Transcript Graphs_charts

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
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UNIT 2 :
Graphs, Charts
& Tables
Please choose a question to attempt from the following:
1
Stem
& Leaf
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2
Dot
Plot
3
Cum
Freq
Table
4
Dot to
boxplot
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Unit 2 Menu
5
Stem to
boxplot
6
Piechart
GRAPHS, CHARTS, TABLES : Question 1
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
19
1
2
8
20
1
5
5
21
8
6
7
n = 25
17 4 = $174
(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?
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Get hint
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Reveal answer
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GRAPHS, CHARTS, TABLES : Question 1
The following stem & leaf diagram shows the distribution of wages for
employees in a small factory …..
16
2
3
6
17
1
1
1
18
2
3
3
19
1
2
8
20
1
5
5
21
8
(a) Use this information to find the
9
Use median Q1 is midpoint
position
8
8= (n+1)9/ from start to
2 to find median
median
5
6
7
7
6
Q3 is midpoint
nfrom
= 25median to
end
17 4 = $174
(i) median
(ii) lower & upper quartiles
What would you like to do now?
(iii) the semi-interquartile range
(b)What is the probability that someone chosen at random earns
less than $180?
Graphs etc Menu
Go to full solution
EXIT
Reveal answer
Go to Comments
GRAPHS, CHARTS, TABLES : Question 1
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
19
1
2
8
20
1
5
5
21
8
6
7
n = 25
17 4 = $174
(a) Use this information to find the
What would you like to do now?
median = $183
(ii) lower & upper quartilesQ1 = $171
(i) median
Q3 = $195
(iii) the semi-interquartile range
= $12
(b)What is the probability that someone chosen at random earns
less than $180? = 2/5
Go to full solution
EXIT
Graphs etc Menu
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Question 1
16
17
18
19
20
21
2
1
2
1
1
8
3
1
3
2
5
6
1
3
8
5
1. Use median = (n+1) / 2 to find
median
9
8
5
6
(a)(i) Since n = 25 then the median is
8
6
9
7
7
n = 25
17 4 = $174
(i) Median
(ii) lower & upper quartiles
(iii) the semi-interquartile range
Begin Solution
Continue Solution
Comments
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13th value
ie median = $183
(NOT 3!!!)
2. There are 12 values before median
so Q1 position = 13 - (12 + 1) / 2
(ii) Both 6th & 7th values are $171
so Q1 = $171
3. There are 12 values after median so
Q3 position = 13 + (12 + 1) / 2
19th is $192 & 20th is $198
so Q3 = $195
What would you like to do now?
Question 1
16
17
18
19
20
21
2
1
2
1
1
8
3
1
3
2
5
6
1
3
8
5
4. Use SIQR = ½ (Q3 – Q1 ) / 2
9
8
5
6
8
6
9
7
(iii) SIQR = ½(Q3 – Q1)
= ($195 - $171)  2
7
n = 25
17 4 = $174
(i) Median
(ii) lower & upper quartiles
(iii) the semi-interquartile range
Begin Solution
Continue Solution
Comments
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= $12
Question 1
16
17
18
19
20
21
2
1
2
1
1
8
3
1
3
2
5
6
1
3
8
5
5. Use P = no of favourable / no of data
9
8
5
6
No of favourable ( under $180) = 10
8
6
9
7
7
n = 25
17 4 = $174
(b)What is the probability that
someone chosen at random
earns less than $180?
Begin Solution
Continue Solution
Comments
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No of data = n = 25
(b) Prob(under $180) =
10/
25
= 2/ 5 .
Comments
1. Use median = (n+1) / 2 to find
median
(a)(i) Since n = 25 then the median is
13th value
ie median = $183
2. There are 12 values before median
so Q1 position = 13 - (12 + 1) / 2
(ii) Both 6th & 7th values are $171
so Q1 = $171
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.
3. There are 12 values after median
so Q3 = 13 + (12 + 1) / 2
19th is $192 & 20th is $198
so Q3 = $195
Next Comment
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Comments
1. Use median = (n+1) / 2 to find
median
To find the upper and lower
quartiles deal with the numbers
on either side of the median
separately.
(a)(i) Since n = 25 then the median is
13th value
ie median = $183
2. There are 12 values before median
so Q1 position = 13 - (12 + 1) / 2
Q1
12 numbers before median.
6 numbers either side of Q1
is midway between
the 6th and 7th number.
(ii) Both 6th & 7th values are $171
so Q1 = $171
3. There are 12 values after median
so Q3 = 13 + (12 + 1) / 2
19th is $192 & 20th is $198
so Q3 = $195
Next Comment
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Comments
1. Use median = (n+1) / 2 to find
median
To find the upper and lower
quartiles deal with the numbers
on either side of the median
separately.
(a)(i) Since n = 25 then the median is
13th value
ie median = $183
2. There are 12 values before median
so Q1 position = 13 - (12 + 1) / 2
Q3
12 numbers after median.
6 numbers either side of Q3
is midway between
the 19th and 20th number.
(ii) Both 6th & 7th values are $171
so Q1 = $171
3. There are 12 values after median
so Q3 = 13 + (12 + 1) / 2
19th is $192 & 20th is $198
so Q3 = $195
Next Comment
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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?
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Charts, Graphs & Tables : Question 2
The weights in grams of 20 bags of crisps were as follows
Establish lowest &
28 29 29 30 31 30 28 30highest
29 28
values and
Plot a dot for
draw
line with use:
scale.
For
probability
each piece of
29 30 30 28 28 29 29 29 29 28
data and label
P = no of favourable
diagram. /
no of data
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?
What would you like to do now?
EXIT
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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.
CLICK
b) What type of distribution does this show?
Tightly clustered
c) If a bag is chosen at random what is the probability it
3/10
will be heavier than the modal weight?
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Question 2
28
30
29
29
29
28
30
29
29
30
30
29
30
29
28
29
31
28
28
28
1. Establish lowest & highest values
and draw line with scale.
(a) Lowest = 28 & highest = 31.
Weights in g
Illustrate this using a dot plot.
26
Begin Solution
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28
30
32
2. Plot a dot for each piece of data and
label diagram.
Question 2
28
30
29
29
29
28
30
29
29
30
30
29
30
29
28
29
31
28
28
28
3. Make sure you know the possible
descriptions of data.
Weights in g
What type of distribution does
this show?
26
28
30
Begin Solution
Continue Solution
Comments
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(b) Tightly clustered distribution.
32
Question 2
28
30
29
29
29
28
30
29
29
30
30
29
4. Use P = no of favourable / no of data
30
29
28
29
31
28
28
28
Mode!
Weights in g
If a bag is chosen at random
what is the probability it will
be heavier than the modal
weight?
26
Begin Solution
28
30
No of favourable ( bigger than 29) = 6
Continue Solution
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32
No of data = n = 20
(c) Prob(W > mode) =
6/
3/
=
20
10 .
What would you like to do now?
Comments
Other types of distribution:
3. Make sure you know the possible
descriptions of data.
Weights in g
26
28
30
(b) Tightly clustered distribution.
32
Next Comment
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Comments
Other types of distribution:
3. Make sure you know the possible
descriptions of data.
Weights in g
26
28
30
(b) Tightly clustered distribution.
32
Next Comment
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Comments
Other types of distribution:
3. Make sure you know the possible
descriptions of data.
Weights in g
26
28
30
(b) Tightly clustered distribution.
32
Next Comment
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Comments
4. Use P = no of favourable / no of data
Mode!
To calculate simple probabilities:
Probability =
Weights in g
Number of favourable outcomes
Number of possible outcomes
2
2
3
3
6
8
0
2
No of favourable ( bigger than 29) = 6
Next Comment
No of data = n = 20
Menu
(c) Prob(W > mode) =
6/
20
= 3/10 .
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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?
Get hint
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Charts, Graphs & Tables : Question 3
The results for a class test were
lowest &
18 14 16 17 14 16 13 11 13 13 16 14Establish
13 18
Use15median =
highest values and
(n+1)
/ 2 to
10 14 17 13 15 15 18 14 17 13 16 10 14draw
13
17
Complete
each
table.
For probability
use:
establish
row
1 step in
at a
which
row
time,
P = no of favourable
lies./
(a) Construct a cumulative frequency table formedian
this
data.
calculating
no of data
running total as
(b) What is the median for this data?
you go.
(c) What is the probability that a pupil selected at random
scored under 14?
What would you like to do now?
EXIT
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Reveal answer
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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. CLICK
(b) What is the median for this data?
Median = 14
(c) What is the probability that a pupil selected at random
scored under 14?
1/3
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Question 3
1. Establish lowest & highest values and
draw a table.
18 14 16 17 14 16 13 11 13
(a) Lowest = 10 & highest = 18
13 16 14 13 18 15 10 14 17
Mark
13 15 15 18 14 17 13 16 10
14 13 17
(a) Construct a cumulative
frequency table for this
data.
Begin Solution
Continue Solution
10
11
12
13
14
15
16
17
18
Frequency Cum Frequency
2
1
0
7
6
3
4
4
3
2
3
3
10
16
19
23
27
30
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2. Complete each row 1 step at a time,
calculating running total as you go.
Question 3
3. Use median = (n+1) / 2 to establish in
which row median lies.
18 14 16 17 14 16 13 11 13
Mark
Frequency Cum Frequency
13 16 14 13 18 15 10 14 17
13 15 15 18 14 17 13 16 10
14 13 17
(b) What is the median for
this data?
What would you like to do now?
Begin Solution
Continue Solution
Comments
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10
11
12
13
14
15
16
17
18
2
1
0
7
6
3
4
4
3
2
3
3
10
16
19
23
27
30
For 30 values median is between
15th & 16th both of which are in row 14.
Median Mark = 14
Question 3
4. Use P = no of favourable / no of data
18 14 16 17 14 16 13 11 13
Mark
Frequency Cum Frequency
13 16 14 13 18 15 10 14 17
13 15 15 18 14 17 13 16 10
14 13 17
(c) What is the probability that
a pupil selected at random
scored under 14?
What would you like to do now?
Begin Solution
Continue Solution
Comments
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10
11
12
13
14
15
16
17
18
2
1
0
7
6
3
4
4
3
2
3
3
10
16
19
23
27
30
No of favourable ( under 14) = 10
No of data = n = 30
(c) Prob(mark<14) =
10/
30
= 1/ 3 .
Comments
Median:
Mark
10
11
12
13
14
15
16
17
18
Freq
2
1
0
7
6
3
4
4
3
Cum Freq
2
3
3
10
16
19
23
27
30
For 30 values median is between
15th & 16th both of which are in row 14.
Median = 14
1 – 15 Q2 16 - 30
Median = 14
Find the mark at which the
cumulative frequency first
reaches between 15th and
16th number.
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Comments
To calculate simple probabilities:
Mark
Freq
Cum Freq
Probability =
10
11
12
13
14
15
16
17
18
2
1
0
7
6
3
4
4
3
2
3
3
10
16
19
23
27
30
Number of favourable outcomes
Number of possible outcomes
No of favourable ( under 14) = 10
No of data = n = 30
(c) Prob(mark<14) =
10/
30
=
1/
3
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.
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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
(a) Find the
50
52
54
56
58
(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?
Get hint
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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.
median
Q1Use
is midpoint
position
= (n+1)
from start
to /
2 remember
tomedian
find median
bigger SIQR
means more
Q3 is midpoint
variation
from median to
(spread) in
end
48
50
52
54
56
58
data.
(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?
What would you like to do now?
Graphs etc Menu
EXIT
Reveal answer
Go to full solution
Go to Comments
Charts, Graphs & Tables : Question 4
The dot plot below shows the number of matches per box in a
sample of 23 boxes.
Median = 50
So Q1 = 49
So Q3 = 52
48
(a) Find the
50
52
54
56
58
(i) median (ii) lower quartile (iii) upper quartile
(b) Construct a boxplot using this data.
CLICK
(c) In a second sample the semi-interquartile range was 2.5.
How does this distribution compare to the above sample?
the data is distributed more widely than (or not as clustered as)
the above data
EXIT
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Full solution
Comments
1. Use median = (n+1) / 2 to find
median
Question 4
(a) (i) Sample size = 23
so median position is 12.
ie (23+1)2
48
50
52
(a) Find the
54
56
(i) median
(ii) lower quartile
(iii) upper quartile
58
Median = 50
2. There are 11 values before median
so Q1 position = 12 - (11 + 1) / 2
(ii) Middle of 1st 11 is position 6.
So Q1 = 49
Begin Solution
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3. There are 11 values after median so
Q3 position = 12 + (11 + 1) / 2
(iii) Middle of 2nd 11 is position 18.
So Q3 = 52
Question 4
4. Draw number line with scale.
Make sure you note highest & lowest
as well as Q1, Q2, Q3.
(b)Lowest = 48, Q1 = 49, Q2 = 50,
Q3 = 52 & Highest = 58.
48
50
52
54
56
58
(b) Construct a boxplot using
this data.
Begin Solution
Continue Solution
Comments
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48
50
52
54
56
58
Question 4
5. Calculate SIQR then compare
remember bigger SIQR means more
variation (spread) in data.
(c) For above sample
SIQR = (52 - 49)  2 = 1.5
48
50
52
54
56
58
(c) In a second sample the
semi-interquartile range was
In a sample where the SIQR is 2.5 the
data is distributed more widely than
(or not as clustered as) the above data
2.5. How does this compare?
What would you like to do now?
Begin Solution
Continue Solution
Comments
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Comments
1. Use median = (n+1) / 2 to find
median
The median:
23 numbers in the list:
(a) (i) Sample size = 23
so median position is 12.
1 - 11 12
13 - 23
ie (23+1)2
Median = 50
2. There are 11 values before median
so Q1 position = 12 - (11 + 1) / 2
Q2
11 numbers on either side
of the median
(ii) Middle of 1st 11 is position 6.
So Q1 = 49
3. There are 11 values after median
so Q3 position = 12 + (11 + 1) / 2
(iii) Middle of 2nd 11 is position 18.
So Q3 = 52
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Comments
1. Use median = (n+1) / 2 to find
median
(a) (i) Sample size = 23
For quartiles:
1 - 5
so median position is 12.
2. There are 11 values before median
so Q1 position = 12 - (11 + 1) / 2
(ii) Middle of 1st 11 is position 6.
So Q1 = 49
3. There are 11 values after median
so Q3 position = 12 + (11 + 1) / 2
(iii) Middle of 2nd 11 is position 18.
So Q3 = 52
7
- 11
12
Q2
12
Q2
Q1
ie (23+1)2
Median = 50
6
13 - 17 18
19 - 23
Q3
Now count through the list until you
reach the 6th, 12th,and 18th number
in the list.
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Comments
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. = 1 (Q3 - Q1)
2
It is a measure of how
spread-out and so how
“consistent” or “reliable”
the data is.
Remember: when asked to
compare data always consider
average and spread.
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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.
Get hint
6
7
8
9
10
11
12
0
1
2
4
5
2
3
2
4
7
5 7 7
2 5 6 6 8 9
6 9 9
7
Reveal answer
Full solution
11 4 = 114kg
1 1 3
Comments
(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.
EXIT
60
70
80
90
100
110
120
Charts, Graphs & Tables : Question 5
What now?
The stem & leaf diagram below shows the weight distribution of 26
people when they joined a slimming club.
Menu
median
Q1Use
is midpoint
6 0 2
position
= (n+1)
from start
to /
7 1 3 5 7 7
When
Reveal answer
2 tomedian
find median
8 2 2 2 5 6 6 8 9comparing
position
9 4 4 6 9 9
two data sets
Q3 is midpoint
Full solution
10 5 7 7
comment on
from11median
to
= 114kg
11
spread4 and
end
12 1 1 3
Comments
average
(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.
EXIT
60
70
80
90
100
110
120
Charts, Graphs & Tables : Question 5
The stem & leaf diagram below shows the weight distribution of 26
people when they joined a slimming club.
6
7
8
9
10
11
12
0
1
2
4
5
2
3
2
4
7
median = 87
5 7 7
2 5 6 6 8 9
6 9 9
7
Menu
Q1 = 77
Q3 = 99
Full solution
11 4 = 114kg
1 1 3
Comments
(a) Find the median, lower & upper quartiles for this data.
(b) Use the data to construct a boxplot.
CLICK
(c) The boxplot below shows the weight distribution for these
people after several months. Compare the two & comment
on the results.
EXIT
60
70
80
90
100
110
120
CLICK
Question 5
6
7
8
9
10
11
12
0
1
2
4
5
2
3
2
4
7
5 7 7
2 5 6 6 8 9
6 9 9
7
11 4 = 114kg
1 1 3
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
2. There are 13 values before median
so Q1 position is 6th value
(a) Find the median, lower &
upper quartiles for this data. (ii)
so
Q1 = 77
3. There are 13 values after median so
Q3 position is 20th position
Begin Solution
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so
Q3 = 99
Question 5
6
7
8
9
10
11
12
0
1
2
4
5
2
3
2
4
7
5 7 7
2 5 6 6 8 9
6 9 9
7
11 4 = 114kg
1 1 3
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.
(b) Use the data to construct
a boxplot.
Begin Solution
Continue Solution
Comments
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60
70
80
90
100
110
120
5. Compare spread and relevant
average.
Question 5
(c) The boxplot below shows
the weight distribution for
(c) Lightest has put on weight –
these people after several
lowest now 65,
months.
heaviest 3 have lost weight –
Compare the two &
highest now 115,
comment on the results.
median same but overall
spread of weights has decreased
as Q3-Q1 was 22
60
70
80
90
100
110
120
but is now only 15.
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Remember:
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,
To draw a boxplot you need
a “five-figure summary”:
Box Plot :
Q3 = 99 & Highest = 123.
Lowest
60
70
80
90
100
110
Q1
Q2 Q3 Highest
120
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.
(a) How many people
Watching TV
went clubbing?
144°
(b) If 84 people went to the
cinema
x° theatre
clubbing
theatre then how big is x°?
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Charts, Graphs & Tables : Question 6
The pie chart below showsangle
the breakdown
of how a sample of
amount
=
630 people spent their Saturday
360° nights. 630
(a) How many people
Watching TV
went clubbing?
144°
(b) If 84 people went to the
cinema
x° theatre
clubbing
theatre then how big is x°?
What would you like to do now?
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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.
(a) How many people
Watching TV
cinema
= 252
went clubbing?
144°
(b) If 84 people went to the
x° theatre
clubbing
theatre then how big is x°?
= 48°
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Question 6
Watching
TV
144°
1. Set up ratio of angles and sectors
and cross multiply.
cinema
x° theatre
clubbin
g
How many people went
clubbing?
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(a) The angle is 144° so …..
angle = amount
360°
630
144° = amount
360°
630
360 x amount = 144 x 630
amount = 144 x 630  360
= 252
Question 6
Watching
TV
144°
2. Set up ratio of angles and sectors
and cross multiply.
cinema
x° theatre
clubbin
g
(b) If 84 people went to the
theatre then how big is x°?
Begin Solution
Continue Solution
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(b) The amount is 84 so …..
angle = amount
360°
630
angle =
84
360°
630
630 x angle = 360° x 84
angle = 360° x 84  630
= 48°
Comments
1. Set up ratio of angles and sectors
and cross multiply.
(a) The angle is 144° so …..
angle = amount
360°
630
144° = amount
360°
630
360 x amount = 144 x 630
Can also be tackled by using
proportion:
amount 
angle at centre
 total sample
360
Amount =
144
360
x 630
amount = 144 x 630  360
= 252
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Comments
2. Set up ratio of angles and sectors
and cross multiply.
(b) The amount is 84 so …..
angle = amount
360°
630
angle =
84
360°
630
Can also be tackled by using
proportion:
amount 
angle at centre
 total sample
360
84 =
x
360
x 630
630 x = 84 x 360
630 x angle = 360° x 84
angle = 360° x 84  630
= 48°
x = 84 x 360
630
End of graphs, charts etc.
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