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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
FS-Ma-L1test-Cha6-7.indd 3 4/30/09 12:03:54 PM
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 ..........................................................................................................................................................................
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.
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.
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.
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.