Absolute, relative and percentage error | Year 9

01 / LEARN

Understand the idea.

Absolute and relative error

Absolute error is the distance between a measurement $m$ and an accepted reference $t$ : $∣ m - t ∣$ . It cannot be negative and keeps the measurement unit. Being above or below the reference does not change this distance.

For a fair comparison across different-sized quantities, divide by the magnitude of the accepted reference. Relative error is $\frac{∣ m - t ∣}{∣ t ∣}$ ; multiplying it by $100 %$ gives percentage error. The reference must be nonzero.

\begin{aligned} e & = ∣ m - t ∣ \\ e_{relative} & = \frac{∣ m - t ∣}{∣ t ∣} \\ e_{percent} & = \frac{∣ m - t ∣}{∣ t ∣} \times 100 % \end{aligned}

Rounded measurements have a range

A rounded reading gives an interval of possible unrounded values. Rounding to the nearest centimetre has a half-step of $0.5 cm$ ; rounding to the nearest $0.1 cm$ has a half-step of $0.05 cm$ .

That rounding allowance describes the recording precision. It does not supply a known true value or measure every possible instrument error. Calculate percentage error only when an accepted reference is provided, keeping this separate from the rounding interval.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find three forms of error

A scale reads 102 g for a certified 100 g mass. Find the absolute, relative and percentage error.

Worked solution

1
Find the absolute difference.
$e = ∣ 102 - 100 ∣ = 2 g$
The reference is the certified $100 g$ mass, and error is a distance.
2
Divide by the accepted value.
$e_{relative} = \frac{2}{100} = 0.02$
Dividing matching units leaves a unitless relative error.
3
Express the relative error as a percentage.
$e_{percent} = 0.02 \times 100 % = 2 %$
A proportion of $0.02$ is two parts per hundred.

Answer

2 g; 0.02; 2%

EXAMPLE 02

A smaller absolute error is not always better

A 40 cm reference length is measured as 39 cm. A 100 cm reference is measured as 98 cm. Compare their percentage errors.

Worked solution

1
Calculate the first absolute error.
$e_{1} = ∣ 39 - 40 ∣ = 1 cm$
The first measurement is one centimetre below its reference.
2
Find the first percentage error.
$p_{1} = \frac{1}{40} \times 100 % = 2.5 %$
The error is compared with the first accepted length.
3
Calculate the second percentage error.
$\begin{aligned} e_{2} & = ∣ 98 - 100 ∣ = 2 cm \\ p_{2} & = \frac{2}{100} \times 100 % = 2 % \end{aligned}$
The second error is larger in centimetres but is divided by a larger reference.
4
Compare relative accuracy.
$2 % < 2.5 %$
The $98 cm$ reading is closer to its reference as a proportion of the reference length.

Answer

The 98 cm measurement is closer in relative terms.

03 / YOUR TURN

Try it yourself.

Check your understanding

A reading is 18 cm and the accepted value is 20 cm. What is the percentage error?

Show answer & reasoning

10%.

The absolute error is $∣ 18 - 20 ∣ = 2 cm$ . Relative to the accepted $20 cm$ , the percentage error is $\frac{2}{20} \times 100 % = 10 %$ .

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6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

A measurement is 78 m and the accepted length is 80 m. Find its absolute, relative and percentage error.
3 marks
A 20 g reference is measured as 20.4 g. Find the percentage error.
2 marks
A 250 L tank capacity is estimated as 240 L. Find the absolute and percentage errors.
2 marks
A ruler reading is rounded to the nearest 0.1 cm. State the greatest possible rounding error.
2 marks
A length is recorded as 6.2 m to the nearest 0.1 m. Give the interval of values that round to this reading.
3 marks
A 100 cm reference is measured as 98 cm. A 50 cm reference is measured as 49 cm. Which has the smaller percentage error?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A measurement is 78 m and the accepted length is 80 m. Find its absolute, relative and percentage error.
3 marks
$∣ 78 - 80 ∣ = 2$ m; $\frac{2}{80} = 0.025$ ; $0.025 \times 100 % = 2.5 %$ .
2 m; 0.025; 2.5%
A 20 g reference is measured as 20.4 g. Find the percentage error.
2 marks
Error $= 0.4$ g; $\frac{0.4}{20} \times 100 % = 2 %$ .
2%
A 250 L tank capacity is estimated as 240 L. Find the absolute and percentage errors.
2 marks
Absolute error $= 10$ L; $\frac{10}{250} \times 100 % = 4 %$ .
10 L and 4%
A ruler reading is rounded to the nearest 0.1 cm. State the greatest possible rounding error.
2 marks
Half of a 0.1 cm step is 0.05 cm.
0.05 cm
A length is recorded as 6.2 m to the nearest 0.1 m. Give the interval of values that round to this reading.
3 marks
The half-step is 0.05 m. Subtract and add it to 6.2 m. The upper endpoint rounds up to 6.3 m.
$6.15 \leq L < 6.25$ m.
A 100 cm reference is measured as 98 cm. A 50 cm reference is measured as 49 cm. Which has the smaller percentage error?
3 marks
$\frac{2}{100} \times 100 % = 2 %$ and $\frac{1}{50} \times 100 % = 2 %$ .
Neither; both have 2% error.

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Curriculum reference & lesson scope

Australian Curriculum v9 / South Australian Curriculum

Year 9 surface area and volume of right prisms and cylinders, capacity, absolute/relative/percentage measurement error, precision and practical measurement problems.

Official curriculum source ↗

In this lesson

Understand the idea.

Absolute and relative error

Rounded measurements have a range

See the method in action.

Find three forms of error

Find the absolute difference.

Divide by the accepted value.

Express the relative error as a percentage.

A smaller absolute error is not always better

Calculate the first absolute error.

Find the first percentage error.

Calculate the second percentage error.

Compare relative accuracy.

Try it yourself.

Check your understanding

Keep practising