The idea, step by step
Bounds describe uncertainty
A reported 12.4 cm to the nearest 0.1 cm means the original length is at least 12.35 cm and less than 12.45 cm. Half the rounding unit gives the maximum rounding error.
Absolute error has the measurement unit. Relative error divides an absolute error by a specified reference value. Using the reported value as reference gives an estimate and should be identified as such.
Combine positive bounds carefully
For a rectangle with positive dimensions, multiply both lower bounds for the smallest area and both upper bounds for the upper limit. Do not add length errors and report the result as an area error.
Bounds show what rounding permits, not a probability distribution. There is no reason to assume that every value is equally likely unless a model explicitly says so.
A common trap
Taking the full rounding unit as the maximum rounding error.
Instead: Nearest-unit rounding has at most half a unit of error.
See the thinking
Nearest centimetre
A 25 cm measurement is rounded to the nearest centimetre. State bounds and estimate percentage error using 25 cm as reference.
- .
- Absolute rounding error≤0.5 cm.
- Estimated percentage=0.5/25×100=2%.
Rectangle bounds
A rectangle is 20 by 10 cm, each to the nearest centimetre. Find area bounds.
- Lower dimensions 19.5 and 9.5; product 185.25.
- Upper limits 20.5 and 10.5; product 215.25.
Check your understanding
A mass is 3.6 kg to the nearest 0.1 kg. Give bounds.
Show answer
3.55≤m<3.65 kg
Half the rounding unit is 0.05 kg.
Your turn
6 questions · 14 marks · Work at your own pace.
Read the questions online
- A length is 8 cm to the nearest centimetre. Give bounds.2 marks
- A mass is 2.30 kg to the nearest 0.01 kg. Give maximum absolute rounding error.2 marks
- A 40 m reported distance has absolute error 0.2 m. Use 40 m as reference to find percentage error.2 marks
- A rectangle is 6 by 4 cm, each to the nearest centimetre. Find lower and upper area limits.3 marks
- Compare 100 g with error 2 g and 500 g with error 5 g. Which has smaller relative error using the readings as reference?3 marks
- A 12.0 s reading is to the nearest 0.1 s. Can its true value be 12.06 s? Explain.2 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- A length is 8 cm to the nearest centimetre. Give bounds.2 marks
The half-unit is 0.5 cm.
7.5≤L<8.5 cm - A mass is 2.30 kg to the nearest 0.01 kg. Give maximum absolute rounding error.2 marks
Half of 0.01 is 0.005.
0.005 kg - A 40 m reported distance has absolute error 0.2 m. Use 40 m as reference to find percentage error.2 marks
0.2/40×100=0.5.
0.5% - A rectangle is 6 by 4 cm, each to the nearest centimetre. Find lower and upper area limits.3 marks
5.5×3.5=19.25.
6.5×4.5=29.25; upper endpoint excluded.
19.25 and 29.25 cm² - Compare 100 g with error 2 g and 500 g with error 5 g. Which has smaller relative error using the readings as reference?3 marks
2/100=2%; 5/500=1%.
The 500 g reading. - A 12.0 s reading is to the nearest 0.1 s. Can its true value be 12.06 s? Explain.2 marks
11.95≤t<12.05.
12.06 lies outside this interval.
No.
Curriculum reference & lesson scope
Australian Curriculum v9 and SA Curriculum
Surface area of composite solids; volume of composite solids; proportion/scaling of objects; related length/area/volume factors; measurement error; the effect of rounding and repeated approximations; practical design models; evaluating assumptions and suitable precision.
Official curriculum source ↗