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M EAN. M ODE most common. M EDIAN middle value. sum of values number of values. R ANGE largest value – smallest value. The Three Averages and Range. There are three different types of average :. The range is not an average, but tells you how the data is spread out:. 20. 15. - PowerPoint PPT Presentation
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The Three Averages and Range
There are three different types of average:
MEDIAN
middle value
The range is not an average, but tells you how the data is spread out:
RANGE
largest value – smallest value
MODE
most common
MEAN
sum of valuesnumber of values
This graph shows pupils’ favourite athletics events.
Favourite athletics event
0
5
10
15
20
Sprint
Long
dist
ance
runn
ing
Hurdle
s
High ju
mp
Long
jum
p
Triple
jump
Shot
Discus
Jave
lin
Fre
quen
cy
Which is the most popular event? How do you know?
The school athletics team take part in an inter-schools competition. James’s shot results (in metres) are below.
Outliers and their effect on the mean
9.46 9.25 8.77 10.25 10.35 9.59 4.02
A data item that is significantly higher or lower than the other items is called an outlier. Outliers affect the mean, by reducing or increasing it.
A data item that is significantly higher or lower than the other items is called an outlier. Outliers affect the mean, by reducing or increasing it.
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?
Here are some 1500 metre race results in minutes.
Outliers and their effect on the mean
It may be appropriate in research or experiments to remove an outlier before carrying out analysis of results.
Discuss:
6.26 6.28 6.30 6.39 5.38 4.54 10.59 6.35 7.01
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?
Calculating the mean from a frequency table
26
3
9
10
15
17
20
Frequency Number of sports
× frequency
4
5
3
2
1
0
Numbers of sports played
TOTAL
0 × 20 = 0
1 × 17 = 17
2 × 15 = 30
3 × 10 = 30
4 × 9 = 36
5 × 3 = 15
6 × 2 = 12
Mean = 140 ÷ 76 =
14076
1.84
2 sports (to the nearest whole)
Starter
Draw up a frequency table and find the mean number of warts per witch.
The mean number of warts is 43 ÷ 16 = 2.69 (2 d.p.)
What is the mode number of warts?
Calculating the median from a frequency table
26
3
9
10
15
17
20
Frequency Number of sports
× frequency
4
5
3
2
1
0
Numbers of sports played
TOTAL
0 × 20 = 0
1 × 17 = 17
2 × 15 = 30
3 × 10 = 30
4 × 9 = 36
5 × 3 = 15
6 × 2 = 12
Median = middle number =
14076
76 ÷ 2 = 38This occurs in the 2 category so we would say the median is 2
Because the data is grouped, we do not know individual scores. It is not possible to add up the scores.
Grouped data
Javelin distances in
metres
Frequency
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
36
Here are the Year Ten boys’ javelin scores.
How could you calculate the mean from this data?
How is the data different from the previous examples you
have calculated with?
It is possible to find an estimate for the mean.
This is done by finding the midpoint of each group.
To find the midpoint of the group 10 ≤ d < 15:
10 + 15 = 25
25 ÷ 2 =
Midpoints
12.5 m
Javelin distances in
metres
Frequency
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
Find the midpoints of the other groups.
135 ≤ d < 40
1
3
10
12
8
1
Frequency Midpoint
30 ≤ d < 35
Frequency × midpoint
25 ≤ d < 30
20 ≤ d < 25
15 ≤ d < 20
10 ≤ d < 15
5 ≤ d < 10
Javelin distances in
metres
Estimating the mean from grouped data
1 × 7.5 = 7.5
8 × 12.5 = 100
12 × 17.5 = 210
10 × 22.5 = 225
3 × 27.5 = 82.5
1 × 32.5
Estimated mean = 695 ÷ 36
1 × 37.5
= 32.5
7.5
12.5
17.5
22.5
27.5
32.5
37.5 = 37.5
36 695TOTAL
= 19.3 m (to 1 d.p.)
How accurate is the estimated mean?
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
Here are the javelin distances thrown by Year 10 before the data was grouped.
Work out the mean from the original data above and compare it with the estimated mean found from the grouped data.
How accurate was the estimated mean?
The estimated mean is 19.3 metres (to 1 d.p.).
The actual mean is 18.7 metres (to 1 d.p.).