Errors, precision, accuracy and uncertainty

Key idea: A complete H2 Physics lesson on SI quantities, estimation, errors, uncertainty and coplanar vectors.

  • GCE A-Level H2 Physics 2027

Learn the idea

Big question: How should a measurement report both its best value and its limits?

Random scatter limits precision, while systematic effects such as zero error shift results and limit accuracy. Repetition can reduce the uncertainty in a mean but cannot remove a shared bias. For derived quantities, add absolute uncertainties through sums and differences, and relative uncertainties through products, quotients and powers; numerical substitution is useful for less convenient expressions.

Separate precision from accuracy

Random error produces unpredictable scatter and mainly limits precision. Repeated readings reveal that scatter and can improve an estimate of the mean. Systematic error shifts readings consistently and mainly limits accuracy; a zero error is one example.

Averaging does not remove a systematic offset. Apply a known zero correction and discuss remaining systematic limitations separately from random uncertainty.

Check your understanding: Why can a tightly grouped set still be inaccurate?

A systematic error can shift every reading together, giving high precision around the wrong value.

Match the propagation rule to the operation

For sums and differences, add absolute uncertainties. For products, quotients and powers, add fractional or percentage uncertainties, multiplying a fractional uncertainty by the magnitude of its power where required.

For a less convenient expression, numerical substitution is allowed: calculate the result at suitable upper and lower input limits and use the resulting spread. A rigorous statistical treatment is not required.

Check your understanding: For y = x³, what approximate percentage uncertainty follows from a 2% uncertainty in x?

About 6%, because the fractional uncertainty is multiplied by the power 3.

Report only justified precision

Round an uncertainty to one significant figure, or sometimes two when its first digit is small, then quote the measured value to the same decimal place. Units belong on both the value and its absolute uncertainty.

Extra calculator digits do not improve the experiment. A result should communicate what was measured and the limits supported by the data.

Check your understanding: How should 2.3471 ± 0.0836 m normally be reported?

About (2.35 ± 0.08) m, with value and uncertainty at the same decimal place.

Random scatter and systematic biasTwo target plots compare scattered readings centred on the accepted value with tightly grouped readings displaced from it.Random scattermean near accepted valuerepeat and averageSystematic biasreadings grouped away from accepted valuecalibrate, correct or redesign
Scroll diagram horizontally to read all labels.
Repeats reveal scatter and improve the estimate of a mean, but they do not remove a systematic offset. Calibration, zero checks or a redesigned method are needed for bias.

Key ideas to keep

  • More decimal places do not remove uncertainty.
  • Apply a known zero correction before reporting the result.
  • Quote the uncertainty to sensible significant figures and match the value's decimal place.

Worked example

Correct a zero error and propagate uncertainty

Question: A rectangle has measured sides (2.40 ± 0.02) m and (1.20 ± 0.01) m. The first instrument has a +0.03 m zero error already included in its reading. Find the corrected area and its uncertainty.

  1. Step 1: Correct the systematic offset

    Why: A positive zero error makes the first indicated length too large.

    Working: L = 2.40 − 0.03 = 2.37 m; its random uncertainty remains ±0.02 m.

  2. Step 2: Calculate the central area

    Why: Use corrected central values before propagating uncertainty.

    Working: A = 2.37(1.20) = 2.844 m².

  3. Step 3: Add relative uncertainties

    Why: Area is a product.

    Working: ΔA/A = 0.02/2.37 + 0.01/1.20 = 0.0168, so ΔA = 0.0478 m².

Answer: A = (2.84 ± 0.05) m².

Check: The zero correction changes the central value; it is not added as another random uncertainty after being corrected.

Question

A rectangle has L = (2.40 ± 0.02) m and W = (1.20 ± 0.01) m. Find its area and absolute uncertainty.

Check the worked solution

A = 2.88 m². For multiplication, add relative uncertainties: 0.02/2.40 + 0.01/1.20 = 0.0167. Thus ΔA ≈ 0.0167 × 2.88 = 0.048 m², so A = (2.88 ± 0.05) m².

Practise with support

Try this

For Q = (a − b)c, state the uncertainty procedure when a, b and c each have absolute uncertainties.

Hint: Use the addition rule before the multiplication rule.

Check your answer

First find d = a − b and add absolute uncertainties: Δd = Δa + Δb. Then Q = dc and add relative uncertainties: ΔQ/Q = Δd/|d| + Δc/|c|.

Practise independently

Your turn

A speed is calculated from d = (5.00 ± 0.05) m and t = (2.00 ± 0.02) s. Find the speed and its percentage uncertainty.

Check your answer

v = 2.50 m s⁻¹. Percentage uncertainty = (0.05/5.00 + 0.02/2.00) × 100% = 2.0%, so v = (2.50 ± 0.05) m s⁻¹.

Common mistakes

Common mistake

More decimal places remove uncertainty and improve accuracy.

What is wrong with this reasoning?

Show better thinking

Displayed digits do not remove random scatter or systematic offset. Quote precision justified by the instrument and propagate the stated uncertainty.

Exam guidance

State the source and type of uncertainty before propagating it; do not present unsupported precision.

Exam-style practice [7 marks]

A speed is calculated from d = (5.00 ± 0.05) m and t = (2.00 ± 0.02) s. Find the speed and its absolute uncertainty. Explain why repeating both readings does not remove a shared timer zero error.

Plan before you answer

  • Calculate the central value.
  • Add fractional uncertainties for the quotient.
  • Distinguish random scatter from a shared offset.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

v = 5.00/2.00 = 2.50 m s⁻¹. The fractional uncertainty is 0.05/5.00 + 0.02/2.00 = 0.020, or 2.0%, so Δv = 0.020(2.50) = 0.05 m s⁻¹. Hence v = (2.50 ± 0.05) m s⁻¹. A timer zero error shifts every time in the same direction, so repetition and averaging do not remove it.

Check what stayed with you

Recall question 1

Which error type mainly limits precision?

Check the answer

Random error.

Recall question 2

How are uncertainties combined for a quotient?

Check the answer

Add fractional or percentage uncertainties.

Recall question 3

What is numerical substitution used for?

Check the answer

Estimating the uncertainty in a derived quantity when simple addition rules are inconvenient.

Try this next

Continue to the next lesson in this topic.

Scalars and coplanar vector operations

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes.

  • GCE A-Level H2 PhysicsTopic 1(f) / Topic 1(g) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027