Chapter 17 - Continuous Data

Continuous data
17Contents:
A The mean of continuous data [11.5]
B Histograms [11.6]
C Cumulative frequency [11.7]
Opening problem
#endboxedheading
Rainfall (r mm) Frequency
50 6 r < 60 7
60 6 r < 70 20
70 6 r < 80 32
80 6 r < 90 22
90 6 r < 100 9
Total 90
Andriano collected data for the rainfall from the last month for 90
towns in Argentina. The results are displayed in the frequency table
alongside:
Things to think about:
² Is the data discrete or continuous?
² What does the interval 60 6 r < 70 actually mean?
² How can the shape of the distribution be described?
² Is it possible to calculate the exact mean of the data?
Examples of continuous numerical variables are:
The height of year
10 students:
the variable can take any value from about 100 cm to 200 cm.
The speed of cars on
a stretch of highway:
the variable can take any value from 0 km/h to the fastest speed that a car can
travel, but is most likely to be in the range 50 km/h to 150 km/h.
In we saw that a can theoretically take any value on part of
the number line. A continuous variable often has to be so that data can be recorded.
Chapter 13 continuous numerical variable
measured
IGCSE01
magentacyan yellow black
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100
y:HAESEIGCSE01IG01_17353IGCSE01_17.CDR Wednesday, 8 October 2008 9:31:01 AM PETER

354 Continuous data (Chapter 17)
Continuous data is placed into class intervals which are usually represented by inequalities.
For example, for heights in the 140s of centimetres we could write 140 6 h < 150.
To find the mean of continuous data, we use the same method as for grouped discrete data described in
Chapter 13 section F.
Since the data is given in intervals, our answer will only be an estimate of the mean.
Example 1 Self Tutor
Height (h cm) Frequency (f)
130 6 h < 140 2
140 6 h < 150 4
150 6 h < 160 12
160 6 h < 170 20
170 6 h < 180 9
180 6 h < 190 3
The heights of students (h cm) in a hockey
training squad were measured and the results
tabled:
a Estimate the mean height.
b State the modal class.
Height (h cm) Mid-value (x) Frequency (f) fx
130 6 h < 140 135 2 270
140 6 h < 150 145 4 580
150 6 h < 160 155 12 1860
160 6 h < 170 165 20 3300
170 6 h < 180 175 9 1575
180 6 h < 190 185 3 555
Total 50 8140
EXERCISE 17A
Weight (kg) Frequency
75 6 w < 80 2
80 6 w < 85 5
85 6 w < 90 8
90 6 w < 95 7
95 6 w < 100 5
100 6 w < 105 1
1 A frequency table for the weights of a volleyball squad is given
alongside.
a Explain why ‘weight’ is a continuous variable.
b What is the modal class? Explain what this means.
c Describe the distribution of the data.
d Estimate the mean weight of the squad members.
THE MEAN OF CONTINUOUS DATA [11.5]A
a Mean =
P
fx
P
f
¼
8140
50
¼ 163 cm
b The modal class is
160 6 h < 170:
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100
Y:HAESEIGCSE01IG01_17354IGCSE01_17.CDR Monday, 20 October 2008 2:31:24 PM PETER

Continuous data (Chapter 17) 355
Height (mm) Frequency
20 6 h < 40 4
40 6 h < 60 17
60 6 h < 80 15
80 6 h < 100 8
100 6 h < 120 2
120 6 h < 140 4
2 A plant inspector takes a random sample of ten week old plants from
a nursery and measures their height in millimetres.
The results are shown in the table alongside.
a What is the modal class?
b Estimate the mean height.
c How many of the seedlings are 40 mm or more?
d What percentage of the seedlings are between 60 and 80 mm?
e If the total number of seedlings in the nursery is 857, estimate
the number which measure:
i less than 100 mm ii between 40 and 100 mm.
0 6 d < 1 1 6 d < 2 2 6 d < 3 3 6 d < 4 4 6 d < 5
Number of students 76 87 54 23 5
Time (t) 5 6 t < 10 10 6 t < 15 15 6 t < 20 20 6 t < 25 25 6 t < 30 30 6 t < 35
Frequency 2 8 16 20 24 10
When data is recorded for a continuous variable there are likely to be many different values. This data is
therefore organised using class intervals. A special type of graph called a histogram is used to display the
data.
A histogram is similar to a bar chart but, to account for the continuous nature of the variable, the bars are
joined together.
HISTOGRAMS [11.6]B
discrete data continuous data
Column Graph Histogram
3 The distances travelled to school by a random sample of students were:
a What is the modal class?
b Estimate the mean distance travelled by the students.
c What percentage of the students travelled at least 2 km to school?
d If there are 28 students in Josef’s class, estimate the number who travelled less than 1 km to school.
4
a What percentage of the players finished the game in less than 20 minutes?
b Estimate the mean time to finish the game.
c If 2589 other people play the game, estimate the number who will complete it in less than 25
minutes.
The times taken in minutes for players to finish a computer game were:
Distance ( m)k
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100
Y:HAESEIGCSE01IG01_17355IGCSE01_17.CDR Tuesday, 18 November 2008 11:47:22 AM PETER

Height (cm) Frequency (f)
100 6 h < 110 1
110 6 h < 120 1
120 6 h < 130 4
130 6 h < 140 19
140 6 h < 150 25
150 6 h < 160 24
160 6 h < 170 13
170 6 h < 180 5
180 6 h < 190 1
Consider the continuous data opposite which summarises the heights
of girls in year 9. The data is continuous, so we can graph it using a
histogram.
In this case the width of each class interval is the same, so we can
construct a frequency histogram. The height of each column is the
frequency of the class, and the modal class is simply the class with the
highest column.
In some situations we may have class intervals with different widths. There are a couple of reasons why
this may happen:
Height (cm) Frequency (f)
100 6 h < 130 6
130 6 h < 140 19
140 6 h < 150 25
150 6 h < 160 23
160 6 h < 170 13
170 6 h < 190 6
² We may wish to collect small numbers of data at the extremities
of our data range.
For example, to make the table of girls’ heights easier to display,
we may combine the smallest three classes and also the tallest two
classes:
² The data may be naturally grouped in the context of the problem.
For example, a post office will charge different rates depending on the weight of the parcel being sent.
The weight intervals will probably not be equally spaced, but it makes sense for the post office to record
the number of parcels sent in each class.
So, the post office may collect the following data of parcels sent over a week:
Mass (m kg) 0 6 m < 1 1 6 m < 2 2 6 m < 5 5 6 m < 20
Number of parcels 18 22 24 11
In either case, the histogram we draw is not a frequency histogram. The frequency of each class is not
represented by the height of its bar, but rather by its area. The height of each bar is called the frequency
density of the class.
Since frequency = frequency density £ class interval width,
frequency density =
frequency
class interval width
0
10
20
30
100 110 120 130 140 150 160 170 180 190
height (cm)
frequency
The modal class is the class with the highest frequency density, and so it is the highest bar on the histogram.
It is not necessarily the class with the highest frequency.
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100

The table below shows masses of parcels received by a company during one week.
Mass (kg) 0 6 m < 1 1 6 m < 2 2 6 m < 3 3 6 m < 6 6 6 m < 10
Number of parcels 20 30 15 60 40
a Draw a histogram to represent the data.
b Find the modal class interval.
c Use a graphics calculator to estimate the mean mass of the parcels received.
Mass (m kg) Frequency CIW Frequency density
0 6 m < 1 20 1 20
1 6 m < 2 30 1 30
2 6 m < 3 15 1 15
3 6 m < 6 60 3 20
6 6 m < 10 40 4 10
a Histogram showing masses of parcels b
c Middle value (x) Frequency (f)
0:5 20
1:5 30
2:5 15
4:5 60
8 40
mean ¼ 4:14 kg (calculator)
The times taken for students to complete a cross-country run were measured. The results were:
Time (t min) 20 6 t < 23 23 6 t < 26 26 6 t < 31
Frequency 27 51 100
a Draw a histogram to represent the data.
b Find the modal class.
c Estimate the mean time for students to run the event.
Time (t min) Frequency CIW Frequency density
20 6 t < 23 27 3 9
23 6 t < 26 51 3 17
26 6 t < 31 100 5 20
75 10 7:5
CIW means
.class interval width
2 4 6 8 10 12
0
10
20
30
frequency
density
mass ( kg)m
key
represents
5 parcels
31 6 t < 41
75
31 6 t < 41
The modal class is 1 6 m < 2.
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100

a Histogram showing cross-country times b The modal class is 26 6 t < 31.
c Middle value (x) Frequency (f)
21:5 27
24:5 51
28:5 100
36 75
The mean ¼ 29:2 mins.
EXERCISE 17B
1 Students were asked to draw a sketch of their favourite fruit. The time taken (t min) was measured for
each student, and the results summarised in the table below.
Time (t min) 0 6 t < 2 2 6 t < 4 4 6 t < 8 8 6 t < 12
Frequency 10 15 30 60
a Construct a histogram to illustrate the data.
b State the modal class.
c Use a graphics calculator to estimate the mean time.
2
a Represent this data on a histogram.
c
3
Distance (d m) 20 6 d < 25 25 6 d < 35 35 6 d < 45 45 6 d < 55 55 6 d < 85
Frequency 15 30 35 25 15
a Draw a histogram of the data.
b What is the modal class?
c
4
10 20 30 40 50 60
time ( min)t
frequencydensity
Travel times to work
0
Use a graphics calculator to estimate
the mean distance thrown.
The histogram shows the times spent travelling to
work by a sample of employees of a large corporation.
Given that people took between and minutes
to get to work, find the sample size used.
60 10 20
Use technology to estimate the mean mass.
A group of students was asked to throw a baseball as far as they could in a given direction. The results
were recorded and tabled. They were:
Mass (m kg) 40 6 m < 50 50 6 m < 60 60 6 m < 65 65 6 m < 70 70 6 m < 80
Frequency 25 75 60 70 30
When the masses of people in a Singapore fitness club were measured, the results were:
0
5
10
15
20
20 22 24 26 28 30 32 34 36 38 40
time ( min)t
frequencydensity
key
represents
10 students
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100

Sometimes it is useful to know the number of scores that lie above or below a particular value. In
such situations it is convenient to construct a cumulative frequency distribution table and a cumulative
frequency graph to represent the data.
The cumulative frequency gives a running total of the scores up to a particular value.
It is the total frequency up to a particular value.
From a frequency table we can construct a cumulative frequency column and then graph this data on a
cumulative frequency curve. The cumulative frequencies are plotted on the vertical axis.
From the cumulative frequency graph we can find:
² the median Q2
² the quartiles Q1 and Q3
² percentiles
¾
These divide the ordered data into quarters.
The median Q2 splits the data into two halves, so it is 50% of the way through the data.
The first quartile Q1 is the score value 25% of the way through the data.
The third quartile Q3 is the score value 75% of the way through the data.
The nth percentile Pn is the score value n% of the way through the data.
So, P25 = Q1, P50 = Q2 and P75 = Q3.
Weight (w kg) Frequency
65 6 w < 70 1
70 6 w < 75 2
75 6 w < 80 8
80 6 w < 85 16
85 6 w < 90 21
90 6 w < 95 19
95 6 w < 100 8
100 6 w < 105 3
105 6 w < 110 1
110 6 w < 115 1
The data shown gives the weights of 80 male basketball players.
a Construct a cumulative frequency distribution table.
b Represent the data on a cumulative frequency graph.
c Use your graph to estimate the:
i median weight
ii number of men weighing less than 83 kg
iii number of men weighing more than 92 kg
iv 85th percentile.
a Weight (w kg) frequency cumulative frequency
65 6 w < 70 1 1
70 6 w < 75 2 3
75 6 w < 80 8 11
80 6 w < 85 16 27
85 6 w < 90 21 48
90 6 w < 95 19 67
95 6 w < 100 8 75
100 6 w < 105 3 78
105 6 w < 110 1 79
110 6 w < 115 1 80
this is 1 + 2
this is 1 + 2 + 8
CUMULATIVE FREQUENCY [11.7]C
this 48 means that there are 48
players who weigh less than 90 kg,
so (90, 48) is a point on the
cumulative frequency graph.
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100

b
Cumulative frequency graphs are very useful for comparing two distributions of unequal sizes. In such
cases we use percentiles on the vertical axis. This effectively scales each graph so that they both range from
0 to 100 on the vertical axis.
The heights of 100 14-year-old girls and 200 14-year-old boys were measured and the results tabled.
Frequency (girls) Height (h cm) Frequency (boys)
5 140 6 h < 145 4
10 145 6 h < 150 10
15 150 6 h < 155 20
30 155 6 h < 160 26
20 160 6 h < 165 40
10 165 6 h < 170 60
8 170 6 h < 175 30
2 175 6 h < 180 10
a Draw on the same axes the
cumulative frequency curve for both
the boys and the girls.
b Estimate for both the boys and the
girls:
i the median
ii the interquartile range (IQR).
c Compare the two distributions.
0
10
20
30
40
50
60
70
70 80 90 100 110
cumulative frequency
weight (kg)
60 120
80
6868
Cumulative frequency graph of basketballers’ weights
5656
8383
9696
median is kg¡»¡88
c i
ii
iii
iv
of ,
median kg
There are men who
weigh less than kg.
There are
men who weigh more
than kg.
of , so the
th percentile kg.
50% 80 = 40
88
20
83
80 56=24
92
85% 80 = 68
85 96
) ¼
¡
¼
¡ ¡ ¡ ¡
9292
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100
Y:HAESEIGCSE01IG01_17360IGCSE01_17.CDR Monday, 20 October 2008 2:37:38 PM PETER

b For girls: i median ¼ 158 cm ii
For boys: i median ¼ 165 cm ii
c The two distributions are similar in shape, but the boys’ heights are further right than the girls’.
The median height for the boys is 7 cm more than for the girls. They are considerably taller.
As the IQRs are nearly the same, the spread of heights is similar for each gender.
a
14
200 = 7%,
CF (girls) Freq. (girls) Height (h cm) Freq. (boys) CF (boys)
5 5 140 6 h < 145 4 4
15 10 145 6 h < 150 10 14
30 15 150 6 h < 155 20 34
60 30 155 6 h < 160 26 60
80 20 160 6 h < 165 40 100
90 10 165 6 h < 170 60 160
98 8 170 6 h < 175 30 190
100 2 175 6 h < 180 10 200
so , is a
point on the boys’
cumulative
frequency graph.
(150 7)
0
10
20
30
40
50
60
70
80
90
100
140 145 150 155 160 165 170 175 180
height ( cm)h
percentiles Cumulative frequency graph of
boys’ and girls’ heights
IQR ¼ 163:5 ¡ 153:7 ¼ 10 cm
IQR ¼ 169 ¡ 158 ¼ 11 cm
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100
y:HAESEIGCSE01IG01_17361IGCSE01_17.CDR Friday, 10 October 2008 9:27:15 AM PETER

EXERCISE 17C
Times (min) Frequency
20 6 t < 25 18
25 6 t < 30 45
30 6 t < 35 37
35 6 t < 40 33
40 6 t < 45 19
45 6 t < 50 8
1 In a running race, the times (in minutes) of 160 competitors were
recorded as follows:
Draw a cumulative frequency graph of the data and use it to estimate:
a the median time
b the approximate number of runners whose time was not more than
32 minutes
c the approximate time in which the fastest 40 runners completed
the course
d the interquartile range.
2
3 The lengths of 30 trout (l cm) were measured. The following data was obtained:
Length (cm) 306 l <32 326 l <34 346 l <36 366 l <38 386 l <40 406 l <42 426 l <44
Frequency 1 1 3 7 11 5 2
a Construct a cumulative frequency curve for the data.
b Estimate the percentage of trout with length less than 39 cm.
c Estimate the median length of trout caught.
d Estimate the interquartile range of trout length and explain what this represents.
e Estimate the 35th percentile and explain what this represents.
f Use a calculator to estimate the mean of the data.
g Comment on the shape of the distribution of trout lengths.
0
20
40
60
80
100
120
0 10 20 30 40 50 60 70
weight ( kg)w
The cumulative frequency
curve shows the weights of
Sam’s goat herd in kilograms.
How many goats does
Sam have?
Estimate the median goat
weight.
Any goats heavier than
the th percentile will
go to market. How many
goats will go to market?
What is the IQR for
Sam’s herd?
60
a
b
c
d
10
30
50
70
90
110
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100
y:HAESEIGCSE01IG01_17362IGCSE01_17.CDR Thursday, 23 October 2008 12:19:05 PM PETER

5 The times taken for trampers to climb Ben Nevis were recorded and the results tabled.
Time (t min) 175 6 t < 190 190 6 t < 205 205 6 t < 220 220 6 t < 235 235 6 t < 250
Frequency 11 35 74 32 8
a Construct a cumulative frequency curve for the walking times.
b Estimate the median time for the walk.
c Estimate the IQR and explain what it means.
d Guides on the walk say that anyone who completes the walk in 3 hours 15 min or less is extremely
fit. Estimate the number of extremely fit trampers.
e Comment on the shape of the distribution of walking times.
6
0
5
10
15
20
25
30
35
40
0 2 4 6 8 10
weight ( kg)w
Cumulative frequency curve of watermelon weight data
Frequency
(Alan)
Weight
(w grams)
Frequency
(John)
4 400 6 w < 550 5
32 550 6 w < 700 60
44 700 6 w < 850 70
52 850 6 w < 1000 60
44 1000 6 w < 1150 35
24 1150 6 w < 1300 20
200 totals 250
4 The weights of cabbages grown by two brothers on
separate properties were measured for comparison.
a Draw, on the same axes, cumulative frequency
curves for both cabbage samples.
b Estimate for each brother:
i the median weight
ii the IQR
c Compare the 60th percentile weights.
d Compare the two distributions.
The given graph describes the
weight of watermelons.
Estimate the:
median weight
IQR
for the weight of the
watermelons.
Construct a cumulative
frequency table for the
data including a
frequency column.
Estimate the mean
weight of the
watermelons.
40
a
i
ii
b
c
The results are shown in the table:
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100

3 A selection of measuring bottles were examined and their capacities were noted. The results are
given in the table below:
a Draw a histogram to illustrate this information. b What is the modal class?
4
a Represent this data on a histogram.
c Estimate the mean height of the plants.
5 The weekly wages of employees in a factory are recorded in the table below.
a Draw a cumulative frequency graph to illustrate this information.
b Use the graph to estimate:
i the median wage
ii the wage that is exceeded by 20% of the employees.
Capacity (C litres) 0 6 C < 0:5 0:5 6 C < 1 1 6 C < 2 2 6 C < 3 3 6 C < 5
Frequency 13 18 24 18 16
Height (h cm) Frequency
0 6 h < 10 11
10 6 h < 20 14
20 6 h < 30 20
30 6 h < 40 15
40 6 h < 60 18
60 6 h < 100 10
Weekly wage ($w) 06w<400 4006w<800 8006w<1200 12006w<1600 16006w<2000
Frequency 20 60 120 40 10
The heights of plants in a field were measured and the
results recorded alongside:
Review set 17A
#endboxedheading
1 A frequency table for the masses of eggs (m grams) in a carton
marked ‘50 g eggs’ is given below.
a Explain why ‘mass’ is a continuous variable.
b What is the modal class? Explain what this means.
c Estimate the mean of the data.
d Describe the distribution of the data.
2 The speeds of vehicles (v km/h) travelling along a stretch of road
are recorded over a 60 minute period. The results are given in
the table alongside.
a Estimate the mean speed of the vehicles.
c What percentage of drivers exceeded the speed limit of
60 km/h?
d Describe the distribution of the data.
Speed (v km/h) Frequency
40 6 v < 45 14
45 6 v < 50 22
50 6 v < 55 35
55 6 v < 60 38
60 6 v < 65 25
65 6 v < 70 10
Mass (g) Frequency
48 6 m < 49 1
49 6 m < 50 1
50 6 m < 51 16
51 6 m < 52 4
52 6 m < 53 3
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100

Review set 17B
1 The table alongside summarises the masses of 50 domestic cats
chosen at random.
a What is the length of each class interval?
c Find the approximate mean.
d Draw a frequency histogram of the data.
e From a random selection of 428 cats, how many would you
expect to weigh at least 8 kg?
2 The table alongside summarises the best times of 100 swimmers who
swim 50 m.
a Estimate the mean time.
Mass (m kg) Frequency
0 6 m < 2 5
2 6 m < 4 18
4 6 m < 6 12
6 6 m < 8 9
8 6 m < 10 5
10 6 m < 12 1
Frequency
25 6 t < 30 5
30 6 t < 35 17
35 6 t < 40 34
40 6 t < 45 29
45 6 t < 50 15
Time ( sec)t
6
0
10
20
30
40
50
60
70
80
90
0 10 20 30 40 50 60
time ( minutes)t
The cumulative frequency curve
shows the time spent by people
in a supermarket on a given day.
Construct a cumulative
frequency table for the
data, using the intervals
, , and
so on.
Use the graph to estimate:
the median time
the IQR
the th percentile.
Copy and complete:
of the people
spent less than ......
minutes in the
supermarket
of the people
spent at least ......
minutes in the
supermarket.
0 5 5 10
80
60%
80%
¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡6 6t < t <
a
b
i
ii
iii
c
i
ii
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100

6 In a one month period at a particular hospital the lengths of newborn
babies were recorded. The results are shown in the table given.
a Represent the data on a frequency histogram.
b How many babies are 52 cm or more?
c What percentage of babies have lengths in the interval
50 cm 6 l < 53 cm?
d Construct a cumulative frequency distribution table.
e Represent the data on a cumulative frequency graph.
f Use your graph to estimate the:
i median length
ii number of babies with length less than 51:5 cm.
Frequency
48 6 l < 49 1
49 6 l < 50 3
50 6 l < 51 9
51 6 l < 52 10
52 6 l < 53 16
53 6 l < 54 4
54 6 l < 55 5
55 6 l < 56 2
Length ( cm)l
3 The table below displays the distances jumped by 50 year 10 students in a long jump competition:
a Display this information on a histogram.
4 The histogram alongside shows the areas of land
blocks on a street. If 20 land blocks were between
300 m2
to 500 m2
in size:
a construct a frequency table for the data
b
5 The percentage scores in a test were recorded. The
results were categorised by gender.
a Draw the cumulative frequency graphs for boys
and girls on the same set of axes. Use percentiles
on the vertical axis.
b Estimate the median and interquartile range of
each data set.
c Compare the distributions.
300 500 700 900 1100
frequencydensity
area (mX)
Distance (d m) 3 6 d < 4 4 6 d < 5 5 6 d < 5:5 5:5 6 d < 6 6 6 d < 7
Frequency 8 16 12 9 5
Frequency
(boys)
Percentage
score (s)
Frequency
(girls)
5 0 6 s < 10 0
8 10 6 s < 20 4
12 20 6 s < 30 8
10 30 6 s < 40 10
30 40 6 s < 50 15
50 50 6 s < 60 25
20 60 6 s < 70 40
10 70 6 s < 80 10
5 80 6 s < 90 5
0 90 6 s < 100 3
estimate the mean area.
IGCSE01
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100

ANSWERS 703
2 a
3b + 2a
ab
b
4d + 3a
ad
c
5b ¡ 3a
ab
d
a
m
e
3a + b
3y
f
3
2a
g
3b ¡ 2
ab
h
dc + ab
ad
i
4 + a
b
j
2d ¡ ac
ad
k
15 + x2
3x
l
pd ¡ 12
6d
m
m2 + 3n
3m
n
2mn ¡ mp
np
o
17b
20
p
16b
15
3 a
x + 6
3
b
m ¡ 2
2
c
4a
3
d
b ¡ 10
5
e
x ¡ 18
6
f
12 + x
4
g
30 ¡ x
6
h
2x + 3
x
i
6x ¡ 3
x
j
b2 + 3
b
k
5 + x2
x
l ¡
11y
6
4 a
14x
15
b
11x
35
c
11
2a
d
21
4y
e
3c + 4b
bc
f
5b ¡ 24a
4ab
g
x + 30
10
h
12 ¡ x
3
EXERCISE 16D.1
1 a
9x ¡ 4
20
b
5x + 10
6
c
20x ¡ 7
42
d
a + 5b
6
e
13x ¡ 9
20
f
5x + 11
14
g
x + 15
30
h
x ¡ 7
42
i
2 ¡ 3x
10
j
3x ¡ 1
12
k
2x + 1
15
l
25x + 5
24
2 a
5x ¡ 1
(x + 1)(x ¡ 2)
b
12x + 17
(x + 1)(x + 2)
c
x + 14
(x ¡ 1)(x + 2)
d
¡6
(x + 2)(2x + 1)
e
7x + 8
(x ¡ 1)(x + 4)
f
3(5x + 2)
(1 ¡ x)(x + 2)
g
4x + 3
x(x + 1)
h
3(x + 5)
x(x + 3)
i
x2 ¡ x + 6
(x + 2)(x ¡ 4)
j
2(x ¡ 1)
x ¡ 3
k
2(x ¡ 1)
x + 2
l
2x2 + 4x ¡ 3
(x + 3)(x + 2)
m
7
(2x ¡ 1)(x + 3)
n
17x ¡ 7
x(3x ¡ 1)
o
¡5x ¡ 2
(x + 2)(x ¡ 2)
p
x + 2
x(x + 1)
q
x2 + 1
x(x ¡ 1)(x + 1)
r
4x2 ¡ x ¡ 9
(x + 1)(x ¡ 1)(x + 2)
s
2x3 ¡ x2 + 1
x(x + 1)(x ¡ 1)
3 a
6
x + 3
b 11
4
c
4
x ¡ 4
d 5
e
x ¡ 1
x + 2
f
x ¡ 2
x
g
x ¡ 1
3x + 5
h
2x + 3
x ¡ 1
4 a cos µ =
x + 2
x2
, sin µ =
x ¡ 1
x2
, tan µ =
x ¡ 1
x + 2
b sin µ¥cos µ =
x ¡ 1
x2
¥
x + 2
x2
=
x ¡ 1
x2
£
x2
x + 2
=
x ¡ 1
x + 2
EXERCISE 16D.2
1 a
2 + x
x(x + 1)
b
2 + x2
x(x + 1)
c
2(x2 + 2x + 2)
(x + 2)(x ¡ 3)
d
2(x + 5)
x + 2
e
x2 ¡ 2x + 3
(x ¡ 2)(x + 3)
f
x ¡ 5
x ¡ 2
g
2(x ¡ 5)
x ¡ 1
h
x + 14
x + 7
2 a
2(x + 1)2
(x + 2)(x ¡ 3)
b i x = ¡2 or 3 ii x = ¡1
3 a
¡2
x ¡ 2
b
2
x + 4
c
x ¡ 2
x + 2
d
x ¡ 6
2 ¡ x
e
¡(x + 2)
4x2
f
12 ¡ x
16x2
4 a
x + 3
3(x + 1)
b i x = ¡1 or 2 ii x = ¡3
REVIEW SET 16A
1 a 3x b 3n c
x
6
d
2
x
2 a
2
c + 3
b cannot be simplified c x + 2 d
x
3(x + 2)
3 a
19x
15
b
2x2
5
c 10
9
d
x
15
4 a 4 b ¡5 c 2x
5 a
11x + 1
12
b
16x ¡ 9
14
c
3x + 2
x(x + 2)
6 a
¡2
x + 4
b
x + 3
x
c
2x + 1
3x + 2
7 a
2(x ¡ 2)
x + 1
b i x = ¡3 or ¡1 ii x = 2
REVIEW SET 16B
1 a 2
3
b 2x c 3n d 2x
2 a cannot be simplified b x + 5 c
2
a + 4
d
b
2(b ¡ a)
3 a
11x
4
b
¡5x
4
c
3x2
2
d 3
8
4 a ¡1 b 5
2
c
3x
a
5 a
13x ¡ 5
15
b ¡
5x + 3
6
c
x + 6
2x(x + 2)
6 a 2(x ¡ 2) b
x ¡ 7
x ¡ 2
c
3x + 1
4x + 1
7 a
¡3(x ¡ 4)
x ¡ 2
b i x = §2 ii x = 4
8 a
¡10
x ¡ 1
b
x + 3
x
EXERCISE 17A
1 a The variable can take any value in the continuous range 75
to 105.
b 85 6 w < 90. This class has the highest frequency.
c symmetrical d ¼ 89:5 kg
2 a 40 6 h < 60 b ¼ 69:6 mm c 46 of them d 30%
e i ¼ 754 ii ¼ 686
3 a 1 6 d < 2 b ¼ 1:66 km c 33:5% d 9 students
4 a 32:5% b ¼ 22:9 min c ¼ 1490 people
IB MYP_3 ANS
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100
Y:HAESEIGCSE01IG01_an703IB_IGC1_an.CDR Wednesday, 19 November 2008 9:12:59 AM PETER

704 ANSWERS
EXERCISE 17B
1 a b 8 6 t < 12
c ¼ 7:26 min
2 a b 65 6 m < 70
c ¼ 61:4 kg
3 a b 35 6 d < 45
c ¼ 41:1 m
4 430 employees
EXERCISE 17C
1
a 32 min b 80 c 28 min d IQR = 10 min
2 a 120 b 29 kg c 48 d 17:5 kg
3 a
b 57% c 38:6 cm d 3:0 cm
e 37:6 cm. 35% of the lengths are less than or equal to this
value.
f ¼ 38:3 cm g negatively skewed
4 a
b i Alan, ¼ 910; John, ¼ 830
ii Alan, ¼ 310; John, ¼ 290
c Alan, ¼ 970; John, ¼ 890
d Alan’s cabbages are generally heavier than John’s. The spread
of each data set is about the same.
5 a
b ¼ 210 min
c 171
2
min. This is the length of time in which the middle 50%
of the data lies.
d ¼ 20 e symmetrical
6 a i 4 kg
ii 2:0 kg
c ¼ 4:25 kg
b Weight (w grams) Freq. Cum. Freq.
0 6 w < 1 1 1
1 6 w < 2 2 3
2 6 w < 3 5 8
3 6 w < 4 12 20
4 6 w < 5 8 28
5 6 w < 6 6 34
6 6 w < 7 3 37
7 6 w < 8 2 39
8 6 w < 9 1 40
0
5
10
15
20
0 2 4 6 8 10 12
t (min)
frequencydensity
m (kg)0
5
10
15
20 30 40 50 60 70 80
frequencydensity
0
1
2
3
4
20 30 40 50 60 70 80 90
d (m)
frequencydensity
0
20
40
60
80
100
120
140
160
180
20 25 30 35 40 45 50
time ( min)t
cumulativefrequency
Cumulative frequency graph of race data
0
5
10
15
20
25
30
35
30 32 34 36 38 40 42 44
cumulativefrequency
length (cm)
Cumulative frequency graph of trout length
0
20
40
60
80
100
120
400 550 700 850 1000 1150 1300
John
Alan
weight ( grams)w
percentiles
Cumulative frequency graph of cabbage weight data
0
20
40
60
80
100
120
140
160
180
175 190 205 220 235 250
Cumulative frequency graph of Ben Nevis climb data
cumulativefrequency
time (min)
IB MYP_3 ANS
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100
Y:HAESEIGCSE01IG01_an704IB_IGC1_an.CDR Friday, 21 November 2008 9:25:46 AM PETER

0
0.5
1
1.5
2
2.5
0 20 40 60 80 100
height ( cm)h
frequencydensity ANSWERS 705
REVIEW SET 17A
1 a The variable can take any value in the continuous range
48 6 m < 53 grams.
b 50 6 m < 51 c ¼ 50:8 grams
d slightly negatively skewed
2 a ¼ 54:9 km/h b 55 6 v < 60 c ¼ 24:3%
d symmetrical
3 a b 0:5 6 C < 1
4 a b 20 6 h < 30
c ¼ 34:0 cm
5 a
b i ¼ $950 ii ¼ $1200
6 a Time (t min) Freq. Cum. Freq.
0 6 t < 5 2 2
5 6 t < 10 3 5
10 6 t < 15 5 10
15 6 t < 20 10 20
20 6 t < 25 20 40
25 6 t < 30 15 55
30 6 t < 35 5 60
35 6 t < 40 10 70
40 6 t < 45 6 76
45 6 t < 50 4 80
b i 25 min
ii 15 min
iii 37 min
c i 27
ii 18
REVIEW SET 17B
1 a 2 kg b 2 6 m < 4 c ¼ 4:76
d e 377 of them
2 a ¼ 39:1 sec b 35 6 t < 40
3 a
b 5 6 d < 5:5
4 a Area (A m2) Frequency
300 6 A < 500 20
500 6 A < 600 20
600 6 A < 700 35
700 6 A < 800 25
800 6 A < 1100 45
b ¼ 712 m2
5 a Percentiles
UE point CF (boys) CF (girls) B G
10 5 0 3:1 0:0
20 3 4 8:1 3:3
30 25 12 15:6 10:0
40 35 22 21:9 18:3
50 75 37 46:9 30:8
60 125 62 78:1 51:7
70 145 102 90:6 85
80 155 112 96:9 93:3
90 160 117 100:0 97:5
100 160 120 100:0 100:0
b For boys: medium ¼ 52, IQR ¼ 19
For girls: medium ¼ 59, IQR ¼ 22
c As the girls graph is further to the right of the boys graph,
the girls are outperforming the boys. Both distributions are
negatively skewed.
0
10
20
30
40
0 1 2 3 4 5
capacity ( litres)C
frequencydensity
0
50
100
150
200
250
300
400 800 1200 1600 2000
cumulativefrequency
Cumulative frequency graph of wage data
wage ( )$
0
10
20
30
3 4 5 6 7
d (m)
frequencydensity
0
20
40
60
80
100
0 10 20 30 40 50 60 70 80 90 100
scores (%)
percentiles
boys
girls
0
5
10
15
20
0 2 4 6 8 10 12
Histogram of masses of cats
m (kg)
frequency
Histogram of long jump data
Cumulative frequency graph of test scores
IB MYP_3 ANS
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100
Y:HAESEIGCSE01IG01_an705IB_IGC1_an.CDR Thursday, 20 November 2008 4:17:48 PM PETER

0
5
10
15
20
48 50 52 54 56
length (cm)
frequency
706 ANSWERS
6 a b 27 babies
c 70% of them
d Upper end point Cumulative frequency
49 1
50 4
51 13
52 23
53 39
54 43
55 48
56 50
e
f i ¼ 52:1 cm ii 18 babies
EXERCISE 18A
1 a x = 0:8 b x ¼ 1:13 c x ¼ 0:706
2 a x = 8 b x = 8:75 c x = 4:8 d x ¼ 3:18
3 a 6 cm b k = 1:5
4 a true b false, e.g.,
c true d false, e.g.,
5 They are not similar.
Comparing lengths; k = 16
12
= 4
3
Comparing widths; k = 12
8
= 3
2
and 4
3
6= 3
2
.
6 FG = 2:4 m
7 a
b
and
8 No, any two equiangular triangles are similar.
EXERCISE 18B.1
1 All of these figures have triangles which can be shown to be
equiangular and therefore are similar.
For example, in a, ¢CBD is similar to ¢CAE as they share an
equal angle at C and CbBD = CbAE = 90o.
EXERCISE 18B.2
1 a x = 2:4 b x = 2:8 c x ¼ 3:27 d x = 9:6
e x = 11:2 f x = 5 g x ¼ 6:67 h x = 7
i x = 7:2
EXERCISE 18C
1 a 7 m b 7:5 m 2 1:8 m 3 2:67 m 4 1:52 m
5 1:44 m 6 9 seconds 7 ¼ 117 m 8 1013 m
9 a SU = 5:5 m, BC = 8:2 m
b No, the ball’s centre is ¼ 11 cm on the D side of C.
EXERCISE 18D
1 a x = 18 b x = 6 c x = 5 d x ¼ 4:38
2 a k = 4 b 20 cm and 24 cm c area A : area B = 1 : 16
3 a k = 2:5 b 100 cm2 c 84 cm2 4 ED ¼ 1:45
5 a V = 80 b V = 40:5 c x ¼ 5:26 d x = 8
6 6750 cm3 7 a 4 cm b 648 cm3 8 6 cm2
9 a k = 2
3
b 14 850 cm3 c 280 cm2
10 No. Comparing capacities, k ¼ 1:37
Comparing lengths, k = 1:6
These values should be the same if the containers are similar.
11 a 2:5 m b 16 : 25 c 64 : 125
REVIEW SET 18A
1
and
2 a x ¼ 1:71 b x ¼ 1:83 3 x = 2:8
4 Hint: Carefully show that triangles are equiangular, giving
reasons.
5 a x ¼ 6:47 b x = 2
p
6 ¼ 4:90
6 a A = 7 b x ¼ 8:14 7 a x = 15 b y = 32
8 ¼ 66:7 m wide 9 a k = 4 b 0:99 m c 0:5 m3
REVIEW SET 18B
1
2 a k = 7
5
b 49 : 25
3 a x = 3 b x = 4 c x = 12
4 a BbAC = NbMC = 90o fgiveng
]C is common to both ) ¢s ABC and MNC are
equiangular, i.e., similar.
b
x
8
=
6
15
) x = 48
15
= 3:2 c 6:4 cm
5 a x = 4 b x ¼ 42:7 6 a x = 3:6 b y = 6:4
7 a Hint: Explain carefully, with reasons, why they are
equiangular.
b CD = 7:2 cm c 22:4 cm2
8 2
p
13 ¼ 7:21 cm by 3
p
13 ¼ 10:8 cm 9 648 cm3
CHALLENGE
1 ¼ 17:1 m 2 3:75 m
0
5
10
15
20
25
30
35
40
45
50
55
48 49 50 51 52 53 54 55 56
length (cm)
frequency
Cumulative frequency graph of lengths of newborns
Histogram of lengths of newborn babies
4
4
22
3
3
1Qw_ 1Qw_
8 cm
2 cm
8 cm
4 cm
12 m
3 m
12 m
9 m
and
8 cm
4 cm
8 cm
4 cm
IB MYP_3 ANS
0
0
5
5
25
25
75
75
50
50
95
95
100
100
0
0
5
5
25
25
75
75
50
50
95
95
100
100
Y:HAESEIGCSE01IG01_an706IB_IGC1_an.CDR Thursday, 20 November 2008 4:19:17 PM PETER

Chapter 17 - Continuous Data

Recommended

Recommended

More Related Content

Viewers also liked

Viewers also liked (14)

Similar to Chapter 17 - Continuous Data

Similar to Chapter 17 - Continuous Data (20)

More from Sarah Sue Calbio

More from Sarah Sue Calbio (20)

Recently uploaded

Recently uploaded (20)

Chapter 17 - Continuous Data