1. Mean from frequency table

2. ABC Motoring
Here are the number of tests taken before successfully passing a driving test by 40 students of “ABC Motoring”
2, 4, 1, 2, 1, 3, 7, 1, 1, 2, 2, 3, 5, 3, 2, 3, 4, 1, 1, 2, 5, 1, 2, 3, 2, 6, 1, 2, 7, 5, 1, 2, 3, 6, 4, 4, 4, 3, 2, 4
It is difficult to analyse the data in this form. We can group the results into a frequency table.

3. ABC Motoring Number of tests taken Number of People 1 9 2 11 3 7 4 6 5
That’s better!

4. Finding the Mean Number of tests taken Number of People 1 9 2 11 3 7 4
Remember, when finding the mean of a set of data, we add together all the pieces of data.
Number of tests taken
Number of People 1 9 2 11 3 7 4 6 5
This tells us that there were nine 1’s in our list. So we would do
= 9
It is simpler to use 1x9!!
We can do this for every row.

5. Finding the Mean Number of tests taken Number of People 1 9 2 11 3 7 46 5
hhhhhhhhhh
1 x 9 = 9
2 x 11 = 22
3 x 7 = 21
4 x 6 = 24
5 x 3 = 15
6 x 2 = 12
7 x 2 = 14 40
107
So we have now added all the values up. What do we do now?
So we divide by the total number of people.
We divide by how many values there were.
We now need to add these together

6. The Mean is 107 ÷ 40 = 2.7 (1dp) Number of tests taken
Number of People 1 9 2 11 3 7 4 6 5
hhhhhhhhhh
1 x 9 = 9
2 x 11 = 22
3 x 7 = 21
4 x 6 = 24
5 x 3 = 15
6 x 2 = 12
7 x 2 = 14 40
107
The Mean is
107 ÷ 40 = 2.7 (1dp)

7. ABC Motoring
Students who learn to drive with ABC motoring, pass their driving test after a mean number of 2.7 tests.

8. Mean from Grouped frequency table

9. 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.

10. 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.

11. Estimating the mean from grouped data1
35 ≤ d < 40 3 10 12 8
Frequency f
Midpoint x
30 ≤ d < 35
Frequency × midpoint fx
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.)