Accuracy of Measuring Instruments
The absolute error of a measurement is half the smallest unit on the instrument, and the percentage error compares it with the measurement: \(\dfrac{\text{absolute error}}{\text{measurement}} \times 100\%\). This NSW Year 9 Advanced maths subtopic uses both to judge accuracy (NESA outcome MA5‑MAG‑C‑01).
You will find absolute and percentage errors for lengths, masses, times and temperatures, and compare a tape measure with a steel rule. You will choose a syringe accurate enough for a dose, see how errors add up along a fence line, compare rounding before and after squaring, and find the possible error in an area or a density. It is part of Year 9 Advanced Surface Area and Volume.
Theory
Every instrument has a precision, its smallest division, and every reading has an absolute error of half that. Percentage error compares that error with the size of the measurement, so it tells you how accurate a reading really is. This Year 9 Advanced maths page shows how to calculate and use both.
The precision of a measuring instrument is the smallest unit it can show: the gap between marks on a scale, or the last decimal place on a digital display. A reading of \(18.42\) km has precision \(0.01\) km.
Zeros that the instrument displays count. A digital scale showing \(2.050\) kg reads to the nearest \(0.001\) kg, so \(7.30\) and \(7.3\) are equal numbers but different measurements.
The absolute error (also called the greatest possible error) is half the precision. A crowd of \(70\,000\) to the nearest \(10\,000\) has absolute error \(5000\).
The percentage error is the absolute error as a percentage of the measurement. The same absolute error matters much more on a short length than on a long one.
When measurements are added, their absolute errors add. When they are multiplied, their percentage errors roughly add. Rounding part-way through a calculation adds error too, so round only at the end.
How to find a percentage error
- Find the precision: the gap between marks on the instrument, the unit in 'to the nearest …', or the place value of the last digit shown (including a final zero).
- Halve it to get the absolute error.
- Divide by the measurement and multiply by \(100\%\). Keep both in the same units.
- Round only the final answer.
Using percentage error
- Comparing accuracy: the reading with the smaller percentage error is more accurate, even if its absolute error is larger.
- Choosing an instrument: find the percentage error each instrument would give and keep those under the target.
- Totals and perimeters: add the absolute errors of every length, then find the percentage error of the total.
- Areas and quotients: find the greatest and least possible results from the limits of each measurement, and use the one further from the stated value.
- Calculations: keep full calculator values. Rounding a length before squaring it can change the answer noticeably.
Read the precision from the last digit shown. \(64.3\) shows tenths of a kilogram; \(85\) shows whole grams.
| \(\text{bathroom: abs. error}\) | \(=\) | \(\dfrac{1}{2} \times 0.1 = 0.05~\text{kg}\) |
| \(\%\text{ error}\) | \(=\) | \(\dfrac{0.05}{64.3} \times 100\% \approx \mathbf{0.08\%}\) |
| \(\text{kitchen: abs. error}\) | \(=\) | \(\dfrac{1}{2} \times 1 = 0.5~\text{g}\) |
| \(\%\text{ error}\) | \(=\) | \(\dfrac{0.5}{85} \times 100\% \approx \mathbf{0.59\%}\) |
The bathroom reading is more accurate. Its absolute error (\(50\) g) is far bigger, but it is a tiny fraction of \(64.3\) kg.
(i) Find the percentage error when it is used to measure \(400\) mL.
(ii) For which volumes is the percentage error less than \(2\%\)?
The precision comes from the markings, not from the number of digits in the reading.
| \(\text{abs. error}\) | \(=\) | \(\dfrac{1}{2} \times 25 = 12.5~\text{mL}\) |
| \(\%\text{ error}\) | \(=\) | \(\dfrac{12.5}{400} \times 100\% = \mathbf{3.125\%}\) |
(ii) The absolute error stays \(12.5\) mL, so the percentage error falls as the volume grows. Solve for the volume \(V\):
| \(\dfrac{12.5}{V} \times 100\) | \(<\) | \(2\) |
| \(1250\) | \(<\) | \(2V\) |
| \(V\) | \(>\) | \(\mathbf{625}~\text{mL}\) |
Any volume above \(625\) mL meets the target. Below that, a jug with finer markings is needed.
(i) the perimeter
(ii) the greatest possible error in the perimeter
(iii) the percentage error of the perimeter, correct to \(2\) decimal places.
| \(P\) | \(=\) | \(3.60 + 4.25 + 5.12 = \mathbf{12.97}~\text{m}\) |
(ii) Each side could be out by \(0.005\) m. In the worst case all three errors go the same way.
| \(3 \times 0.005\) | \(=\) | \(\mathbf{\pm 0.015}~\text{m}\) |
(iii)
| \(\%\text{ error}\) | \(=\) | \(\dfrac{0.015}{12.97} \times 100\%\) |
| \(\) | \(=\) | \(0.1156\ldots\%\) |
| \(\) | \(\approx\) | \(\mathbf{0.12\%}\) |
(i) Find the percentage error in each side.
(ii) Find the greatest percentage error in the area, correct to \(1\) decimal place.
(i)
| \(\dfrac{0.5}{6} \times 100\%\) | \(\approx\) | \(8.3\%\) |
| \(\dfrac{0.5}{4} \times 100\%\) | \(=\) | \(12.5\%\) |
(ii) Find the largest and smallest possible areas and see which is further from \(24\).
| \(A_{\max}\) | \(=\) | \(6.5 \times 4.5 = 29.25\) |
| \(A_{\min}\) | \(=\) | \(5.5 \times 3.5 = 19.25\) |
| \(29.25 - 24\) | \(=\) | \(5.25\) |
| \(24 - 19.25\) | \(=\) | \(4.75\) |
| \(\%\text{ error}\) | \(=\) | \(\dfrac{5.25}{24} \times 100\% \approx \mathbf{21.9\%}\) |
The area's error is roughly the two side errors added together (\(8.3\% + 12.5\% \approx 20.8\%\)). Multiplying measurements makes percentage errors grow.
Common pitfalls
Frequently asked questions
How do you calculate percentage error?
Divide the absolute error by the measurement and multiply by \(100\%\). The absolute error is half the smallest unit the instrument can measure.
What is the absolute error of a measurement?
It is half of the precision, the smallest unit used. A length of \(3.7\) m measured to the nearest \(0.1\) m has an absolute error of \(\pm 0.05\) m.
Which measurement is more accurate?
The one with the smaller percentage error. A large object measured roughly can be more accurate than a small object measured finely, because the error is a smaller part of the whole.
What is greatest possible error?
Greatest possible error is another name for absolute error: half of the unit the measurement was rounded to. For a value rounded to the nearest \(10\) it is \(5\).
Is 2.50 the same measurement as 2.5?
They are the same number but not the same measurement. \(2.50\) was measured to the nearest hundredth and \(2.5\) only to the nearest tenth, so \(2.50\) has one tenth of the absolute error.
Why does the percentage error get bigger when you calculate an area?
Area multiplies two measurements, and each has its own error. The percentage errors roughly add, so the area's percentage error is about the sum of the side errors.