Theory makes sense right up until you sit down with your own numbers. Then it all goes vague do I add these or square them first, and why does the answer have so many decimal places?
So this page skips the theory. It’s eleven worked examples, start to finish, covering the situations you’re most likely to meet: single readings, repeated readings, adding, subtracting, multiplying, dividing, powers, and a full uncertainty budget with five sources.
How to Read These Examples
Every example follows the same four beats:
- The situation – what’s being measured and with what
- The working – every line shown, nothing skipped
- The result – rounded and written properly
- What it tells you – the practical takeaway
Two conventions used throughout: uncertainties are rounded to one significant figure, and the measured value is then rounded to match its decimal place.
Example 1: Single Reading From an Analogue Scale
The situation: you measure the width of a steel plate with a rule graduated in millimetres. It reads 8.4 cm.
The working:
Smallest division = 1 mm = 0.1 cm Uncertainty = half the smallest division = 0.1 ÷ 2 = 0.05 cm
Percentage uncertainty = (0.05 ÷ 8.40) × 100 = 0.60%
The result: 8.40 ± 0.05 cm
What it tells you: notice the value is written 8.40, not 8.4. Since the uncertainty reaches the second decimal place, the value must too. Writing 8.4 ± 0.05 is inconsistent and loses marks.
Example 2: Single Reading From a Digital Instrument
The situation: a digital thermometer displays 36.7 °C.
The working:
The last digit is in the tenths place, so the uncertainty is ±1 in that place. Uncertainty = 0.1 °C
Percentage uncertainty = (0.1 ÷ 36.7) × 100 = 0.27%
The result: 36.7 ± 0.1 °C
What it tells you: this is resolution uncertainty only the floor, not the whole story. If the thermometer’s manufacturer quotes ±0.5 °C accuracy, that larger figure is what you should use. A crisp digital display is not the same as a correct one.
Example 3: Repeated Readings Three Methods Compared
The situation: you measure a resistor six times with a digital multimeter and get 47.2, 47.5, 47.3, 47.4, 47.2, 47.4 Ω.
Three different methods are in common use. Here’s what each gives for the same data.
Find the mean first:
Sum = 283.0 Ω Mean = 283.0 ÷ 6 = 47.17 Ω
Method A – Range method
Uncertainty = (largest − smallest) ÷ 2 = (47.5 − 47.2) ÷ 2 = 0.15 Ω
Method B – Mean absolute error (common in physics and chemistry syllabuses)
Deviations from the mean: 0.03, 0.33, 0.13, 0.23, 0.03, 0.23 Sum = 1.00 Mean absolute error = 1.00 ÷ 6 = 0.17 Ω
Method C – Standard uncertainty of the mean (used in professional work)
Sample standard deviation, s = 0.22 Ω u = s ÷ √n = 0.22 ÷ √6 = 0.22 ÷ 2.449 = 0.09 Ω
The results:
| Method | Uncertainty | Reported as |
|---|---|---|
| Range | 0.15 Ω | 47.2 ± 0.2 Ω |
| Mean absolute error | 0.17 Ω | 47.2 ± 0.2 Ω |
| Standard uncertainty of the mean | 0.09 Ω | 47.17 ± 0.09 Ω |
What it tells you: the third method gives roughly half the uncertainty of the other two, and it isn’t cheating. The range method uses only your two extreme readings and ignores the four in between. The standard-deviation method uses all six, and the ÷√n step reflects that you’re reporting a mean, which is more reliable than any single reading.
Use whichever your syllabus specifies just say which one you used.
Example 4: Combining Type A and Type B Uncertainty
The situation: a digital caliper measuring a bore. Six readings: 30.11, 30.13, 30.12, 30.14, 30.12, 30.13 mm. The caliper has 0.01 mm resolution and a calibration certificate stating U = 0.02 mm at k = 2.
Type A – from the readings:
Mean = 180.75 ÷ 6 = 30.125 mm Sample standard deviation, s = 0.0105 mm u_A = 0.0105 ÷ √6 = 0.0043 mm
Type B – from the instrument:
Resolution: half-width 0.005 mm, rectangular distribution u = 0.005 ÷ √3 = 0.0029 mm
Calibration certificate: U = 0.02 mm at k = 2 u = 0.02 ÷ 2 = 0.0100 mm
Combine in quadrature:
u꜀ = √(0.0043² + 0.0029² + 0.0100²) u꜀ = √(0.0000185 + 0.0000084 + 0.0001000) u꜀ = √0.0001269 = 0.0113 mm
Expand:
U = 2 × 0.0113 = 0.0225 ≈ 0.02 mm
The result: 30.13 ± 0.02 mm (k = 2)
What it tells you: the calibration certificate contributes 0.0100 mm more than the other two sources combined. Your repeatability is excellent and your resolution is fine. If you need a better result, you need a better-calibrated instrument, not a steadier hand.
That ranking is the real output of the calculation. The ± figure just summarises it.
Example 5: Adding Measurements
The situation: three components stacked end to end, each measured with the same instrument.
A = 45.2 ± 0.1 mm B = 30.5 ± 0.1 mm C = 12.8 ± 0.1 mm
The working:
For sums, add the absolute uncertainties.
Total length = 45.2 + 30.5 + 12.8 = 88.5 mm Total uncertainty = 0.1 + 0.1 + 0.1 = 0.3 mm
Percentage uncertainty = (0.3 ÷ 88.5) × 100 = 0.34%
The result: 88.5 ± 0.3 mm
What it tells you: the absolute uncertainty tripled, but the percentage uncertainty barely moved component C alone was 0.78%, and the stack is 0.34%. Adding several similar measurements is comparatively forgiving.
Subtraction is not, as the next example shows.
Example 6: Subtracting Measurements the Trap
The situation: finding the wall thickness of a tube by measuring the outer and inner dimensions.
Outer = 100.0 ± 0.1 mm (0.10%) Inner = 98.0 ± 0.1 mm (0.10%)
The working:
For differences, you still add the absolute uncertainties. They never cancel.
Thickness = 100.0 − 98.0 = 2.0 mm Uncertainty = 0.1 + 0.1 = 0.2 mm
Percentage uncertainty = (0.2 ÷ 2.0) × 100 = 10%
The result: 2.0 ± 0.2 mm
What it tells you: two measurements each accurate to 0.1% produced a result accurate to only 10%. That’s a hundredfold degradation, and it happened purely because the two numbers were close together.
Subtracting two similar quantities is the fastest way to ruin an experiment’s precision. If you can measure the thickness directly instead of by difference, do it. This one principle will save you more grief than any other on this page.
Example 7: Multiplying Area of a Sheet
The situation: a rectangular sheet measured as 120.0 ± 0.5 mm by 80.0 ± 0.5 mm.
The working:
For products, add the percentage uncertainties.
Length: (0.5 ÷ 120.0) × 100 = 0.42% Width: (0.5 ÷ 80.0) × 100 = 0.63%
Area = 120.0 × 80.0 = 9600 mm² Combined percentage = 0.42% + 0.63% = 1.05% Absolute uncertainty = 0.0105 × 9600 = 101 mm²
The result: 9600 ± 100 mm²
What it tells you: the same ±0.5 mm hurts more on the shorter dimension. This is why percentage uncertainty is the form worth working in it exposes which measurement is actually limiting you, and here it’s the width.
Example 8: Dividing Calculating Density
The situation: mass = 250.0 ± 0.5 g, volume = 32.0 ± 0.5 cm³.
The working:
Division follows the same rule as multiplication add the percentages.
Mass: (0.5 ÷ 250.0) × 100 = 0.20% Volume: (0.5 ÷ 32.0) × 100 = 1.56%
Density = 250.0 ÷ 32.0 = 7.8125 g/cm³ Combined percentage = 0.20% + 1.56% = 1.76% Absolute uncertainty = 0.0176 × 7.8125 = 0.14 g/cm³
The result: 7.8 ± 0.1 g/cm³
What it tells you: volume contributes almost eight times as much uncertainty as mass. Weighing more precisely would achieve nothing here you’d need a better way to measure volume.
This pattern shows up constantly in density work, and it’s why displacement methods and volumetric flasks matter more than analytical balances for this particular measurement.
Example 9: Powers Volume of a Sphere
The situation: a ball bearing with diameter 25.0 ± 0.1 mm.
The working:
Volume of a sphere from diameter: V = (π/6) × d³
The diameter is cubed, so its percentage uncertainty is multiplied by 3.
Diameter percentage uncertainty = (0.1 ÷ 25.0) × 100 = 0.40% Volume percentage uncertainty = 3 × 0.40% = 1.20%
V = (π/6) × 25.0³ = 0.5236 × 15625 = 8181 mm³ Absolute uncertainty = 0.012 × 8181 = 98 mm³
The result: 8180 ± 100 mm³
What it tells you: a 0.4% uncertainty in a length became a 1.2% uncertainty in a volume, simply because of the cube. Any quantity raised to a power amplifies its uncertainty by that power which is why, when a formula contains a cube, that measurement is almost always the one to improve first.
A square root works the other way, halving the percentage uncertainty.
Example 10: Ohm’s Law Resistance From Voltage and Current
The situation: V = 12.4 ± 0.1 V, I = 0.250 ± 0.005 A.
The working:
R = V ÷ I, so add the percentages.
Voltage: (0.1 ÷ 12.4) × 100 = 0.81% Current: (0.005 ÷ 0.250) × 100 = 2.00%
R = 12.4 ÷ 0.250 = 49.6 Ω Combined percentage = 0.81% + 2.00% = 2.81% Absolute uncertainty = 0.0281 × 49.6 = 1.4 Ω
The result: 50 ± 1 Ω
(Some courses would keep two significant figures in the uncertainty and write 49.6 ± 1.4 Ω. Either is defensible be consistent.)
What it tells you: the ammeter is responsible for over two-thirds of the total uncertainty. A more precise voltmeter would be a waste of money.
Small currents are a recurring weak point in electrical measurement. The absolute uncertainty stays fixed while the reading shrinks, so the percentage climbs fast.
Example 11: A Complete Uncertainty Budget
The situation: verifying a digital pressure gauge at 5 bar. Six readings: 5.02, 5.04, 5.03, 5.02, 5.04, 5.03 bar. Resolution 0.01 bar. Calibration certificate U = 0.010 bar at k = 2. Manufacturer quotes a temperature effect of ±0.005 bar over the working range. Drift since last calibration is estimated at ±0.008 bar.
Type A:
Mean = 30.18 ÷ 6 = 5.03 bar s = 0.0089 bar u_A = 0.0089 ÷ √6 = 0.0037 bar
Type B – convert each source:
| Source | Value | Distribution | Divisor | Standard uncertainty |
|---|---|---|---|---|
| Repeatability | 0.0089 bar | Normal | √6 | 0.0037 bar |
| Calibration certificate | 0.010 bar (k=2) | Normal | 2 | 0.0050 bar |
| Resolution | 0.005 bar | Rectangular | √3 | 0.0029 bar |
| Temperature effect | 0.005 bar | Rectangular | √3 | 0.0029 bar |
| Drift since calibration | 0.008 bar | Rectangular | √3 | 0.0046 bar |
Combine:
u꜀ = √(0.0037² + 0.0050² + 0.0029² + 0.0029² + 0.0046²) u꜀ = √0.0000762 = 0.0087 bar
Expand:
U = 2 × 0.0087 = 0.0175 ≈ 0.02 bar
The result: 5.03 ± 0.02 bar (k = 2, approximately 95% confidence)
What it tells you: drift is the second-largest contributor at 0.0046 bar larger than resolution and temperature combined. That’s a source you can fix for free by shortening the calibration interval. No new equipment required.
This is what a budget is for. Not the number at the bottom, but the ranked list above it.
What These Examples Have in Common
Read across all eleven and three patterns repeat:
One source usually dominates. In almost every example above, the largest contributor was several times bigger than the smallest. Because uncertainties combine as squares, small contributors barely register. Find the big one and everything else is noise.
The rule depends on the operation, not the quantity. Adding or subtracting? Absolute uncertainties. Multiplying or dividing? Percentages. Powers? Multiply the percentage by the power. It doesn’t matter whether you’re measuring pressure, resistance or length.
Percentages reveal what absolutes hide. Example 6 looked harmless in absolute terms ±0.2 mm and catastrophic in percentage terms at 10%. Convert to percentages before you decide anything.
Practice Problems With Answers
Work through these before looking at the answers.
1. A rule with 1 mm divisions reads 6.7 cm. State the result with its uncertainty and percentage uncertainty.
2. A digital balance displays 4.32 g. State the result with its uncertainty.
3. Four timings: 10.2, 10.4, 10.3, 10.3 s. Use the range method.
4. Add 25.0 ± 0.2 cm and 14.0 ± 0.3 cm.
5. Subtract 47.0 ± 0.2 mm from 50.0 ± 0.2 mm. Give the percentage uncertainty in the answer.
6. Multiply 8.0 ± 0.1 mm by 5.0 ± 0.1 mm.
7. A sample has mass 45.0 ± 0.1 g and volume 6.0 ± 0.2 cm³. Find the density.
8. A cube has side 10.0 ± 0.2 mm. Find its volume with uncertainty.
9. A certificate states U = 0.006 mm at k = 2. Find the standard uncertainty.
10. An instrument has 0.001 mm resolution. Find its standard uncertainty contribution.
Answers
1. 6.70 ± 0.05 cm; 0.75%
2. 4.32 ± 0.01 g; 0.23%
3. Mean 10.30 s; (10.4 − 10.2) ÷ 2 = 0.10 → 10.30 ± 0.10 s
4. 39.0 ± 0.5 cm
5. 3.0 ± 0.4 mm → 13% (from inputs of 0.4% each the subtraction trap again)
6. 1.25% + 2.00% = 3.25% of 40 mm² = 1.3 → 40 ± 1 mm²
7. 0.22% + 3.33% = 3.56% of 7.5 = 0.27 → 7.5 ± 0.3 g/cm³
8. 2.00% × 3 = 6.00% of 1000 mm³ = 60 → 1000 ± 60 mm³
9. 0.006 ÷ 2 = 0.003 mm
10. (0.001 ÷ 2) ÷ √3 = 0.00029 mm
Common Mistakes These Examples Reveal
Mismatched decimal places. Example 1 gave 8.40 ± 0.05, not 8.4 ± 0.05. The value must reach the same decimal place as the uncertainty.
Adding percentages for a sum. Example 5 added absolute uncertainties. Adding percentages there would be wrong.
Expecting subtraction to reduce uncertainty. Example 6. Uncertainties accumulate under subtraction exactly as they do under addition.
Forgetting the exponent. Example 9 multiplied by 3 because the diameter was cubed. Missing that underestimates the result by a factor of three.
Dropping a certificate value straight into the budget. Example 4 divided 0.02 by 2 first. Skipping that step doubles the contribution.
Using the full resolution instead of half. Examples 4 and 11 both used half the resolution as the half-width before dividing by √3.
Adding standard uncertainties instead of combining in quadrature. Example 11 would have given 0.019 bar by straight addition, against 0.0087 by quadrature — more than double.
Reporting too many figures. Example 8’s raw answer was 7.8125 ± 0.1377. Reporting it that way claims precision the calculation just disproved.
Frequently Asked Questions
How do you calculate uncertainty in a measurement?
For a single reading, use half the smallest scale division on an analogue instrument or ±1 in the last digit on a digital one. For repeated readings, use the range method, the mean absolute error, or the standard deviation divided by √n. Examples 1 to 3 show all of these.
What is an example of absolute uncertainty?
In 8.40 ± 0.05 cm, the absolute uncertainty is 0.05 cm a plain number in the same units as the measurement.
What is an example of percentage uncertainty?
For 8.40 ± 0.05 cm: (0.05 ÷ 8.40) × 100 = 0.60%.
How do you combine uncertainties when adding measurements?
Add the absolute uncertainties. Example 5: three components each ±0.1 mm give a total of ±0.3 mm.
How do you combine uncertainties when multiplying?
Add the percentage uncertainties, then convert back to absolute using the calculated result. Example 7 works through this.
Why does subtraction increase percentage uncertainty so much?
The absolute uncertainties still add, but the result gets smaller. Example 6 shows two 0.1% measurements producing a 10% result.
What happens to uncertainty when a quantity is squared or cubed?
Its percentage uncertainty is multiplied by the power. Cubing triples it, as in Example 9. A square root halves it.
How many readings should I take?
Five or six is usually enough. Because the standard uncertainty falls as √n, going from 6 readings to 24 only halves that contribution.
Should the uncertainty be one or two significant figures?
One is standard for coursework, and many laboratories use two. Example 10 shows both forms. Pick one and apply it consistently.
Which uncertainty source should I try to reduce?
The largest one. Because contributions combine as squares, reducing anything but the dominant term has almost no effect on the total.
Conclusion
Eleven examples, one repeating lesson: the arithmetic is easy, and the useful part is the ranking it produces.
The rules, condensed:
- Single analogue reading → half the smallest division. Single digital reading → ±1 in the last digit.
- Repeated readings → range ÷ 2, mean absolute error, or s ÷ √n. The last is the smallest and the most defensible.
- Adding or subtracting → add the absolute uncertainties. They never cancel.
- Multiplying or dividing → add the percentage uncertainties.
- Powers → multiply the percentage uncertainty by the power.
- Several sources at once → combine in quadrature, then multiply by k = 2.
- Round the uncertainty to one significant figure, then match the value’s decimal place to it.
