The idea, step by step
Centre and spread answer different questions
Continuous variables measure quantities such as time, height and mass. Recorded measurements are often rounded, but the underlying variable can take values between the recorded values. The mean uses every recorded value, the median describes the middle ordered position, and the range is a simple measure of spread.
Always order the data before finding the median. For an even number of values average the middle pair. Attach the measurement unit to each result, and state whether an unusually large or small value makes the mean less representative.
Grouped measurements give estimates
A grouped table keeps counts for intervals rather than each original measurement. Use each interval midpoint as a representative value, multiply it by the interval frequency, add these products, and divide by the total frequency.
The result is an estimated mean because the original values may lie anywhere inside their intervals. A class boundary may identify the median class, but it does not by itself reveal the exact median. State what the data support instead of giving invented precision.
A common trap
The interval midpoint is the actual value of every observation.
Instead: A midpoint is a working representative. It makes the grouped mean an estimate, not an exact raw-data mean.
See the thinking
Find all three summaries
Race times are 12.4, 12.8, 13.1, 13.1, 14.6 and 15.0 seconds. Find the mean, median and range.
- The sum is 81.0 and there are 6 values, so the mean is 81.0÷6=13.5.
- The middle values are 13.1 and 13.1, so the median is 13.1.
- The range is 15.0−12.4=2.6.
Estimate a mean from intervals
Estimate the mean journey time using the table.
| Time interval(min) | Frequency |
|---|---|
| 0≤t<10 | 2 |
| 10≤t<20 | 5 |
| 20≤t<30 | 3 |
- Use midpoints 5, 15 and 25 minutes.
- The weighted total is 5×2+15×5+25×3=160.
- There are 10 journeys, so the estimated mean is 160÷10=16.
Check your understanding
Times are 4.2, 4.7, 5.1 and 6.0 minutes. What is their median?
Show answer
4.9 minutes
The middle pair is 4.7 and 5.1; their average is 4.9.
Your turn
6 questions · 15 marks · Work at your own pace.
Read the questions online
- Find the mean of the lengths 2.4, 2.8, 3.2 and 3.6 cm.2 marks
- Find the median and range of 8.2, 7.5, 9.0, 8.4 and 7.9 seconds.2 marks
- Find the mean and median of 1.2, 1.4, 1.5, 1.7 and 9.2 kg. Which is more typical of the four smaller masses?3 marks
- Estimate the mean waiting time for the table. Explain why it is not exact.3 marks
Waiting times Interval(min) Frequency 0≤t<10 7 10≤t<20 2 20≤t<30 1 - One class has mean height 1.55 m. Another has median height 1.55 m. Does this prove their height distributions are the same?2 marks
- A runner's five times have mean 14.2 s. Four times total 55.9 s. Find the fifth time.3 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- Find the mean of the lengths 2.4, 2.8, 3.2 and 3.6 cm.2 marks
The total is 12.0 cm.
12.0÷4=3.0 cm.
3.0 cm - Find the median and range of 8.2, 7.5, 9.0, 8.4 and 7.9 seconds.2 marks
Order 7.5, 7.9, 8.2, 8.4, 9.0.
The middle is 8.2 and 9.0−7.5=1.5.
Median 8.2 s; range 1.5 s - Find the mean and median of 1.2, 1.4, 1.5, 1.7 and 9.2 kg. Which is more typical of the four smaller masses?3 marks
The sum is 15.0, so the mean is 3.0.
The middle value is 1.5; 9.2 pulls the mean upward.
Mean 3.0 kg; median 1.5 kg; median is more typical. - Estimate the mean waiting time for the table. Explain why it is not exact.3 marks
Waiting times Interval(min) Frequency 0≤t<10 7 10≤t<20 2 20≤t<30 1 Midpoints are 5, 15, 25.
(5×7+15×2+25×1)÷10=90÷10=9.
Estimated mean 9 minutes; interval midpoints replace unknown individual times. - One class has mean height 1.55 m. Another has median height 1.55 m. Does this prove their height distributions are the same?2 marks
The summaries measure different features.
Different data sets can share a summary without sharing spread or shape.
No. - A runner's five times have mean 14.2 s. Four times total 55.9 s. Find the fifth time.3 marks
The five times total 5×14.2=71.0 s.
71.0−55.9=15.1 s.
15.1 s
Curriculum reference & lesson scope
Australian Curriculum v 9 and SA Curriculum Mathematics
Year 10 continuous distributions, bivariate data, two-way categorical tables, bias and investigations, with the approved percentile and standard-deviation content included in the normal chapter.
Official curriculum source ↗