1. One or two middle numbers?
If there are 9 numbers in a list, will there be 1 or 2 middle numbers?
If there are 10 numbers in a list, will there be 1 or 2 middle numbers?
Discuss what the median is in each case.
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.

2. Two middle values so take the mean.
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
Ordered data
Two middle values so take the mean.
Median drive = 90 yards

3. 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).
Pupils should first predict whether there will be one or two middle numbers. They should see a connection between the fact that an odd number divided by 2 will always give an answer ending in .5 so that there will be two medians. Remind pupils that these numbers are not the medians - they are just the positions of the medians in the list.
There are 37 numbers in a list. Where is the median?
38 ÷ 2 = 19th number in the list.

4. Questions

5. Answers

6. Mean from Frequency table
LO:- to be able to calculate the mean from grouped data.

7. 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 whole list of results, what would it look like?
Numbers of sports played
Frequency 20 1 17 2 15 3 10 4 9 5 6
Pupils may benefit from calculating the mean from the list before they are ready to appreciate the multiplication method.
What do you think the mean would be?

8. Calculating the mean from a frequency table2 6 3 9 10 15 17 20
Frequency
Number of sports
× frequency 4 5 1
Numbers of sports played
0 × 20
= 0
1 × 17
= 17
2 × 15
= 30
3 × 10
= 30
4 × 9
= 36
Ask pupils to estimate the mean first. Discuss a suitable level of accuracy for rounding off in the context of discrete data.
5 × 3
= 15
6 × 2
= 12
TOTAL 76
140
Mean = 140 ÷ 76 =
2 sports (to the nearest whole)

9. Grouped Data

10. Grouped data Here are the Year Ten boys’ javelin scores.
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
35 ≤ d < 40 36
How could you calculate the mean from this data?
How is the data different from the previous examples you have calculated with?
The data has been grouped.
Because the data is grouped, we do not know individual scores. It is not possible to add up the scores.

11. Midpoints Javelin distances in metres Frequency 5 ≤ d < 10 18
15 ≤ d < 20 12
20 ≤ d < 25 10
25 ≤ d < 30 3
30 ≤ d < 35
35 ≤ d < 40
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 ≤ d < 15:
= 25
25 ÷ 2 =
The other midpoints are displayed on the next page. Point out the link between the midpoint and the median/ mean. Discuss the fact that it is likely that the scores within a group are evenly distributed i.e. half above and half below the midpoint. This is the best assumption to make, although it is obviously not always true. (The greater the data set, the more likely this is to be the case.) Some pupils may point out that 15 is not included in the group 10 ≤ d < 15; however, since it can get very close to 15 (e.g ) this will make no difference to an estimated mean.
12.5 m
Find the midpoints of the other groups.

12. Estimating the mean from grouped data1
35 ≤ d < 40 3 10 12 8
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
7.5
1 × 7.5
= 7.5
12.5
8 × 12.5
= 100
17.5
12 × 17.5
= 210
22.5
10 × 22.5
= 225
27.5
3 × 27.5
= 82.5
Ask pupils to estimate the mean first. Discuss a suitable level of accuracy for rounding off in the context of continuous data.
32.5
1 × 32.5
= 32.5
37.5
1 × 37.5
= 37.5
TOTAL 36
695
Estimated mean = 695 ÷ 36
= 19.3 m (to 1 d.p.)

13. 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
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
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.
Emphasise that although the estimated mean can be quite accurate it is preferable to use the original data if this is available. This can be particularly relevant in GCSE coursework.
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?

14. Questions

15. Answers