Accuracy and precision | Year 11

01 / LEARN

Understand the idea.

Understand the idea

Accuracy describes agreement with a trusted reference. Repeatability describes agreement between repeated readings. An instrument can give nearly identical readings while consistently reading too high, so repeated agreement alone does not establish accuracy.

Resolution is the smallest marked or displayed change. Rounding a reading to a stated unit hides some detail: the reported value represents an interval extending half a rounding unit on either side, using the usual round-half-up convention.

Choose and apply a method

For a positive length recorded as $8.4 cm$ to the nearest $0.1 cm$ , the half-unit margin is $0.05 cm$ . Thus $8.35 \leq L < 8.45$ . Include the lower endpoint and exclude the upper one, which rounds to the next reported value.

Absolute error measures the size of the difference from a reference and retains the original unit. Percentage error divides that difference by the nonzero reference magnitude. These comparisons and a rounding interval describe different things; rounding alone does not identify every source of uncertainty.

\begin{array}{c} percentage error \\ = \frac{∣ measured - reference ∣}{∣ reference ∣} \times 100 % \end{array}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find an interval

A parcel mass is 2.36 kg to the nearest 0.01 kg. State its interval.

Worked solution

1
Find half the rounding unit.
$\frac{0.01}{2} = 0.005 kg$
Masses within half a rounding unit give the same reported value.
2
Calculate the two boundaries.
$\begin{aligned} 2.36 - 0.005 & = 2.355 \\ 2.36 + 0.005 & = 2.365 \end{aligned}$
Subtract and add the same half-unit margin.
3
State the rounding interval.
$2.355 \leq m < 2.365 (kg)$
The upper boundary rounds to 2.37 kg under the usual convention, so it is excluded.

Answer

$2.355 \leq m < 2.365$ kg.

EXAMPLE 02

Compare with a reference

A scale reads 502 g for a verified 500 g mass. Find absolute and percentage error.

Worked solution

1
Find the absolute error.
$e = ∣ 502 - 500 ∣ = 2 g$
Absolute value measures the size of the error independently of its direction.
2
Compare with the verified mass.
$\frac{e}{500} = \frac{2}{500} = 0.004$
The reference mass, not the scale reading, supplies the comparison base.
3
Express the ratio as a percentage.
$0.004 \times 100 % = 0.4 %$
A proportion is converted to a percentage by multiplying by 100%.

Answer

2 g; 0.4%.

03 / YOUR TURN

Try it yourself.

Check your understanding

A length is 43 cm to the nearest centimetre. Give its interval.

Show answer & reasoning

$42.5 \leq L < 43.5$ cm.

Nearest-centimetre rounding has a half-unit margin of $0.5 cm$ . Subtract and add it to obtain $42.5 \leq L < 43.5 (cm)$ . The upper boundary would round to $44 cm$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Write the interval for 7.2 m to the nearest 0.1 m.
2 marks
Write the interval for 340 g to the nearest 10 g.
2 marks
A thermometer reads 21.4 degrees Celsius when a reference is 20.0 degrees Celsius. Find absolute error.
2 marks
A measured length is 98 cm and its reference is 100 cm. Find percentage error.
2 marks
A scale repeatedly reads 1.05 kg for a verified 1.00 kg mass. Describe its readings.
2 marks
A rectangle is recorded as 4.0 cm by 3.0 cm, each to the nearest 0.1 cm. Find minimum possible area.
3 marks

Read the worked solutions

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

Write the interval for 7.2 m to the nearest 0.1 m.
2 marks
Half a rounding unit is 0.05 m.
$7.15 \leq L < 7.25$ m.
Write the interval for 340 g to the nearest 10 g.
2 marks
Half a 10 g unit is 5 g.
$335 \leq m < 345$ g.
A thermometer reads 21.4 degrees Celsius when a reference is 20.0 degrees Celsius. Find absolute error.
2 marks
$∣ 21.4 - 20.0 ∣ = 1.4$ . No Celsius percentage error is required.
1.4 degrees Celsius.
A measured length is 98 cm and its reference is 100 cm. Find percentage error.
2 marks
$\frac{∣ 98 - 100 ∣}{100} \times 100 = 2 %$ .
2%.
A scale repeatedly reads 1.05 kg for a verified 1.00 kg mass. Describe its readings.
2 marks
The readings agree with one another, not with the reference.
Repeatable but inaccurate by 0.05 kg.
A rectangle is recorded as 4.0 cm by 3.0 cm, each to the nearest 0.1 cm. Find minimum possible area.
3 marks
Lower dimensions are 3.95 cm and 2.95 cm.
$3.95 \times 2.95 = 11.6525$ cm $2$ .
11.6525 cm $2$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics measurement skills and applications; original explanatory lessons and papers.

Official curriculum source ↗

In this lesson

Understand the idea.

Understand the idea

Choose and apply a method

See the method in action.

Find an interval

Find half the rounding unit.

Calculate the two boundaries.

State the rounding interval.

Compare with a reference

Find the absolute error.

Compare with the verified mass.

Express the ratio as a percentage.

Try it yourself.

Check your understanding

Keep practising