• Study Resource
• Explore

Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Transcript
```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
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.
11
Averages (The Median)
The median is the middle value of a set of data once
the data has been ordered.
Example 1. Robert hit 12 balls at Grimsby driving range.
The recorded distances of his drives, measured in yards,
are given below. Find the median distance for his drives.
85, 125, 130, 65, 100, 70, 75, 50, 140, 135, 95, 70
50, 65, 70, 70, 75, 85, 95, 100, 125, 130, 135, 140
Two middle values so
take the mean.
Ordered data
Median drive = 90 yards
When there are two middle numbers
To find out where a middle number in a very long list, call the
number of numbers n. Then 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.
Questions
Mean from Frequency table
LO:- to be able to calculate the
mean from grouped data.
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 would be?
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)
Grouped Data
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
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?
Because the data is grouped, we
do not know individual scores. It is
not possible to add up the scores.
Midpoints
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
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 = 12.5 m
Find the midpoints of the other groups.
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.)
How accurate is the estimated mean?
Here are the javelin distances thrown by Year 10 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?
Questions