle \sum x\;=58+59+60+60+61+61+62+63=484
add all the values
mean  =∑xn=4848\displaystyle \text{mean}\;=\dfrac{\sum x}{n}=\dfrac{484}{8}
total divided by the number of values
=60.5\displaystyle =60.5
(ii)
∑x  =42+50+57+60+63+67+72+73=484\displaystyle \sum x\;=42+50+57+60+63+67+72+73=484
add all the values
mean  =∑xn=4848\displaystyle \text{mean}\;=\dfrac{\sum x}{n}=\dfrac{484}{8}
total divided by the number of values
=60.5\displaystyle =60.5
(iii)
n  =8, ∑x=484, ∑x2=3364+3481+3600+3600+3721+3721+3844+3969=29300\displaystyle n\;=8,\allowbreak\ \qquad \sum x=484,\allowbreak\ \qquad \sum x^{2}=3364+3481+3600+3600+3721+3721+3844+3969=29300
add the values, and add the squares of the values
mean  =4848=60.5\displaystyle \text{mean}\;=\dfrac{484}{8}=60.5
σ  =∑x2n−(∑xn)2=293008−(60.5)2\displaystyle \sigma\;=\sqrt{\dfrac{\sum x^{2}}{n}-\left(\dfrac{\sum x}{n}\right)^{2}}=\sqrt{\dfrac{29300}{8}-(60.5)^{2}}
population standard deviation: the formula on the exam paper
=2.25\displaystyle =\sqrt{2.25}
variance first, then the root
=1.5\displaystyle =1.5
(iv)
n  =8, ∑x=484, ∑x2=1764+2500+3249+3600+3969+4489+5184+5329=30084\displaystyle n\;=8,\allowbreak\ \qquad \sum x=484,\allowbreak\ \qquad \sum x^{2}=1764+2500+3249+3600+3969+4489+5184+5329=30084
add the values, and add the squares of the values
mean  =4848=60.5\displaystyle \text{mean}\;=\dfrac{484}{8}=60.5
σ  =∑x2n−(∑xn)2=300848−(60.5)2\displaystyle \sigma\;=\sqrt{\dfrac{\sum x^{2}}{n}-\left(\dfrac{\sum x}{n}\right)^{2}}=\sqrt{\dfrac{30084}{8}-(60.5)^{2}}
population standard deviation: the formula on the exam paper
=100.25\displaystyle =\sqrt{100.25}
variance first, then the root
=4012≈10.0\displaystyle =\frac{\sqrt{401}}{2}\approx10.0
three significant figures
(v)
The means are equal, so on average the classes did equally well.
compare the averages first
σA  =1.50 < σB=10.0\displaystyle \sigma_{A}\;=1.50\ <\ \sigma_{B}=10.0
a smaller standard deviation means the marks are closer together
So Class A is more consistent; Class B has a wider spread of marks, with both weaker and stronger students.
answer in the context of the question
Final answer
(i)mean=60.5\text{mean}=60.5
(ii)mean=60.5\text{mean}=60.5
(iii)σ=1.5\sigma=1.5
(iv)σ=4012≈10.0\sigma=\frac{\sqrt{401}}{2}\approx10.0
(v)The two classes have the same mean mark, but Class A has the smaller standard deviation, so its marks are more consistent. Class B's marks are more spread out.

A chart that exaggerates

A bar chart is honest only if the bar heights are in proportion to the values. If the vertical axis does not start at zero, the picture is misleading.

Example 1.3

Is the advert fair?

A shop's advert compares weekly sales (in hundreds of dollars) of two branches, 6262 for the Orchard branch and 7070 for the Tampines branch. The bar chart's vertical axis starts at 50, not at zero.

  • (i)How many times taller does the taller bar look than the shorter bar?
  • (ii)How many times larger are the taller branch's sales than the shorter branch's?
  • (iii)Explain why the chart is misleading.
Show solutionHide solution
Colour keyround, estimate or compare
(i)
visual ratio  =70−5062−50=2012\displaystyle \text{visual ratio}\;=\dfrac{70-50}{62-50}=\dfrac{20}{12}
bar heights are measured from the start of the axis
=53≈1.67\displaystyle =\frac{5}{3}\approx1.67
three significant figures
(ii)
actual ratio  =7062\displaystyle \text{actual ratio}\;=\dfrac{70}{62}
the sales themselves, measured from zero
=3531≈1.13\displaystyle =\frac{35}{31}\approx1.13
three significant figures
(iii)
The axis does not start at zero, so the bar heights are not in proportion to the sales.
a truncated axis
1.67 (looks) ≫ 1.13 (actual)\displaystyle 1.67\ \text{(looks)}\ \gg\ 1.13\ \text{(actual)}
the picture exaggerates the difference
Advert comparing two branchesBar chart whose vertical axis starts at fifty505560657075Weekly sales (hundreds of dollars)OrchardTampines
The advert's chart. The vertical axis does not start at zero.
Final answer
(i)The taller bar looks about 1.671.67 times as tall.
(ii)The sales are about 1.131.13 times as large.
(iii)The vertical axis starts above zero, so the bars are cut off and the difference looks much bigger than it really is.
Think About It
Class A and Class B have the same mean mark. Does that mean the two classes are equally good at the subject?
Show the worked answerHide the worked answer
mean A  =60.5, mean B=60.5\displaystyle \text{mean A}\;=60.5,\allowbreak\ \qquad \text{mean B}=60.5
the same centre
range A  =5, range B=31\displaystyle \text{range A}\;=5,\allowbreak\ \qquad \text{range B}=31
but the spread is very different
Equal means only say the centres match. The range (and the standard deviation) show the groups are different.
a measure of centre and a measure of spread together
Final answerNot necessarily. The mean only gives the centre. Class B's marks are much more spread out, so it has some very weak and some very strong students, while Class A's students are all close to the mean. To compare the groups you need the spread as well as the mean.
Supports S1.1 · S1.2.a · S1.4

Collecting, classifying and tabulating data

Data are numbers or categories collected to answer a question. Before drawing anything, classify the data, because the type decides which diagram is sensible. Then tabulate them, so that you count every value exactly once.

Table 1: types of data
TypeWhat it isExample
Categoricallabels or categories, not numbersfavourite kind of lunch
Discrete numericalnumbers you count, taking separate valuesnumber of brothers and sisters
Continuous numericalnumbers you measure, taking any value in a rangeheight in centimetres, time in seconds
The Core Idea

Tally first, then total.

Go through the data once, tick each value, count the ticks to get the frequency, and then check that the frequencies add up to the number of values.

Example 2.1

From raw data to a frequency table

The number of brothers and sisters for 20 students is: 1, 0, 2, 1, 3, 1, 2, 0, 1, 2, 4, 1, 2, 1, 0, 3, 2, 1, 1, 2.

  • (i)Make a frequency table.
  • (ii)How many of them had at least 2?
  • (iii)Write down the mode.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
Go through the data once, tick each value, then count the ticks in each row.
a tally keeps the count tidy
xf0318263241\displaystyle \begin{array}{c|c}x & f\\ \hline 0 & 3\\ 1 & 8\\ 2 & 6\\ 3 & 2\\ 4 & 1\end{array}
the frequency table
∑f  =3+8+6+2+1=20\displaystyle \sum f\;=3+8+6+2+1=20
check: the frequencies must add up to the number of values
(ii)
values from 2 upwards: 6+2+1  =9\displaystyle \text{values from 2 upwards: }6+2+1\;=9
add the frequencies of the rows you need
(iii)
The value 1 appears 8 times, more than any other.
mode = the most frequent value
Final answer
(i)xf0318263241\begin{array}{c|c}x & f\\ \hline 0 & 3\\ 1 & 8\\ 2 & 6\\ 3 & 2\\ 4 & 1\end{array}
(ii)99
(iii)mode=1\text{mode}=1

Grouping data

For continuous data, or discrete data with many different values, we put the values into classes. Classes must not overlap and must leave no gaps, so each value belongs to exactly one class. We write a class as 10≤x<2010\le x<20: it includes its lower boundary and excludes its upper boundary.

Example 2.2

Grouping continuous data

The heights, in centimetres, of 24 plants, in the order measured, are: 25, 38, 47, 35, 15, 24, 43, 31, 30, 34, 52, 34, 21, 28, 44, 49, 12, 18, 36, 33, 22, 40, 41, 27. Use classes of width 10, starting at 10.

  • (i)Make a grouped frequency table.
  • (ii)Write down the width and the mid-value of the third class.
  • (iii)Write down the modal class.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
Put each value in the class that includes its lower boundary and excludes its upper boundary.
a value equal to an upper boundary goes in the next class
classf10≤x<20320≤x<30630≤x<40840≤x<50650≤x<601\displaystyle \begin{array}{c|c}\text{class} & f\\ \hline 10\le x<20 & 3\\ 20\le x<30 & 6\\ 30\le x<40 & 8\\ 40\le x<50 & 6\\ 50\le x<60 & 1\end{array}
the grouped frequency table
∑f  =3+6+8+6+1=24\displaystyle \sum f\;=3+6+8+6+1=24
check the total
(ii)
width  =40−30=10\displaystyle \text{width}\;=40-30=10
upper boundary minus lower boundary
mid-value  =30+402=35\displaystyle \text{mid-value}\;=\dfrac{30+40}{2}=35
half way between the boundaries
(iii)
highest frequency  =8\displaystyle \text{highest frequency}\;=8
the class with the largest frequency
Final answer
(i)classf10≤x<20320≤x<30630≤x<40840≤x<50650≤x<601\begin{array}{c|c}\text{class} & f\\ \hline 10\le x<20 & 3\\ 20\le x<30 & 6\\ 30\le x<40 & 8\\ 40\le x<50 & 6\\ 50\le x<60 & 1\end{array}
(ii)Width 1010, mid-value 3535.
(iii)30≤x<4030\le x<40
Think About It
Why must a value of 20 go in only one class, and which one if the classes are 10≤x<2010\le x<20 and 20≤x<3020\le x<30?
Show the worked answerHide the worked answer
Colour keyadd, subtract, multiply, divide or simplify
20 satisfies 20≤x<30\displaystyle 20\ \text{satisfies}\ 20\le x<30
the lower boundary is included
20 does not satisfy 10≤x<20\displaystyle 20\ \text{does not satisfy}\ 10\le x<20
the upper boundary is excluded
So the value is counted once, in the second class.
exactly one class per value
Final answerIf a value could belong to two classes it would be counted twice and the total frequency would be wrong. With these classes, a value of 20 goes in the class that includes its lower boundary: 20≤x<3020\le x<30.

Reading tables

A table shows data in rows and columns. To interpret one, find the row and column you need, and add along a row or down a column to get a total.

Example 2.3

Reading a two-way table

A survey asks 70 students how they travel to school. Secondary Three: bus 12, MRT 15, walking 8. Secondary Four: bus 10, MRT 18, walking 7.

  • (i)Find the total number of students for each way of travelling.
  • (ii)What fraction of all the students walk to school? Give the fraction in its simplest form.
  • (iii)What percentage of the Secondary Four students take the MRT?
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
bus  =12+10=22,MRT=15+18=33,walking=8+7=15\displaystyle \text{bus}\;=12+10=22,\quad \text{MRT}=15+18=33,\quad \text{walking}=8+7=15
add down each column
(ii)
1570\displaystyle \dfrac{15}{70}
walkers over everybody
=314\displaystyle =\frac{3}{14}
simplify by dividing top and bottom by the same number
(iii)
1835×100\displaystyle \dfrac{18}{35}\times100
the Secondary Four row total is the denominator
=51.4\displaystyle =51.4
Final answer
(i)Bus 2222, MRT 3333, walking 1515.
(ii)314\frac{3}{14}
(iii)51.4%51.4\%
Supports S1.2.b · S1.2.c · S1.2.d · S1.2.e · S1.2

Bar graphs, pictograms, line graphs and pie charts

Four diagrams show amounts by size. A bar graph uses bars of equal width. A pictogram uses symbols, each worth the same amount (see the key). A line graph joins points to show change over time. A pie chart splits a circle into sectors in proportion to the data.

The Core Idea

Read the scale and the key before the picture.

Check what the vertical axis counts (or what one symbol is worth), and where the axis starts. Then read, then calculate.

Customers each dayBar chart of customers on five days0510152025303540Number of customersMonTueWedThuFriDay
Customers at a food stall on five days. The bar heights are the frequencies, read against the vertical scale.
Example 3.1

Reading a bar graph

The bar chart shows the number of customers at a food stall on five days.

  • (i)Which day has the highest bar, and what is its value?
  • (ii)Find the total of all the bars.
  • (iii)How many more are there on the busiest day than on the quietest day?
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
tallest bar: Thu  =35\displaystyle \text{tallest bar: }\text{Thu}\;=35
read the top of the bar against the vertical scale
(ii)
total  =15+25+20+35+30=125\displaystyle \text{total}\;=15+25+20+35+30=125
add every bar
(iii)
35−15  =20\displaystyle 35-15\;=20
tallest minus shortest: Thu and Mon
Customers each dayBar chart of customers on five days0510152025303540Number of customersMonTueWedThuFriDay
Customers at a food stall on five days. The bar heights are the frequencies, read against the vertical scale.
Final answer
(i)Thu, with 3535.
(ii)125125
(iii)2020
Library books borrowedPictogram of books borrowed on four daysMonTueWedThu= 20 books
Books borrowed on four days. One symbol stands for 20 books, so a half symbol stands for half of that.
Example 3.2

Reading a pictogram

In a pictogram, one symbol stands for 20 books. The number of symbols for each category is: Mon 3, Tue 4.5, Wed 2.5, Thu 3.5.

  • (i)How many books does Tue stand for?
  • (ii)Which category has the most, and how many is that?
  • (iii)Find the total of all the categories.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
4.5×20  =90\displaystyle 4.5\times20\;=90
number of symbols times the value of one symbol; a half symbol is half the key value
(ii)
4.5×20  =90\displaystyle 4.5\times20\;=90
work out the largest bar-equivalent
(iii)
total  =60+90+50+70=270\displaystyle \text{total}\;=60+90+50+70=270
add the values, not the symbols
Library books borrowedPictogram of books borrowed on four daysMonTueWedThu= 20 books
Books borrowed on four days. One symbol stands for 20 books, so a half symbol stands for half of that.
Final answer
(i)9090 books
(ii)Tue, with 9090 books.
(iii)270270 books
SGSchoolKaki Tip

In a pictogram, a half symbol is half of the key value. Count the whole symbols and the half symbols, then multiply by the key.

Monthly salesLine graph of monthly sales over six months0510152025Sales (hundreds of dollars)JanFebMarAprMayJunMonth
Monthly sales of a small shop. Joining the points shows the trend.
Example 3.3

Reading a line graph

A line graph shows the monthly sales of a shop, in hundreds of dollars. The points are: Jan 12, Feb 15, Mar 14, Apr 19, May 23, Jun 21.

  • (i)In which month is the value highest?
  • (ii)Between which two consecutive months was the increase the greatest, and how big was it?
  • (iii)Is it correct to say the value rose every month? Give a reason.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
highest point: May  =23\displaystyle \text{highest point: }\text{May}\;=23
the highest point on the line
(ii)
increases: 3, −1, 5, 4, −2\displaystyle \text{increases: }3,\allowbreak\ -1,\allowbreak\ 5,\allowbreak\ 4,\allowbreak\ -2
subtract each value from the next
largest: 5 from Mar to Apr\displaystyle \text{largest: }5\ \text{from Mar to Apr}
(iii)
changes: 3, −1, 5, 4, −2\displaystyle \text{changes: }3,\allowbreak\ -1,\allowbreak\ 5,\allowbreak\ 4,\allowbreak\ -2
look at the sign of each change
At least one change is negative, so the line goes down somewhere.
Monthly salesLine graph of monthly sales over six months0510152025Sales (hundreds of dollars)JanFebMarAprMayJunMonth
Monthly sales of a small shop. Joining the points shows the trend.
Final answer
(i)May, with 2323.
(ii)From Mar to Apr, an increase of 55.
(iii)No: at least one month shows a fall.
SGSchoolKaki Tip

Do not read values between the points of a line graph as facts. The points are measured at fixed times; the line in between is a guide to the trend, not a measurement.

Favourite lunchPie chart of favourite lunch among fifty studentsRiceNoodlesSandwichSalad
Favourite lunch among 50 students. Each slice is drawn with the angle (frequency÷total×360\text{frequency}\div\text{total}\times360 degrees).
The Core Idea

Pie chart: angle =frequencytotal×360=\frac{\text{frequency}}{\text{total}}\times360 degrees

All the angles add up to a full turn. The same fraction of the circle is the same fraction of the people, so you can go from angle to frequency, or back.

Example 3.4

Pie charts: angles and fractions

A survey of 50 students asks for their favourite kind of lunch: Rice 20, Noodles 15, Sandwich 10, Salad 5. The results are shown in a pie chart.

  • (i)Find the angle of each sector.
  • (ii)What fraction of the students chose Noodles? Give the fraction in its simplest form.
  • (iii)A sector of angle 108 degrees is drawn for one lunch type. How many students does it represent?
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
total  =20+15+10+5=50\displaystyle \text{total}\;=20+15+10+5=50
add all the frequencies
angle  =frequency50×360\displaystyle \text{angle}\;=\dfrac{\text{frequency}}{50}\times360
each slice is its fraction of a full turn
angles: 144, 108, 72, 36 degrees\displaystyle \text{angles: }144,\allowbreak\ 108,\allowbreak\ 72,\allowbreak\ 36\allowbreak\ \text{degrees}
sum of angles  =144+108+72+36=360\displaystyle \text{sum of angles}\;=144+108+72+36=360
the angles must add up to a full turn
(ii)
108360  =1550\displaystyle \dfrac{108}{360}\;=\dfrac{15}{50}
an angle over a full turn, or the frequency over the total
=310\displaystyle =\frac{3}{10}
simplify
(iii)
108360×50  =15\displaystyle \dfrac{108}{360}\times50\;=15
fraction of a full turn, times the total number of students
Favourite lunchPie chart of favourite lunch among fifty studentsRiceNoodlesSandwichSalad
Favourite lunch among 50 students. Each slice is drawn with the angle (frequency÷total×360\text{frequency}\div\text{total}\times360 degrees).
Final answer
(i)144, 108, 72, 36144,\allowbreak\ 108,\allowbreak\ 72,\allowbreak\ 36 degrees
(ii)310\frac{3}{10}
(iii)1515 students
Think About It
A pie chart has a sector of 90 degrees. What fraction of the whole is that, in its simplest form?
Show the worked answerHide the worked answer
Colour keyadd, subtract, multiply, divide or simplify
90360\displaystyle \dfrac{90}{360}
angle over a full turn
=14\displaystyle =\frac{1}{4}
divide the top and the bottom by 90
Final answerA full circle is 360 degrees, so the fraction is 90 out of 360, which simplifies to 14\frac{1}{4} of the whole, one quarter.
Supports S1.2.f · S1.2.g · S1.2.h · S1.2 · S1.6

Dot diagrams, histograms and stem-and-leaf diagrams

These three diagrams show how the values of one variable are spread out. A dot diagram has one dot per value. A histogram shows grouped continuous data with bars that touch. A stem-and-leaf diagram keeps every value while showing the shape.

Goals scoredDot diagram of goals scored by sixteen students012345Goals scored
One dot for each student. A tall pile means many students scored that number of goals.
Example 4.1

Reading a dot diagram

A dot diagram shows the number of goals scored by each of 16 students. The values, one dot each, are: 0, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5.

  • (i)What is the mode?
  • (ii)Find the range.
  • (iii)How many students scored at or below the value halfway between the lowest and highest scores?
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
most dots: 3, with 5 dots\displaystyle \text{most dots: 3, with 5 dots}
the tallest pile of dots
(ii)
range  =5−0=5\displaystyle \text{range}\;=5-0=5
largest minus smallest
(iii)
middle of range  =0+52=2.5\displaystyle \text{middle of range}\;=\dfrac{0+5}{2}=2.5
halfway between the smallest and largest
dots at or below 2.5  =7\displaystyle \text{dots at or below 2.5}\;=7
count the dots
Goals scoredDot diagram of goals scored by sixteen students012345Goals scored
One dot for each student. A tall pile means many students scored that number of goals.
Final answer
(i)33
(ii)55
(iii)77 students
The Core Idea

Histogram with equal classes: height = frequency.

The bars touch, there are no gaps, and every class has the same width. Read the height against the vertical scale to get the frequency.

Time to complete a puzzleHistogram of puzzle completion times in classes of ten minutes02468101214Frequency010203040Time in minutes
Equal class widths, so the height of each bar is its frequency. The bars touch, because the data are continuous.
Example 4.2

Reading a histogram

The histogram shows the times taken to complete a puzzle for 30 students, in classes of equal width 10 minutes.

  • (i)Which class has the highest frequency?
  • (ii)How many students are in the classes 10≤x<3010\le x<30 together?
  • (iii)Check the total of the frequencies.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
tallest bar: 12\displaystyle \text{tallest bar: 12}
equal class widths, so the height of a bar is its frequency
20≤x<30\displaystyle 20\le x<30
(ii)
10+12  =22\displaystyle 10+12\;=22
add the two bars that make up that range
(iii)
∑f  =6+10+12+2=30\displaystyle \sum f\;=6+10+12+2=30
the bars must add up to the number of items
Time to complete a puzzleHistogram of puzzle completion times in classes of ten minutes02468101214Frequency010203040Time in minutes
Equal class widths, so the height of each bar is its frequency. The bars touch, because the data are continuous.
Final answer
(i)20≤x<3020\le x<30
(ii)2222
(iii)3030
SGSchoolKaki Tip

Do not read a bar height as a frequency unless the classes have equal widths. Every histogram in this chapter has equal widths; with unequal widths the heights would no longer be the frequencies, and the picture would mislead (see the section on misleading diagrams).

Marks in a quizStem and leaf diagram of fifteen quiz marksStemLeaf213373144484257525612 | 3 means 23
Stem-and-leaf diagram of quiz marks. Leaves are in order. The key tells you how to read it.
Example 4.3

Reading a stem-and-leaf diagram

A stem-and-leaf diagram shows the quiz marks of fifteen students. The stems are tens and the leaves are units, so the key is 2∣32\mid3 means 23. The diagram: stem 2: leaves 1, 3, 3, 7; stem 3: leaves 1, 4, 4, 4, 8; stem 4: leaves 2, 5, 7; stem 5: leaves 2, 5; stem 6: leaves 1.

  • (i)How many values are there?
  • (ii)Write down the mode.
  • (iii)Find the median.
  • (iv)Find the range.
Show solutionHide solution
Colour keyadd, subtract, multiply, divide or simplify
(i)
count the leaves: 4+5+3+2+1  =15\displaystyle \text{count the leaves: }4+5+3+2+1\;=15
every leaf is one value
(ii)
The value 34 appears 3 times, more than any other.
mode = the most frequent value
(iii)
ordered: 21, 23, 23, 27, 31, 34, 34, 34, 38, 42, 45, 47, 52, 55, 61\displaystyle \text{ordered: }21,\allowbreak\ 23,\allowbreak\ 23,\allowbreak\ 27,\allowbreak\ 31,\allowbreak\ 34,\allowbreak\ 34,\allowbreak\ 34,\allowbreak\ 38,\allowbreak\ 42,\allowbreak\ 45,\allowbreak\ 47,\allowbreak\ 52,\allowbreak\ 55,\allowbreak\ 61
put the data in order first
n  =15, position=n+12=8\displaystyle n\;=15,\allowbreak\ \qquad \text{position}=\dfrac{n+1}{2}=8
the middle position
median  =the 8th value=34\displaystyle \text{median}\;=\text{the 8th value}=34
an odd number of values: one value sits in the middle
(iv)
range  =61−21=40\displaystyle \text{range}\;=61-21=40
largest value minus smallest value, read from the first and last leaves
Marks in a quizStem and leaf diagram of fifteen quiz marksStemLeaf213373144484257525612 | 3 means 23
Stem-and-leaf diagram of quiz marks. Leaves are in order. The key tells you how to read it.
Final answer
(i)1515
(ii)mode=34\text{mode}=34
(iii)median=34\text{median}=34
(iv)4040
SGSchoolKaki Tip

In a stem-and-leaf diagram, put the leaves in order and always write the key. Without the key, a reader cannot tell whether a stem of two means twenty or two.

Think About It
In a stem-and-leaf diagram with the key 2∣32\mid3 means 23, what does 5∣05\mid0 stand for?
Show the worked answerHide the worked answer
5∣0 ⇒ 5 tens and 0 units\displaystyle 5\mid0\ \Rightarrow\ 5\ \text{tens and}\ 0\ \text{units}
read the stem as tens, the leaf as units
5×10+0  =50\displaystyle 5\times10+0\;=50
Final answerThe stem is the tens digit and the leaf is the units digit, so 5∣05\mid0 stands for 50.
Locked · Chapter 17

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  • Choosing and criticising diagrams
  • Mean, median and mode
  • Quartiles and the cumulative frequency curve
  • Range, IQR and standard deviation
  • Comparing data sets
  • Formula panel
  • Exercise

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Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 17 of 19