Transcript D2 Averages and range

Maths Age 14-16
D2 Averages and range
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Contents
D2 Averages and range
A D2.1 The mode
A D2.2 The mean
A D2.3 Calculating the mean from frequency tables
A D2.4 The median
A D2.5 Comparing data
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The three averages and range
There are three different types of average:
MODE
MEAN
MEDIAN
most common
sum of values
number of values
middle value
The range is not an average, but tells you how the data is
spread out:
RANGE
largest value – smallest value
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The mode
The most common item is called the mode.
The mode is the item that occurs the most often in a data set.
In the graph the mode is sprint because it is represented by
the highest bar.
We could also say “The modal athletic event is sprint.”
Is it possible to have more than one modal value?
Is it possible to have no modal value?
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Yes
Yes
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The mode
These figures show the number of pupils that attended a
school athletics club each week.
14
13
15
14
15
11
13
16
12
14
14
15
15
9
0
10
11
12
Discuss :
Over how many weeks were the results collected?
What is the modal number of pupils attending?
Are there any unusual results in the data set?
Very unusual results are called outliers. Can you think of
any possible reasons for the outlier in this data set?
If the data set were very large, what would be the best way
to find the mode?
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How many sports do you play?
A group of pupils were asked how many sports they played.
This graph shows the results.
14
Frequency
12
10
8
6
4
2
0
0
1
2
3
4
5
6
Numbers of sports played
How many pupils play more than two sports?
What is the modal number of sports played?
How many pupils took part in the survey?
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Grouped data
This graph represents girls’ times for a 100 m sprint race.
10
8
6
4
2
0
12
13
14
15
16
17
18
19
20
What is the modal time interval?
How many girls are in this interval?
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When the mode is not appropriate
Another survey is carried out among university students.
The results are represented in this table:
Numbers of
sports played
Frequency
0
20
1
17
2
15
3
10
4
9
5
3
6
2
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A newspaper reporter writes:
“You may be surprised to learn
that the average number of
sports played by university
students is 0.”
Do you think this is a fair
representation of the data?
Why is the mode a misleading
average in this example?
Should the reporter say which
average has been used?
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Skewed data
Data that is heavily weighted towards one end of the data
set is said to be skewed.
When data is skewed, the mode is not an appropriate
average.
25
14
12
20
Frequency
Frequency
10
15
10
8
6
4
5
2
0
0
1
2
3
4
5
6
Numbers of sports played
Positively skewed data
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7
1
2
3
4
5
6
7
Numbers of sports played
Negatively skewed data
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Contents
D2 Averages and range
A D2.1 The mode
A D2.2 The mean
A D2.3 Calculating the mean from frequency tables
A D2.4 The median
A D2.5 Comparing data
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Comparing data
St Clement Danes School holds an inter-form athletics
competition. Each class must select their five best boys and
five best girls for each event.
Here are the times in seconds for the 100 m sprint for the two
best classes.
10B girls
12.8
14.7
15.3
15.4
15.4
10B boys
11.5
12.9
13.4
13.8
14.3
10C girls
14.9
15.2
15.9
16.4
16.5
10C boys
12.0
12.2
12.8
13.1
13.1
Which class should win and why?
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The mean
The mean is the most commonly used average.
To calculate the mean of a set of values we add together the
values and divide by the total number of values.
Sum of values
Mean =
Number of values
For example, the mean time for Class 10B girls is:
12.8 + 14.7 + 15.3 + 15.4 + 15.4
73.6
=
= 14.72
5
5
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The mean
Calculate the mean times for the other three groups.
10B girls
mean time
10B boys 10C girls
14.72
13.18
15.78
10C boys
12.64
Now calculate means for Class 10B and Class 10C
(with girls and boys combined).
mean time
Class 10B
Class 10C
13.95
14.21
Based on these results, who should win?
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Calculating the mean
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Calculating a missing data item
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Outliers and their effect on the mean
The school athletics team takes part in an inter-school
competition. James’s shot results (in metres) are below.
9.46
9.25
8.77
10.25
10.35
9.59
4.02
Discuss:
What is the mean throw?
Is this a fair representation of James’s ability? Explain.
What would be a fair way for the competition to operate?
A data item that is significantly higher or lower
than the other items is called an outlier. Outliers
can increase or reduce the mean dramatically,
making it a less accurate measure of the data.
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Outliers and their effect on the mean
Here are some 1500 metre race results in minutes.
6.26
6.28
6.30
6.39
5.38
4.54
10.59
6.35
7.01
Discuss:
Are there any outliers?
Will the mean be increased or reduced by the outlier?
Calculate the mean with the outlier.
Now calculate the mean without the outlier. How much
does it change?
It may be appropriate in research or experiments to remove
an outlier before carrying out analysis of results.
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Contents
D2 Averages and range
A D2.1 The mode
A D2.2 The mean
A D2.3 Calculating the mean from frequency tables
A D2.4 The median
A D2.5 Comparing data
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Calculating the mean from a frequency table
Here are the results of a survey carried out among university
students.
If you were to write out the
Numbers of Frequency
whole list of results, what
sports played
would it look like?
0
20
1
17
2
15
3
10
4
9
5
3
6
2
What do you think the mean will be?
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Calculating the mean from a frequency table
Numbers of Frequency
sports played
0
1
2
3
4
5
6
TOTAL
20
17
15
10
9
3
2
76
Number of sports
× frequency
0 × 20
1 × 17
2 × 15
3 × 10
4×9
5×3
6×2
=0
= 17
= 30
= 30
= 36
= 15
= 12
140
Mean = 140 ÷ 76 = 2 sports (to the nearest whole number)
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Grouped data
Javelin
Frequency
distances in
metres
5 ≤ d < 10
1
10 ≤ d < 15
8
15 ≤ d < 20
12
20 ≤ d < 25
10
25 ≤ d < 30
3
30 ≤ d < 35
35 ≤ d < 40
1
1
36
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Here are the boys’ javelin
scores.
How is the data different from
the previous examples?
Because the data is grouped, we
do not know individual scores. It is
not possible to add up the scores.
How could you calculate the
mean from this data?
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Midpoints
Javelin
Frequency
distances in
metres
5 ≤ d < 10
1
10 ≤ d < 15
8
15 ≤ d < 20
12
20 ≤ d < 25
10
It is possible to find an estimate
for the mean.
25 ≤ d < 30
30 ≤ d < 35
35 ≤ d < 40
10 + 15 = 25
3
1
1
This is done by finding the
midpoint of each group.
To find the midpoint of the group
10 ≤ d < 15:
25 ÷ 2 = 12.5 m
Find the midpoints of the other groups.
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Estimating the mean from grouped data
Javelin
Frequency
distances in
metres
5 ≤ d < 10
1
10 ≤ d < 15
8
15 ≤ d < 20
12
20 ≤ d < 25
10
25 ≤ d < 30
3
30 ≤ d < 35
1
35 ≤ d < 40
1
TOTAL
36
Midpoint
7.5
12.5
17.5
22.5
27.5
32.5
37.5
Frequency × midpoint
1 × 7.5
8 × 12.5
12 × 17.5
10 × 22.5
3 × 27.5
1 × 32.5
1 × 37.5
= 7.5
= 100
= 210
= 225
= 82.5
= 32.5
= 37.5
695
Estimated mean = 695 ÷ 36 = 19.3 m (to 1 d.p.)
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How accurate is the estimated mean?
Here are the javelin distances thrown before the data was
grouped.
35.00 31.05 28.89 25.60 25.33
24.11
23.50 21.82 21.78
21.77 21.60 21.00 20.70 20.20 20.00 19.50 19.50 18.82
17.35 17.31 16.64 15.79 15.75 15.69 15.52 15.25 15.00
14.50 12.80 12.50 12.00 12.00 12.00
11.85
10.00
9.50
Work out the mean from the original data above and compare it
with the estimated mean found from the grouped data.
The estimated mean is 19.3 metres (to 1 d.p.).
The actual mean is 18.7 metres (to 1 d.p.).
How accurate was the estimated mean?
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Contents
D2 Averages and range
A D2.1 The mode
A D2.2 The mean
A D2.3 Calculating the mean from frequency tables
A D2.4 The median
A D2.5 Comparing data
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The median
The median is the middle number when all
numbers are in order.
Calculate the median of the 1500 m results.
6.26
6.28
6.30
6.39
5.38
4.54
10.59
6.35
7.01
Write the results in order and find the middle value:
4.54
5.38
6.26
6.28
6.30
6.35 6.39
7.01 10.59
Why is this a more appropriate average than
the mean for these results?
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Choosing the most appropriate average
What are the mean and median for these sets of attendance
figures for three lunchtime activities?
Drama club
20
20
20
20
20
20
20
Choir
17
18
19
20
21
22
23
Orchestra
18
18
19
20
25
28
29
Explain your answers.
Which average is the best one to use when deciding
which of the three activities is the most popular? Why?
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Outliers and the median and mean
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When there are two middle numbers
2.15 2.21 2.40 2.55 2.80 3.32 3.46 3.63 3.83 4.74
Here are 10B girls’ long jump results in metres.
How could you work out the median jump?
If there are two middle numbers, you need to find what is
halfway between them.
Add the two numbers together and then divide by two.
2.80 m + 3.32 m = 6.12 m
6.12 m ÷ 2 = 3.06 m
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Finding halfway between two numbers
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One or two middle numbers?
If there are 9 numbers in a list, will there be 1
or 2 middle numbers?
2
3
4
6
6
7
8
9
11
If there are 10 numbers in a list, will there
be 1 or 2 middle numbers?
1
2
3
4
6
6
7
8
9
11
If there is an even number of numbers in a list, there
will be two middle numbers.
If there is an odd number of numbers in a list, there
will be one middle number.
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When there are two middle numbers
To find out where the middle number is in a very long list, call
the number of numbers n. The middle number is then:
(n + 1) ÷ 2
For example,
There are 100 numbers in a list. Where is the median?
101 ÷ 2 = 50.5th number in the list (halfway between
the 50th and the 51st).
There are 37 numbers in a list. Where is the median?
38 ÷ 2 = 19th number in the list.
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Where is the median?
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Contents
D2 Averages and range
A D2.1 The mode
A D2.2 The mean
A D2.3 Calculating the mean from frequency tables
A D2.4 The median
A D2.5 Comparing data
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The range
The highest and lowest scores can be useful in deciding
who is more consistent.
The lowest score subtracted from the highest score is
called the range.
Remember that the range is not an average, but a measure of
spread.
If the scores are spread out then the range will be higher and
the scores less consistent.
If the scores are close together then the range will be lower
and the scores more consistent.
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The range
Here are the high jump scores in metres for two girls in five
different competitions:
Joanna
Kirsty
1.62
1.59
1.41
1.45
1.35
1.41
1.20
1.30
1.15
1.30
Find the range for each girl’s results and use
this to find out who is the most consistent.
Joanna’s range = 1.62 – 1.15 = 0.47
Kirsty’s range = 1.59 – 1.30 = 0.29
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The range
Joanna
Kirsty
1.62
1.59
1.41
1.45
1.35
1.41
1.20
1.30
1.15
1.30
Now calculate the mean for each girl.
Range
Mean
Joanna
0.47 m
Kirsty
0.29 m
1.35 m
1.41 m
Use these results to decide which one you would
enter into the athletics competition and why.
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Calculating the mean, median and range
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Comparing sets of data
Here is a summary of Chris and Rob’s performance in the 200
metres over a season. They each ran 10 races.
Mean
Range
Chris
24.8 seconds
1.4 seconds
Rob
25.0 seconds
0.9 seconds
Which of these conclusions are correct?
Rob is more reliable.
Rob is better because his mean is higher.
Chris is better because his range is higher.
Chris must have run a better time for his quickest race.
On average, Chris is faster but he is less consistent.
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Comparing sets of data
Mean
Range
Chris
Rob
24.8 seconds
1.4 seconds
25.0 seconds
0.9 seconds
Here is the original data for Chris and Rob.
25.8 25.1 25.0 25.0 24.9 24.6 24.5 24.4 24.5 24.4
25.2 25.2 25.1 25.1 25.1 25.0 25.0 25.0 24.9 24.3
Use the summary table above to decide which
data set is Chris’s and which is Rob’s?
Who has the best time?
Who has the worst time?
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Comparing hurdles scores
Age 14
12.1
14.0
15.3
15.4
15.4
15.6
15.7
15.7
16.1
16.7
17.0
Age 15
12.3
13.7
15.5
15.5
15.6
15.9
16.0
16.1
16.1
17.1
22.9
Here are the top eleven hurdles scores in
seconds for pupils aged 14 and 15.
Work out the mean and range.
Mean
Age 14
15.4
Age 15
16.1
Range
4.9
10.6
Which age group do you think
is better and why?
Why might the 15 year olds feel the
comparison is unfair?
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Finding the interquartile range
The time of 22.9 seconds is an outlier.
When there are outliers in the data, it is more appropriate to
calculate the interquartile range.
The interquartile range is the range of the middle half of
the data.
The lower quartile is the data value that is quarter of the way
along the list.
The upper quartile is the data value that is three quarters of
the way along the list.
interquartile range = upper quartile – lower quartile
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Locating the upper and lower quartiles
Age 14
Age 15
12.1
14.0
15.3
15.4
15.4
15.6
15.7
15.7
16.1
16.7
17.0
12.3
13.7
15.5
15.5
15.6
15.9
16.0
16.1
16.1
17.1
22.9
There are 11 times in each list.
Where is the median in each list?
Where is the lower quartile in each list?
Where is the upper quartile in each list?
Interquartile range for 14 year olds:
16.1 – 15.3 = 0.8
Interquartile range for 15 year olds:
16.1 – 15.5 = 0.6
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The location of quartiles in an ordered data set
When there are n values in an ordered data set:
n+1
The lower quartile =
The median =
4
n+1
The upper quartile =
th value
th value
2
3(n + 1)
4
th value
The interquartile range = the upper quartile – the lower quartile
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Finding the interquartile range
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Review
To review the work you have covered in this topic:
1) Play “Guess the word”.
Write out the key words on cards.
Shuffle the cards.
Describe the word on each card to your partner.
Your partner must guess the word.
Do as many as you can in one minute, then swap over.
2) Make up challenges involving sets of data for your partner,
such as working out the mean.
3) Make a list of possible mistakes to avoid in this topic.
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