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hscscience Maths Std · Y11
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Module 2 · L13 of 22 50–55 min ⚡ +95 XP available

Errors and Limits of Accuracy

Every measurement is an approximation. The absolute error is always half the smallest unit of the instrument, and when measurements are combined, errors compound. Knowing this prevents catastrophic calculation mistakes.

Today's hook, A ruler is marked in millimetres. You measure a piece of timber as 45 cm. What is the largest it could actually be? What is the smallest? How does this matter if you are cutting 20 of these pieces from a single plank?
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Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

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01
Recall, your gut answer first
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A ruler is marked in millimetres. You measure a piece of timber as 45 cm.

Without calculating what is the largest it could actually be? What is the smallest? How does this matter if you are cutting 20 of these pieces from a single plank?

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The three formulas you need to own
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Error calculations start with three core formulas. Every bounds and percentage-error question is just a rearrangement of these.

Absolute error is always half the smallest unit, it captures the maximum possible difference between the measured value and the true value. Upper and lower bounds put a bracket around the true value. Percentage error tells you how significant the error is relative to the measurement.

The same absolute error is a larger percentage of a small measurement than of a large one.
$$\text{AE} = \tfrac{1}{2} \times \text{smallest unit}$$
Absolute error
Always half the smallest graduation. Ruler in mm: AE = 0.5 mm. Scale in 0.1 kg: AE = 0.05 kg.
Upper & lower bounds
Upper = value + AE; Lower = value − AE. True value lies in $[x-e,\; x+e]$.
Percentage error
$\% \text{ error} = \frac{\text{AE}}{\text{measurement}} \times 100\%$. Smaller measurement = larger percentage error.
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What you'll master
Know

Key facts

  • Absolute error = ½ × smallest unit of measurement
  • Upper bound = value + absolute error; lower bound = value − absolute error
  • Percentage error = (absolute error ÷ measurement) × 100%
  • Errors compound when measurements are added or multiplied
Understand

Concepts

  • Why every measurement has an inherent uncertainty
  • How measurement precision affects reliability of calculations
  • Why a small measurement has a larger percentage error than a large one
Can do

Skills

  • State the absolute error for any given instrument precision
  • Calculate upper and lower bounds for a measurement
  • Calculate percentage error
  • Find bounds for calculated quantities (area, perimeter)
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Key terms
Absolute errorThe maximum possible difference between the measured value and the true value; equal to half the smallest graduation of the measuring instrument.
BoundsThe upper bound is the largest possible true value; the lower bound is the smallest possible true value; true value lies in $[\text{lower}, \text{upper}]$.
Percentage errorThe absolute error expressed as a percentage of the measured value; indicates the relative size of the error.
PrecisionThe smallest unit that an instrument can measure; a more precise instrument has a smaller unit and therefore a smaller absolute error.
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Every measurement has an error

When you read a measurement from any instrument, you round to the nearest marked graduation.

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Every measurement has an error
core concept

When you read a measurement from any instrument, you round to the nearest marked graduation. This introduces an uncertainty of up to half that graduation on either side.

  • Ruler in mm: precision = 1 mm, absolute error = 0.5 mm
  • Scale in 0.1 kg: precision = 0.1 kg, absolute error = 0.05 kg
  • Thermometer in 1°C: precision = 1°C, absolute error = 0.5°C
  • Odometer in 0.1 km: precision = 0.1 km, absolute error = 0.05 km

A measured value of $x$ with absolute error $e$ means the true value lies in the interval $[x-e, \; x+e]$.

Key point, The absolute error does not depend on the measured value, it depends only on the instrument's smallest unit. So two measurements with the same instrument have the same absolute error, but different percentage errors.

Absolute error = half the smallest graduation of the measuring instrument. Percentage error = (absolute error ÷ measurement) × 100%. Measurements are reported as: value ± absolute error. Smaller instruments give smaller percentage errors.

Pause, copy the absolute error formula (half the smallest graduation of the instrument), the percentage error formula (absolute error ÷ measurement × 100%), and the measurement notation (value ± absolute error) into your book.

Did you get this? True or false: a ruler graduated in millimetres has an absolute error of 1 mm.

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Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · UPPER AND LOWER BOUNDS

A length is measured as 34 cm using a ruler marked in centimetres (precision = 1 cm). (a) State the absolute error. (b) Find the upper and lower bounds of the true length.

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(a) Absolute error $= \dfrac{1}{2} \times 1 \text{ cm} = 0.5 \text{ cm}$
Precision is 1 cm; absolute error is always half the smallest unit.
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PROBLEM 2 · PERCENTAGE ERROR

A mass is recorded as 45 kg using scales with precision 0.5 kg. Find the percentage error, correct to 2 decimal places.

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Absolute error $= \dfrac{1}{2} \times 0.5 = 0.25 \text{ kg}$
Precision = 0.5 kg; AE = half the precision.
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PROBLEM 3 · BOUNDS FOR AREA

A rectangle is measured as 8 m × 5 m using a tape measure with precision 0.1 m. (a) State the bounds for each dimension. (b) Find the maximum and minimum possible area.

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(a) Error $= 0.05$ m; Length: $[7.95,\; 8.05]$ m; Width: $[4.95,\; 5.05]$ m
Absolute error $= \frac{1}{2}(0.1) = 0.05$ m; subtract and add from each measurement.
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Quick check: A distance is measured as 6.0 m with precision 0.1 m. What is the percentage error?

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Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Common errors · the 3 traps that cost marks

Trap 01
Using the full unit as the absolute error
The absolute error is half the smallest unit, not the unit itself. A ruler marked in mm has AE = 0.5 mm, not 1 mm. This is the single most common error in this topic.
Trap 02
Using the measurement in the wrong unit for % error
% error $= \frac{\text{AE}}{\text{measurement}} \times 100\%$, both AE and measurement must be in the same unit. Convert AE to cm if the measurement is in cm before dividing.
Trap 03
Finding maximum area with upper × lower
Maximum area requires both dimensions at their upper bound. Using upper × lower gives a value between the min and max, not the extremes you need.

Fill the gap: A rectangle is measured as 12 cm × 9 cm with precision 1 mm. The absolute error is mm. The upper bound of the 12 cm side is 12. cm.

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Quick-fire practice · 5 calculations

Quick-fire practice · 5 calculations

Quick-fire practice · 5 calculations

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A length is measured as 72 mm using a ruler with 1 mm graduations. State the absolute error and the upper and lower bounds.

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A container holds 2.4 L, measured using a jug marked in 0.1 L divisions. Find the absolute error and bounds.

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A plank is measured as 85 cm using a ruler with 1 mm precision. Find the percentage error (to 3 significant figures).

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A measurement of 0.6 kg is taken with precision 0.1 kg. Find the percentage error.

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Three lengths of rope each measure 2.5 m with precision 0.01 m. They are joined end to end. Find the maximum and minimum total length.

Odd one out: Three of these statements are correct. Which one is wrong?

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Revisit your thinking

Earlier you estimated the largest and smallest possible length of the 45 cm timber piece.

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Revisit your thinking

Earlier you estimated the largest and smallest possible length of the 45 cm timber piece. Let's check:

Ruler in mm → precision = 1 mm = 0.1 cm → AE = 0.5 mm = 0.05 cm

Upper bound: $45 + 0.05 = 45.05$ cm. Lower bound: $45 - 0.05 = 44.95$ cm.

Over 20 pieces: maximum total = $20 \times 45.05 = 901.0$ cm; minimum total = $20 \times 44.95 = 899.0$ cm. That is a possible variation of 2 cm across the full plank which could be critical when fitting timber.

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Multiple choice, then short answer under exam conditions.

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Multiple choice
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Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 43 marks

Q1. A length is measured as 8.5 cm using a ruler with 1 mm graduations. (a) State the absolute error of this measurement. (b) State the upper and lower bounds for the true length. (c) Calculate the percentage error, correct to 2 significant figures. (3 marks)

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ApplyBand 43 marks

Q2. The dimensions of a swimming pool are measured as 25 m × 12 m using a tape measure with precision 0.5 m. (a) State the absolute error and the bounds for the 25 m measurement. (b) Find the maximum and minimum possible area of the pool. (3 marks)

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AnalyseBand 54 marks

Q3. A surveyor measures three sides of a triangular paddock as 120 m, 85 m, and 95 m using a distance wheel with precision 1 m. (a) State the absolute error for each measurement. (b) Find the maximum possible perimeter. (c) Find the minimum possible perimeter. (d) What is the range of possible perimeters? (4 marks)

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📖 Comprehensive answers (click to reveal)

Drill 1: AE = 0.5 mm; bounds: $[71.5,\; 72.5]$ mm  ·  2: AE = 0.05 L; bounds: $[2.35,\; 2.45]$ L  ·  3: AE = 0.5 mm = 0.05 cm; % error $= 0.05/85 \times 100 \approx 0.0588\%$  ·  4: AE = 0.05 kg; % error $= 0.05/0.6 \times 100 = 8.\overline{3}\%$  ·  5: Each: $[2.49,\; 2.51]$ m; Max = $7.53$ m; Min = $7.47$ m

Q1 (3 marks): (a) $\frac{1}{2} \times 1 \text{ mm} = 0.5 \text{ mm}$ [1]. (b) $[8.45,\; 8.55]$ cm [1]. (c) $\frac{0.05}{8.5} \times 100 \approx 0.59\%$ [1].

Q2 (3 marks): (a) AE = 0.25 m; bounds for 25 m: $[24.75,\; 25.25]$ m; bounds for 12 m: $[11.75,\; 12.25]$ m [1]. (b) Max: $25.25 \times 12.25 = 309.3125 \text{ m}^2$ [1]. Min: $24.75 \times 11.75 = 290.8125 \text{ m}^2$ [1].

Q3 (4 marks): (a) AE $= 0.5$ m for each [1]. (b) Max: $(120.5 + 85.5 + 95.5) = 301.5$ m [1]. (c) Min: $(119.5 + 84.5 + 94.5) = 298.5$ m [1]. (d) Range $= 301.5 - 298.5 = 3$ m [1].