intermediate 2 – additional question bank unit 2 : graphs, charts & tables you have chosen to...
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 2 : Graphs, Charts & Tables
You have chosen to study:
Please choose a question to attempt from the following:
1 2 3 4
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5 6
Stem& Leaf
Dot Plot
CumFreqTable
Dot toboxplot
Stem toboxplot
Piechart
(a) Use this information to find the (i) median
(ii) lower & upper quartiles
(iii) the semi-interquartile range
(b)What is the probability that someone chosen at random earns less than £180?
The following stem & leaf diagram shows the distribution of wages for employees in a small factory …..
16 2 3 6 9
17 1 1 1 8 8 9
18 2 3 3 5 6 7 7
19 1 2 8
20 1 5 5 6 n = 25
21 8 17 4 = £174
GRAPHS, CHARTS, TABLES : Question 1
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(a) Use this information to find the (i) median
(ii) lower & upper quartiles
(iii) the semi-interquartile range
(b)What is the probability that someone chosen at random earns less than £180?
The following stem & leaf diagram shows the distribution of wages for employees in a small factory …..
16 2 3 6 9
17 1 1 1 8 8 9
18 2 3 3 5 6 7 7
19 1 2 8
20 1 5 5 6 n = 25
21 8 17 4 = £174
GRAPHS, CHARTS, TABLES : Question 1
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Use median position = (n+1) / 2 to find median
Q1 is midpoint from start to
median
Q3 is midpoint from median to
end
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(a) Use this information to find the (i) median
(ii) lower & upper quartiles
(iii) the semi-interquartile range
(b)What is the probability that someone chosen at random earns less than £180?
The following stem & leaf diagram shows the distribution of wages for employees in a small factory …..
16 2 3 6 9
17 1 1 1 8 8 9
18 2 3 3 5 6 7 7
19 1 2 8
20 1 5 5 6 n = 25
21 8 17 4 = £174
GRAPHS, CHARTS, TABLES : Question 1
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median = £183
Q1 = £171
Q3 = £195= £12
= 2/5
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Question 1
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16 2 3 6 917 1 1 1 8 8 918 2 3 3 5 6 7 719 1 2 820 1 5 5 6 n = 2521 8 17 4 = £174
(i) Median(ii) lower & upper quartiles(iii) the semi-interquartile range
1. Use median = (n+1) / 2 to find median
(a)(i) Since n = 25 then the median is
13th value ie median = £183
(ii) Both 6th & 7th values are £171 so Q1 = £171
2. There are 12 values before median so Q1 position = 13 - (12 + 1) / 2
3. There are 12 values after median so Q3 position = 13 + (12 + 1) / 2
19th is £192 & 20th is £198 so Q3 = £195
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(NOT 3!!!)
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Question 1
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16 2 3 6 917 1 1 1 8 8 918 2 3 3 5 6 7 719 1 2 820 1 5 5 6 n = 2521 8 17 4 = £174
(i) Median(ii) lower & upper quartiles(iii) the semi-interquartile range
4. Use SIQR = ½ (Q3 – Q1 ) / 2
(iii) SIQR = ½(Q3 – Q1)
= (£195 - £171) 2
= £12
Question 1 5. Use P = no of favourable / no of data
16 2 3 6 917 1 1 1 8 8 918 2 3 3 5 6 7 719 1 2 820 1 5 5 6 n = 2521 8 17 4 = £174
(b)What is the probability that
someone chosen at random
earns less than £180?
No of favourable ( under £180) = 10
No of data = n = 25
(b) Prob(under £180) = 10/25 = 2/5 .
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1. Use median = (n+1) / 2 to find median
(a)(i) Since n = 25 then the median is
13th value ie median = £183
(ii) Both 6th & 7th values are £171 so Q1 = £171
2. There are 12 values before median so Q1 position = 13 - (12 + 1) / 2
3. There are 12 values after median so Q3 = 13 + (12 + 1) / 2
19th is £192 & 20th is £198 so Q3 = £195
Median: the middle number in the ordered list. 25 numbers in the list.
1 – 12 13 14 - 25
12 numbers on either side of the median median is the 13th number in order.
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1. Use median = (n+1) / 2 to find median
(a)(i) Since n = 25 then the median is
13th value ie median = £183
(ii) Both 6th & 7th values are £171 so Q1 = £171
2. There are 12 values before median so Q1 position = 13 - (12 + 1) / 2
3. There are 12 values after median so Q3 = 13 + (12 + 1) / 2
19th is £192 & 20th is £198 so Q3 = £195
To find the upper and lower quartiles deal with the numbers on either side of the median separately.
Q1 12 numbers before median.6 numbers either side of Q1
is midway between the 6th and 7th number.
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1. Use median = (n+1) / 2 to find median
(a)(i) Since n = 25 then the median is
13th value ie median = £183
(ii) Both 6th & 7th values are £171 so Q1 = £171
2. There are 12 values before median so Q1 position = 13 - (12 + 1) / 2
3. There are 12 values after median so Q3 = 13 + (12 + 1) / 2
19th is £192 & 20th is £198 so Q3 = £195
To find the upper and lower quartiles deal with the numbers on either side of the median separately.
Q3 12 numbers after median.6 numbers either side of Q3
is midway between the 19th and 20th number.
Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows
28 29 29 30 31 30 28 30 29 28
29 30 30 28 28 29 29 29 29 28
a) Illustrate this using a dot plot.
b) What type of distribution does this show?
c) If a bag is chosen at random what is the probability it will be heavier than the modal weight?
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Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows
28 29 29 30 31 30 28 30 29 28
29 30 30 28 28 29 29 29 29 28
a) Illustrate this using a dot plot.
b) What type of distribution does this show?
c) If a bag is chosen at random what is the probability it will be heavier than the modal weight?
Establish lowest & highest values and
draw line with scale.
Plot a dot for each piece of data and label
diagram.
For probability use:
P = no of favourable / no of data
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Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows
28 29 29 30 31 30 28 30 29 28
29 30 30 28 28 29 29 29 29 28
a) Illustrate this using a dot plot.
b) What type of distribution does this show?
c) If a bag is chosen at random what is the probability it will be heavier than the modal weight?
Tightly clustered
3/10
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Question 2 1. Establish lowest & highest values
and draw line with scale.28 29 29 30 31 30 28 30 29 2831 30 30 28 2829 29 29 29 28
Illustrate this using a dot plot.
26 28 30 32
(a) Lowest = 28 & highest = 31.
Weights in g
2. Plot a dot for each piece of data and label diagram.
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Question 2 3. Make sure you know the possible
descriptions of data.28 29 29 30 31 30 28 30 29 2831 30 30 28 2829 29 29 29 28
26 28 30 32
Weights in g
What type of distribution does
this show?
(b) Tightly clustered distribution.Comments
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Question 2
28 29 29 30 31 30 28 30 29 2831 30 30 28 2829 29 29 29 28
26 28 30 32
Weights in g
4. Use P = no of favourable / no of data
No of favourable ( bigger than 29) = 6
No of data = n = 20
(c) Prob(W > mode) = 6/20 = 3/10 .
If a bag is chosen at random
what is the probability it will
be heavier than the modal
weight?
Mode!
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3. Make sure you know the possible descriptions of data.
26 28 30 32
Weights in g
(b) Tightly clustered distribution.
Other types of distribution:
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3. Make sure you know the possible descriptions of data.
26 28 30 32
Weights in g
(b) Tightly clustered distribution.
Other types of distribution:
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3. Make sure you know the possible descriptions of data.
26 28 30 32
Weights in g
(b) Tightly clustered distribution.
Other types of distribution:
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Comments
Probability =
Number of favourable outcomes Number of possible outcomes
Weights in g
4. Use P = no of favourable / no of data
No of favourable ( bigger than 29) = 6
No of data = n = 20
(c) Prob(W > mode) = 6/20 = 3/10 .
Mode!
26
28
30
32
To calculate simple probabilities:
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Charts, Graphs & Tables : Question 3 The results for a class test were
18 14 16 17 14 16 13 11 13 13 16 14 13 18 15
10 14 17 13 15 15 18 14 17 13 16 10 14 13 17
(a) Construct a cumulative frequency table for this data.
(b) What is the median for this data?
(c) What is the probability that a pupil selected at random
scored under 14?
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Charts, Graphs & Tables : Question 3 The results for a class test were
18 14 16 17 14 16 13 11 13 13 16 14 13 18 15
10 14 17 13 15 15 18 14 17 13 16 10 14 13 17
(a) Construct a cumulative frequency table for this data.
(b) What is the median for this data?
(c) What is the probability that a pupil selected at random
scored under 14?
Establish lowest & highest values and
draw table. Complete each row 1 step at a
time, calculating
running total as you go.
For probability use:
P = no of favourable / no of data
Use median = (n+1) / 2 to establish in which row
median lies.
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Charts, Graphs & Tables : Question 3 The results for a class test were
18 14 16 17 14 16 13 11 13 13 16 14 13 18 15
10 14 17 13 15 15 18 14 17 13 16 10 14 13 17
(a) Construct a cumulative frequency table for this data.
(b) What is the median for this data?
(c) What is the probability that a pupil selected at random
scored under 14?
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Median = 14
1/3
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Question 3 1. Establish lowest & highest values and draw a table.
(a) Lowest = 10 & highest = 1818 14 16 17 14 16 13 11 13 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10
14 13 17
(a)Construct a cumulative frequency table for this data.
Mark Frequency Cum Frequency
10 11 12 13 14 15 16 17 18
2 21 30 3
6 163 194 234 273 30
7 10
2. Complete each row 1 step at a time, calculating running total as you go.
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Question 3 3. Use median = (n+1) / 2 to establish in which row median lies.
18 14 16 17 14 16 13 11 13 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10
14 13 17
Mark Frequency Cum Frequency
10 11 12 13 14 15 16 17 18
2 21 30 3
6 163 194 234 273 30
7 10
(b) What is the median for
this data?
For 30 values median is between
15th & 16th both of which are in row 14.
Median Mark = 14
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Question 3
18 14 16 17 14 16 13 11 13 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10
14 13 17
Mark Frequency Cum Frequency
10 11 12 13 14 15 16 17 18
2 21 30 3
6 163 194 234 273 30
7 10(c) What is the probability that a pupil selected at random scored under 14?
No of favourable ( under 14) = 10
No of data = n = 30
(c) Prob(mark<14) = 10/30 = 1/3 .
4. Use P = no of favourable / no of data
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For 30 values median is between
15th & 16th both of which are in row 14.
Median = 14
Mark Freq Cum Freq
10 11 12 13 14 15 16 17 18
2 21 30 3
6 16
4 234 273 30
7 10
3 19
Median:
1 – 15 Q2 16 - 30
Find the mark at which the cumulative frequency first reaches between 15th and 16th number.
Median = 14
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Mark Freq Cum Freq
10 11 12 13 14 15 16 17 18
2 21 30 3
6 16
4 234 273 30
7 10
3 19
No of favourable ( under 14) = 10
No of data = n = 30
(c) Prob(mark<14) = 10/30 = 1/3 .
Probability =
Number of favourable outcomes Number of possible outcomes
To calculate simple probabilities:
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Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes.
48 50 52 54 56 58
(a) Find the (i) median (ii) lower quartile (iii) upper quartile
(b) Construct a boxplot using this data.
(c) In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample?
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Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes.
48 50 52 54 56 58
(a) Find the (i) median (ii) lower quartile (iii) upper quartile
(b) Construct a boxplot using this data.
(c) In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample?
Use median position = (n+1) / 2 to find median
Q1 is midpoint from start to
median
Q3 is midpoint from median to
end
remember bigger SIQR means more
variation (spread) in
data.
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Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes.
48 50 52 54 56 58
(a) Find the (i) median (ii) lower quartile (iii) upper quartile
(b) Construct a boxplot using this data.
(c) In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample?
Median = 50
So Q1 = 49
So Q3 = 52
the data is distributed more widely than (or not as clustered as)
the above data
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Question 4
48 50 52 54 56 58
(a) Find the (i) median
(ii) lower quartile
(iii) upper quartile
(a) (i) Sample size = 23
so median position is 12.
ie (23+1)2
Median = 50
1. Use median = (n+1) / 2 to find median
2. There are 11 values before median so Q1 position = 12 - (11 + 1) / 2
3. There are 11 values after median so Q3 position = 12 + (11 + 1) / 2
(ii) Middle of 1st 11 is position 6.
So Q1 = 49
(iii) Middle of 2nd 11 is position 18.
So Q3 = 52
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Question 4
48 50 52 54 56 58
(b) Construct a boxplot using
this data.
4. Draw number line with scale.
Make sure you note highest & lowest
as well as Q1, Q2, Q3.
48 50 52 54 56 58
(b)Lowest = 48, Q1 = 49, Q2 = 50,
Q3 = 52 & Highest = 58.
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Question 4
48 50 52 54 56 58
5. Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data.
(c) In a second sample the
semi-interquartile range was
2.5. How does this compare?
(c)For above sample
SIQR = (52 - 49) 2 = 1.5
In a sample where the SIQR is 2.5 the data is distributed more widely than (or not as clustered as) the above data
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(a) (i) Sample size = 23
so median position is 12.
ie (23+1)2
Median = 50
1. Use median = (n+1) / 2 to find median
2. There are 11 values before median so Q1 position = 12 - (11 + 1) / 2
3. There are 11 values after median so Q3 position = 12 + (11 + 1) / 2
(ii) Middle of 1st 11 is position 6.
So Q1 = 49
(iii) Middle of 2nd 11 is position 18.
So Q3 = 52
23 numbers in the list:
1 - 11 12 13 - 23
Q2
11 numbers on either side of the median
The median:
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(a) (i) Sample size = 23
so median position is 12.
ie (23+1)2
Median = 50
1. Use median = (n+1) / 2 to find median
2. There are 11 values before median so Q1 position = 12 - (11 + 1) / 2
3. There are 11 values after median so Q3 position = 12 + (11 + 1) / 2
(ii) Middle of 1st 11 is position 6.
So Q1 = 49
(iii) Middle of 2nd 11 is position 18.
So Q3 = 52
For quartiles:
1 - 5 6 7 - 11 12
Q2Q1
Q3
13 - 17 18 19 - 2312
Q2
Now count through the list until you reach the 6th, 12th,and 18th number in the list.
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5. Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data.
(c)For above sample
SIQR = (52 - 49) 2 = 1.5
In a sample where the SIQR is 2.5 the data is distributed more widely than or not as clustered as the above data
The semi-interquartile range is a measure of the range of the “middle” 50%.
S.I.R. = (Q3 - Q1) 12
Remember: when asked to compare data always consider average and spread.
It is a measure of how spread-out and so how “consistent” or “reliable” the data is.
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Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club.
(a) Find the median, lower & upper quartiles for this data.
(b) Use the data to construct a boxplot.
(c) The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results.
6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3
11 4 = 114kg
60 70 80 90 100 110 120
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Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club.
(a) Find the median, lower & upper quartiles for this data.
(b) Use the data to construct a boxplot.
(c) The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results.
6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3
11 4 = 114kg
60 70 80 90 100 110 120
Use median position = (n+1) / 2 to find median
position
Q1 is midpoint from start to
median
Q3 is midpoint from median to
end
When comparing
two data sets comment on spread and
average
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Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club.
(a) Find the median, lower & upper quartiles for this data.
(b) Use the data to construct a boxplot.
(c) The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results.
6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3
11 4 = 114kg
60 70 80 90 100 110 120
median = 87
Q1 = 77
Q3 = 99 Full solution
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Question 5 1. Use median = (n+1) / 2 to find
median
(a)(i) Since n = 26 then the median is
between 13th & 14th value ie median = 87
(ii) so Q1 = 77
2. There are 13 values before median so Q1 position is 6th value
3. There are 13 values after median so Q3 position is 20th position
so Q3 = 99
6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3
11 4 = 114kg
(a)Find the median, lower & upper quartiles for this data.
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Question 5
6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3
11 4 = 114kg
4. Draw number line with scale.
Make sure you note highest & lowest
as well as Q1, Q2, Q3.
(b)Lowest = 60, Q1 = 77, Q2 = 87,
Q3 = 99 & Highest = 123.
60 70 80 90 100 110 120
(b) Use the data to construct
a boxplot.
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Question 5 5. Compare spread and relevant average.
(c) The boxplot below shows
the weight distribution for
these people after several
months.
Compare the two &
comment on the results.
60 70 80 90 100 110 120
(c) Lightest has put on weight –
lowest now 65,
heaviest 3 have lost weight –
highest now 115,
median same but overall
spread of weights has decreased
as Q3-Q1 was 22
but is now only 15.
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4. Draw number line with scale.
Make sure you note highest & lowest
as well as Q1, Q2, Q3.
(b)Lowest = 60, Q1 = 77, Q2 = 87,
Q3 = 99 & Highest = 123.
60 70 80 90 100 110 120
Box Plot :
Lowest HighestQ1 Q2 Q3
Remember:
To draw a boxplot you needa “five-figure summary”:
five-figure summary
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Charts, Graphs & Tables : Question 6
The pie chart below shows the breakdown of how a sample of
630 people spent their Saturday nights.
clubbing
144°x° theatre
cinemaWatching TV(a) How many people
went clubbing?
(b) If 84 people went to the
theatre then how big is x°?
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Charts, Graphs & Tables : Question 6
The pie chart below shows the breakdown of how a sample of
630 people spent their Saturday nights.
clubbing
144°x° theatre
cinemaWatching TV(a) How many people
went clubbing?
(b) If 84 people went to the
theatre then how big is x°?
angle360°
= amount630
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Charts, Graphs & Tables : Question 6
The pie chart below shows the breakdown of how a sample of
630 people spent their Saturday nights.
clubbing
144°x° theatre
cinemaWatching TV(a) How many people
went clubbing?
(b) If 84 people went to the
theatre then how big is x°?
= 252
= 48°
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Question 6
clubbing
144°x° theatre
cinemaWatching TV
How many people went clubbing?
1. Set up ratio of angles and sectors and cross multiply.
(a) The angle is 144° so …..
angle360°
= amount630
144°360°
= amount630
360 x amount = 144 x 630
amount = 144 x 630 360
= 252
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Question 6
clubbing
144°x° theatre
cinemaWatching TV
2. Set up ratio of angles and sectors and cross multiply.
(b) The amount is 84 so …..
angle360°
= amount630
angle360°
= 84630
630 x angle = 360° x 84
angle = 360° x 84 630
= 48°
(b) If 84 people went to the theatre then how big is x°?
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1. Set up ratio of angles and sectors and cross multiply.
(a) The angle is 144° so …..
angle360°
= amount630
144°360°
= amount630
360 x amount = 144 x 630
amount = 144 x 630 360
= 252
Can also be tackled by using proportion:
Amount = x 630 144360
360
angle at centreamount total sample
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Comments
Can also be tackled by using proportion:
360
angle at centreamount total sample
2. Set up ratio of angles and sectors and cross multiply.
(b) The amount is 84 so …..
angle360°
= amount630
angle360°
= 84630
630 x angle = 360° x 84
angle = 360° x 84 630
= 48°
84 = x 630 x360
630 x = 84 x 360
x = 84 360 630
x
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End of graphs, charts etc.
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