MathematicsExam code: 0580
Averages and Range
Statistics
Why it is only an estimate
- Grouped data records how many values fall in each class interval but not the values themselves
- Without the original values the exact total cannot be found, so the mean can only be estimated
- The estimate assumes every value in a class sits at the midpoint of that class
- Find a midpoint by adding the two endpoints and halving
- The phrase "estimate of the mean" in a question is the signal that the data is grouped
The method
- Add a midpoint column, then a frequency × midpoint column
- Total the frequency column and the frequency × midpoint column
- Divide the second total by the first
- Whichever class holds the largest frequency is the modal class
Worked example
An estimate of the mean from grouped data
The table shows the time t minutes taken by 50 people. Calculate an estimate of the mean, and write down the modal class.
| Time (t minutes) | 0 < t ⩽ 10 | 10 < t ⩽ 20 | 20 < t ⩽ 30 | 30 < t ⩽ 40 |
|---|---|---|---|---|
| Frequency | 8 | 14 | 18 | 10 |
Solution:
- Midpoints are (0 + 10) ÷ 2 = 5, then 15, 25 and 35
- Multiply each frequency by its midpoint
- 8 × 5 = 40, 14 × 15 = 210, 18 × 25 = 450, 10 × 35 = 350
- Total of frequency × midpoint = 40 + 210 + 450 + 350 = 1050
- Total frequency = 8 + 14 + 18 + 10 = 50
- Estimate of the mean = 1050 ÷ 50 = 21 minutes
- A frequency of 18 is the largest in the table, so the modal class is 20 < t ⩽ 30
Exam tip
Work out each midpoint from its own class
An estimate of the mean is set in 9 of the 21 papers, and in 8 of them the classes differ in width and the midpoints rise in uneven steps. Work each one out from the two ends of its own class: 30 < t ⩽ 35 has midpoint 32.5, but 35 < t ⩽ 50 has 42.5. The midpoints earn a method mark of their own, and one scheme gives nothing for class widths used instead.