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 90towns 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

IGCSE01magentacyan yellow black

0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

y:\HAESE\IGCSE01\IG01_17\353IGCSE01_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 =

PfxPf

¼ 8140

50

¼ 163 cm

b The modal class is

160 6 h < 170:


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_17\354IGCSE01_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 25minutes.

The times taken in minutes for players to finish a computer game were:

Distance ( m)k


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_17\355IGCSE01_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.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 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 10 120

10

20

30

freq

uen

cyden

sity

mass ( kg)m

keyrepresents5 parcels

31 6 t < 41

75

31 6 t < 41

The modal class is 1 6 m < 2.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 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 60time ( min)t

freq

uen

cyden

sity

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 40time ( min)t

freq

uen

cyden

sity key

represents10 students


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 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

¾These 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 < 70 170 6 w < 75 275 6 w < 80 880 6 w < 85 1685 6 w < 90 2190 6 w < 95 1995 6 w < 100 8100 6 w < 105 3105 6 w < 110 1110 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 1 170 6 w < 75 2 375 6 w < 80 8 1180 6 w < 85 16 2785 6 w < 90 21 4890 6 w < 95 19 6795 6 w < 100 8 75100 6 w < 105 3 78105 6 w < 110 1 79110 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 48players who weigh less than 90 kg,

so (90, 48) is a point on the

cumulative frequency graph.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 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

83839696

median is kg��

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 = 4088

2083

80 56=24

92

85% 80 = 6885 96

) ¼

¡

¼

� � � �

9292


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_17\360IGCSE01_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

14200 = 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


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

y:\HAESE\IGCSE01\IG01_17\361IGCSE01_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

cumulative frequencyThe 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


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

y:\HAESE\IGCSE01\IG01_17\362IGCSE01_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:


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 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


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 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


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 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

freq

uen

cyden

sity

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.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


ANSWERS 703

2 a3b+ 2a

abb

4d+ 3a

adc

5b¡ 3a

abd

a

m

e3a+ b

3yf

3

2ag

3b¡ 2

abh

dc+ ab

ad

i4 + a

bj

2d¡ ac

adk

15 + x2

3xl

pd¡ 12

6d

mm2 + 3n

3mn

2mn¡mp

npo

17b

20p

16b

15

3 ax+ 6

3b

m¡ 2

2c

4a

3d

b¡ 10

5

ex¡ 18

6f

12 + x

4g

30¡ x

6h

2x+ 3

x

i6x¡ 3

xj

b2 + 3

bk

5 + x2

xl ¡11y

6

4 a14x

15b

11x

35c

11

2ad

21

4y

e3c+ 4b

bcf

5b¡ 24a

4abg

x+ 30

10h

12¡ x

3

EXERCISE 16D.1

1 a9x¡ 4

20b

5x+ 10

6c

20x¡ 7

42d

a+ 5b

6

e13x¡ 9

20f

5x+ 11

14g

x+ 15

30h

x¡ 7

42

i2¡ 3x

10j

3x¡ 1

12k

2x+ 1

15l

25x+ 5

24

2 a5x¡ 1

(x+ 1)(x¡ 2)b

12x+ 17

(x+ 1)(x+ 2)

cx+ 14

(x¡ 1)(x+ 2)d

¡6

(x+ 2)(2x+ 1)

e7x+ 8

(x¡ 1)(x+ 4)f

3(5x+ 2)

(1¡ x)(x+ 2)

g4x+ 3

x(x+ 1)h

3(x+ 5)

x(x+ 3)i

x2 ¡ x+ 6

(x+ 2)(x¡ 4)

j2(x¡ 1)

x¡ 3k

2(x¡ 1)

x+ 2l

2x2 + 4x¡ 3

(x+ 3)(x+ 2)

m7

(2x¡ 1)(x+ 3)n

17x¡ 7

x(3x¡ 1)

o¡5x¡ 2

(x+ 2)(x¡ 2)p

x+ 2

x(x+ 1)

qx2 + 1

x(x¡ 1)(x+ 1)r

4x2 ¡ x¡ 9

(x+ 1)(x¡ 1)(x+ 2)

s2x3 ¡ x2 + 1

x(x+ 1)(x¡ 1)

3 a6

x+ 3b 1 1

4c

4

x¡ 4d 5

ex¡ 1

x+ 2f

x¡ 2

xg

x¡ 1

3x+ 5h

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 a2 + x

x(x+ 1)b

2 + x2

x(x+ 1)c

2(x2 + 2x+ 2)

(x+ 2)(x¡ 3)

d2(x+ 5)

x+ 2e

x2 ¡ 2x+ 3

(x¡ 2)(x+ 3)f

x¡ 5

x¡ 2

g2(x¡ 5)

x¡ 1h

x+ 14

x+ 7

2 a2(x+ 1)2

(x+ 2)(x¡ 3)b i x = ¡2 or 3 ii x = ¡1

3 a¡2

x¡ 2b

2

x+ 4c

x¡ 2

x+ 2

dx¡ 6

2¡ xe

¡(x+ 2)

4x2f

12¡ x

16x2

4 ax+ 3

3(x+ 1)b i x = ¡1 or 2 ii x = ¡3

REVIEW SET 16A

1 a 3x b 3n cx

6d

2

x

2 a2

c+ 3b cannot be simplified c x+ 2 d

x

3(x+ 2)

3 a19x

15b

2x2

5c 10

9d

x

15

4 a 4 b ¡5 c 2x

5 a11x+ 1

12b

16x¡ 9

14c

3x+ 2

x(x+ 2)

6 a¡2

x+ 4b

x+ 3

xc

2x+ 1

3x+ 2

7 a2(x¡ 2)

x+ 1b i x = ¡3 or ¡1 ii x = 2

REVIEW SET 16B

1 a 23

b 2x c 3n d 2x

2 a cannot be simplified b x+ 5 c2

a+ 4d

b

2(b¡ a)

3 a11x

4b

¡5x

4c

3x2

2d 3

8

4 a ¡1 b 52

c3x

a

5 a13x¡ 5

15b ¡5x+ 3

6c

x+ 6

2x(x+ 2)

6 a 2(x¡ 2) bx¡ 7

x¡ 2c

3x+ 1

4x+ 1

7 a¡3(x¡ 4)

x¡ 2b i x = §2 ii x = 4

8 a¡10

x¡ 1b

x+ 3

x

EXERCISE 17A

1 a The variable can take any value in the continuous range 75to 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 ANSmagentacyan yellow black

0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_an\703IB_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 < 70c ¼ 61:4 kg

3 a b 35 6 d < 45c 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 830ii 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 17 12

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 11 6 w < 2 2 32 6 w < 3 5 83 6 w < 4 12 204 6 w < 5 8 285 6 w < 6 6 346 6 w < 7 3 377 6 w < 8 2 398 6 w < 9 1 40

0

5

10

15

20

0 2 4 6 8 10 12t (min)fr

equen

cyden

sity

m (kg)0

5

10

15

20 30 40 50 60 70 80

freq

uen

cyden

sity

0

1

2

3

4

20 30 40 50 60 70 80 90d (m)

freq

uen

cyden

sity

0

20

40

60

80

100

120

140

160

180

20 25 30 35 40 45 50

time ( min)t

cum

ula

tive

freq

uen

cy

Cumulative frequency graph of race data

0

5

10

15

20

25

30

35

30 32 34 36 38 40 42 44

cum

ula

tive

freq

uen

cy

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)wper

centi

les

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

cum

ula

tive

freq

uen

cy

time (min)


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_an\704IB_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 100height ( cm)h

freq

uen

cyden

sity

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 25 6 t < 10 3 510 6 t < 15 5 1015 6 t < 20 10 2020 6 t < 25 20 4025 6 t < 30 15 5530 6 t < 35 5 6035 6 t < 40 10 7040 6 t < 45 6 7645 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 0:020 3 4 8:1 3:330 25 12 15:6 10:040 35 22 21:9 18:350 75 37 46:9 30:860 125 62 78:1 51:770 145 102 90:6 8580 155 112 96:9 93:390 160 117 100:0 97:5100 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 5capacity ( litres)C

freq

uen

cyden

sity

0

50

100

150

200

250

300

400 800 1200 1600 2000

cum

ula

tive

freq

uen

cy

Cumulative frequency graph of wage data

wage ( )$

0

10

20

30

3 4 5 6 7

d (m)

freq

uen

cyden

sity

0

20

40

60

80

100

0 10 20 30 40 50 60 70 80 90 100scores (%)

per

centi

les

boys

girls

0

5

10

15

20

0 2 4 6 8 10 12

Histogram of masses of cats

m (kg)

freq

uen

cy

Histogram of long jump data

Cumulative frequency graph of test scores


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_an\705IB_IGC1_an.CDR Thursday, 20 November 2008 4:17:48 PM PETER

0

5

10

15

20

48 50 52 54 56length (cm)

freq

uen

cy

706 ANSWERS

6 a b 27 babies

c 70% of them

d Upper end point Cumulative frequency

49 150 451 1352 2353 3954 4355 4856 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 = 1612

= 43

Comparing widths; k = 128

= 32

and 436= 3

2.

6 FG = 2:4 m

7 a

band

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 = 23

b 14 850 cm3 c 280 cm2

10 No. Comparing capacities, k ¼ 1:37Comparing 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 = 2p6 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 = 75

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.

bx

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 2p13 ¼ 7:21 cm by 3

p13 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 56length (cm)

freq

uen

cy

Cumulative frequency graph of lengths of newborns

Histogram of lengths of newborn babies

4

4

22

3

3

1\Qw_ 1\Qw_

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


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_an\706IB_IGC1_an.CDR Thursday, 20 November 2008 4:19:17 PM PETER

chapter 17 - continuous data

Education