Measurement error | Year 10

01 / LEARN

Understand the idea.

Bounds describe uncertainty

A rounded measurement represents an interval, not one exact value. For nearest-unit rounding, values within half a unit produce the same reading. The usual bounds include the lower endpoint and exclude the upper endpoint.

Absolute error has the original measurement unit. Relative error divides an error by a reference value; multiplying by $100$ expresses it as a percentage. Using the rounded reading as reference gives an estimate, not a known exact relative error.

relative error = \frac{absolute error}{reference value}

Combine positive bounds carefully

For positive rectangular dimensions, increasing either length increases area. Multiply both lower bounds to obtain the minimum area, and multiply both upper limits to obtain the area's upper limit. Keep the endpoint inequalities consistent.

Adding two length uncertainties cannot produce an area uncertainty because their units differ. Bounds describe all values permitted by rounding; they do not imply that those values have equal probability.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Nearest centimetre

A 25 cm measurement is rounded to the nearest centimetre. State bounds and estimate percentage error using 25 cm as reference.

Worked solution

1
Find half the rounding unit.
$\frac{1 cm}{2} = 0.5 cm$
Rounding to the nearest centimetre allows at most half a centimetre of rounding error.
2
Write the length bounds.
$24.5 \leq L < 25.5 (cm)$
The lower boundary rounds to $25$ under the stated convention; the upper boundary rounds to the next reading.
3
Estimate the maximum percentage error.
$\frac{0.5}{25} \cdot 100 % = 2 %$
The question specifies the reported $25 cm$ value as reference.

Answer

24.5≤L<25.5 cm; about 2%

EXAMPLE 02

Rectangle bounds

A rectangle is 20 by 10 cm, each to the nearest centimetre. Find area bounds.

Worked solution

1
Bound each positive dimension.
$\begin{aligned} 19.5 & \leq l < 20.5 \\ 9.5 & \leq w < 10.5 \end{aligned} (cm)$
Each nearest-centimetre measurement has a half-centimetre rounding interval.
2
Multiply the two lower bounds.
$A_{min} = 19.5 (9.5) = 185.25 {cm}^{2}$
For positive dimensions, the smallest area uses both smallest lengths.
3
Multiply the upper limits and combine.
$\begin{array}{c} A_{upper} = 20.5 (10.5) = 215.25 {cm}^{2}, \\ 185.25 \leq A < 215.25 ({cm}^{2}) . \end{array}$
The upper length endpoints are excluded, so the upper area endpoint is excluded too.

Answer

185.25≤A<215.25 cm²

03 / YOUR TURN

Try it yourself.

Check your understanding

A mass is 3.6 kg to the nearest 0.1 kg. Give bounds.

Show answer & reasoning

3.55≤m<3.65 kg

Half of $0.1 kg$ is $0.05 kg$ , so the permitted mass interval is $3.55 \leq m < 3.65$ in kilograms.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

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.
$7.5 \leq L < 8.5$ 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
$\frac{0.2}{40} \times 100 = 0.5$ .
0.5%
A rectangle is 6 by 4 cm, each to the nearest centimetre. Find lower and upper area limits.
3 marks
$5.5 \times 3.5 = 19.25$ .
$6.5 \times 4.5 = 29.25$ ; 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
$\frac{2}{100} = 2$ %; $\frac{5}{500} = 1$ %.
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
$11.95 \leq t < 12.05$ .
12.06 lies outside this interval.
No.

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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 ↗

In this lesson

Understand the idea.

Bounds describe uncertainty

Combine positive bounds carefully

See the method in action.

Nearest centimetre

Find half the rounding unit.

Write the length bounds.

Estimate the maximum percentage error.

Rectangle bounds

Bound each positive dimension.

Multiply the two lower bounds.

Multiply the upper limits and combine.

Try it yourself.

Check your understanding

Keep practising