Maths Level 1 - moodle.stephensoncoll.ac.uk

EDEXCEL FUNCTIONAL SKILLS PILOT

Maths Level 1Test your skills

Chapters 6 and 7

Investigating election statisticsApplying skills in:

• handling data

• probability

Answer all questions in this task.

Write your answers in the spaces provided.

Where you see this sign you must show how you get your answersas marks may be awarded for your working out.

You may use a calculator.

© Pearson Education 2009 Functional Maths Level 1 • Chapters 6–7 Test • page 1 Draft for Pilot

FS-Ma-L1test-Cha6-7.indd 1 4/30/09 12:03:52 PM

1 The table shows the election results for the past three elections.

a) Who gained the least number of votes for any year? When did this happen?

Candidate Year b) Which candidates entered all three elections?

.....................................................................................................................................................................................

.....................................................................................................................................................................................

c) What was the range in the number of votes in the 2005 election?

Range d) Which election year had the biggest range in votes?

Year


Name of candidateNumber of votes

2002 Election 2005 Election 2008 ElectionAlice Adams 20

Mary Butterworth 47 70 67

Michael Butterworth 89 67 101

Nigel Cooper 122 165

Sajid Hamid 89 77 65

Robina Hassan 29

Henry Jackson 78 58

Lorna Moore 45

David Roberts 63 103

Tim Shacklock 52

Elections for a president of a social clubA social club holds elections for a new president every three years.


Investigating election statist ics 6 – 7Investigating election statist ics 6 – 7


e) What was the mean number of votes per candidate in the 2002 election?

Mean number of votes f) Investigate whether the mean number of votes per candidate has increased or

decreased in the 2005 and 2008 elections, compared with the 2002 election.

.....................................................................................................................................................................................

.....................................................................................................................................................................................

.....................................................................................................................................................................................

A total of 520 people voted in the 2008 election.

g) Use the number of votes Tim Shacklock gained to work out the probability of Tim Shacklock winning a similar election.

Probability h) If the same candidates enter the 2011 election as the 2008 election,

and no other circumstances change, which two candidates do you judge to be likely to win? Why?

Candidate 1

Candidate 2


2 Michael and Mary Butterworth are a married couple who have both entered all three elections.

They want to compare how their individual numbers of votes have changed over the three elections. They decide to plot a line graph showing the number of votes received.

a) Provide the missing information from the graph in the box below.

Title ..........................................................................................................................................................................

Vertical axis label ..............................................................................................................................................

Horizontal axis label ........................................................................................................................................

Key ............................................................................................................................................................................

b) What is the change in voting figures for each candidate from 2002 to 2005, and from 2005 to 2008?

Michael Butterworth

Mary Butterworth


60

30

40

50

20

10

02002 2005 2008

90

100

80

70

110



One of the Butterworths must drop out of the election in 2011 as it is taking up too much time.

c) Who should drop out and why?

.....................................................................................................................................................................................

.....................................................................................................................................................................................

.....................................................................................................................................................................................

3 Many members in the social club do not bother to vote. The club president is concerned about this and asked the club secretary to look at who voted in each election.

The secretary collects information about the voters in the 2008 election and organises this information in a table.

There were 520 voters in total.

a) Work out the probability of a voter chosen at random being an unemployed woman.

Probability


Group Number of votersfemale unemployed 260male unemployed 110female employed 40male employed 110



b) Use the graph paper below to represent this information in a bar chart.

c) How many more unemployed members voted than employed voters in 2008?

Number of members

120

60

80

100

40

20

0female

unemployedmale

unemployedfemale

employedmale

employed

180

200

160

140

220

240

260

280

Pearson

Edexcel: ICT Teacher Notes

S/S

JM30 Apr 2008141086

Ex 04a

8pt ITC Officina Sans

SL18 Jan 2008

EL11 Jan 2007




The secretary collects information about the voters in the 2005 election and represents this in a pie chart.

d) What fraction of the voters in 2005 were female and unemployed?

Fraction e) Who were less likely to vote in 2005, men or women? Why?

.....................................................................................................................................................................................

.....................................................................................................................................................................................

.....................................................................................................................................................................................

.....................................................................................................................................................................................

f) Use the pie chart above and the bar chart you have drawn to say whether the proportion of employed female voters has increased or decreased from 2005 to 2008.

....................................................................................................................................................................................

.....................................................................................................................................................................................

.....................................................................................................................................................................................

.....................................................................................................................................................................................

Groups of votes in 2005 elections

male employment

female employment

male unemployment

female unemployment

Pearson


S/S

JM30 Apr 2008141086

Ex 05a


SL18 Jan 2008

EL11 Jan 2007


A SolutionsAnswer

1a) 20 is the smallest number of votes in the table. Alice Adams 20051b) There are three candidates with votes entered in all three columns. Mary Butterworth

Michael Butterworth Sajid Hamid

1c) 165 – 20 1451d) In 2002 the range is 122 – 47 = 75

In 2005 the range is 145In 2008 the range is 103 – 29 = 74145 is the biggest number

2005

1e) 2002:47 + 89 + 122 + 89 + 78 = 425There are five candidates so divide 425 by 5425 ÷ 5 = 85Check:Show working for inverse operation, e.g. 85 × 5 = 425Or estimation of mean, e.g. 50 + 90 + 120 + 90 + 80 = 430 430 ÷ 5 = 86 which is close to 85

Mean number of votes per candidate in 2002 is 85

1f) 2005:20 + 70 + 67 + 165 + 77 + 63 = 462There are six candidates so divide 462 by 6462 ÷ 6 = 772008:67 + 101 + 65 + 29 + 58 + 45 + 103 + 52 = 520There are eight candidates so divide 520 by 8520 ÷ 8 = 65

Mean number of votes per candidate is decreasing with each election

1g) 52 people voted for Tim Shacklock out of 520This gives an indication of how popular Tim Shacklock is with the voters 52

520 = 110

110

1h) Michael Butterworth and David Roberts received substantially more votes than anyone else.

Michael Butterworth or David Roberts

2a) Title Number of votes received by Mr and Mrs Butterworth Vertical axis label Number of votesHorizontal axis label Election yearKey Red line votes for Michael, blue line votes for Mary

Any similar answers conveying the same meaning are acceptable

2b) Michael Butterworth: votes dropped by 22 from 2002 to 2005Mary Butterworth: votes increased by 23 from 2002 to 2005 but have slightly decreased by 3 in 2008.

Answers must be clear and include the difference between the votes for the three years for the two candidates

2c) Suggest Mary drops out as Michael’s votes have significantly increased whereas Mary’s votes have slightly decreased.

Any similar description is acceptable




Answer1a) 20 is the smallest number of votes in the table. Alice Adams 20051b) There are three candidates with votes entered in all three columns. Mary Butterworth

Michael Butterworth Sajid Hamid

1c) 165 – 20 1451d) In 2002 the range is 122 – 47 = 75

In 2005 the range is 145In 2008 the range is 103 – 29 = 74145 is the biggest number

2005

1e) 2002:47 + 89 + 122 + 89 + 78 = 425There are five candidates so divide 425 by 5425 ÷ 5 = 85Check:Show working for inverse operation, e.g. 85 × 5 = 425Or estimation of mean, e.g. 50 + 90 + 120 + 90 + 80 = 430 430 ÷ 5 = 86 which is close to 85

Mean number of votes per candidate in 2002 is 85

1f) 2005:20 + 70 + 67 + 165 + 77 + 63 = 462There are six candidates so divide 462 by 6462 ÷ 6 = 772008:67 + 101 + 65 + 29 + 58 + 45 + 103 + 52 = 520There are eight candidates so divide 520 by 8520 ÷ 8 = 65

Mean number of votes per candidate is decreasing with each election

1g) 52 people voted for Tim Shacklock out of 520This gives an indication of how popular Tim Shacklock is with the voters =

1h) Michael Butterworth and David Roberts received substantially more votes than anyone else.

Michael Butterworth or David Roberts

2a) Title Number of votes received by Mr and Mrs Butterworth Vertical axis label Number of votesHorizontal axis label Election yearKey Red line votes for Michael, blue line votes for Mary

Any similar answers conveying the same meaning are acceptable

2b) Michael Butterworth: votes dropped by 22 from 2002 to 2005Mary Butterworth: votes increased by 23 from 2002 to 2005 but have slightly decreased by 3 in 2008.

Answers must be clear and include the difference between the votes for the three years for the two candidates

2c) Suggest Mary drops out as Michael’s votes have significantly increased whereas Mary’s votes have slightly decreased.

Any similar description is acceptable

Answer3a) There are 260 unemployed female voters out of a total of 520

Probability (voter being unemployed woman) = 260520 = 1

2

12 or 0.5 or 50%

3b) Bar chart drawn with suitable labels on both axes. For example, vertical axis label ‘Numbers of voters’ (or something similar), horizontal axis ‘Group’ (or something similar).Bars of equal width and gaps between the bars

3c) There are 260 + 110 = 370 unemployed votersThere are 40 + 110 = 150 employed voters370 – 150 = 220

220 voters

3d) A quarter of the circle represents unemployed females. 14

3e) Men are represented by 14 of the pie chart and women are

represented by 34 of the pie chart.

men

3f) From the pie chart, the proportion of employed female voters in 2005 is

12

From the bar chart, there are 40 out of 520 employed female voters in 2008. This is less than half of the voters.

decreased

120

60

80

100

40

20

0female

unemployedmale

unemployedfemale

employedmale

employed

180

200

160

140

220

240

260

280

Pearson


S/S

JM30 Apr 2008141086

Ex 06a


SL18 Jan 2008

EL11 Jan 2007



Maths Level 1 - moodle.stephensoncoll.ac.uk

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