Limits of Accuracy
Every measurement is rounded, so the true value lies within half a unit either side of the recorded value. This NSW Year 9 Advanced maths subtopic finds these limits of accuracy (NESA outcome MA5‑MAG‑C‑01).
You will read the smallest unit and the limit of accuracy from scales, and give the range of true values for measured lengths, temperatures, volumes and masses. You will compare rounding with truncating, find the least and greatest possible perimeters, areas and volumes, decide whether a lift could be overloaded, and find the possible fuel economy of a car. It is part of Year 9 Advanced Surface Area and Volume.
Every question with a fully worked solution.
- Limits of Accuracy - Video - Limits of Accuracy Watch
Theory
No measurement is exact. A reading to the nearest unit could be up to half a unit out either way, so the true value lies between a lower limit and an upper limit. This Year 9 Advanced maths page shows how to find those limits and carry them through perimeters, areas, volumes and speeds.
The precision of a measuring instrument is its smallest division: \(1\) mm on a ruler, \(5\) kg on a scale marked every \(5\) kg. A measurement given 'to the nearest \(50\) m' or 'to \(1\) decimal place' has precision \(50\) m or \(0.1\).
The absolute error, or limit of accuracy, is half the precision. A reading is written as value \(\pm\) absolute error.
The lower limit is the reading minus the absolute error; the upper limit is the reading plus it. The true value \(x\) satisfies lower \(\le x <\) upper. The upper limit itself is excluded, because it would round up.
A truncated reading has its extra digits cut off rather than rounded, so the true value lies from the reading up to one unit above it.
When measurements are combined, the errors combine too. The greatest and least possible results come from choosing each limit carefully.
Choosing limits when measurements are combined:
| To get the | sum \(a + b\) | difference \(a - b\) | product \(ab\) | quotient \(\dfrac{a}{b}\) |
|---|---|---|---|---|
| greatest | upper \(+\) upper | upper \(-\) lower | upper \(\times\) upper | upper \(\div\) lower |
| least | lower \(+\) lower | lower \(-\) upper | lower \(\times\) lower | lower \(\div\) upper |
How to find the limits of a measurement
- Find the precision: the size of one division on the scale (labelled gap \(\div\) number of gaps), or the unit named in 'to the nearest …'.
- Halve it to get the absolute error.
- Subtract and add it to the reading to get the lower and upper limits. Write lower \(\le x <\) upper, with units.
- Truncated? Then the limits are the reading and the reading plus one unit.
Combining measurements
- Write the limits of every measurement first, in the same units.
- Choose limits from the table: for a difference or a quotient, the second number goes the opposite way.
- Calculate and round only at the end.
- Decide questions such as 'can we be sure?' by checking whether the whole interval is on one side of the target.
- Two recordings of the same thing are both possible if their intervals overlap; the overlap is where both are correct.
(i) the size of one division
(ii) the limit of accuracy
(iii) the limits of the true mass when the pointer is on the mark shown.
(i) There are \(4\) equal gaps between labels \(20\) kg apart.
| \(\text{division}\) | \(=\) | \(20 \div 4 = \mathbf{5}~\text{kg}\) |
(ii) The limit of accuracy is half a division.
| \(\pm \dfrac{1}{2} \times 5\) | \(=\) | \(\mathbf{\pm 2.5}~\text{kg}\) |
(iii) The pointer is on \(35\) kg. Go \(2.5\) kg either side.
| \(35 - 2.5 \le\) | \(m\) | \(< 35 + 2.5\) |
| \(\mathbf{32.5}~\text{kg} \le\) | \(m\) | \(< \mathbf{37.5}~\text{kg}\) |
(i) A bridge is \(1300\) m long, correct to the nearest \(50\) m.
(ii) A bag of flour has a mass of \(0.68\) kg, correct to \(2\) decimal places.
(iii) The same reading, \(0.68\) kg, if it was truncated (cut off) after \(2\) decimal places.
(i) The precision is \(50\) m, so the absolute error is \(25\) m.
| \(1300 \pm 25:\ \ 1275~\text{m} \le\) | \(L\) | \(< 1325~\text{m}\) |
(ii) Two decimal places means the precision is \(0.01\) kg, so the absolute error is \(0.005\) kg.
| \(0.675~\text{kg} \le\) | \(m\) | \(< 0.685~\text{kg}\) |
(iii) Truncating only chops off digits. Any mass from \(0.68\) up to \(0.69\) is written as \(0.68\): for example \(0.6879 \to 0.68\).
| \(0.68~\text{kg} \le\) | \(m\) | \(< 0.69~\text{kg}\) |
A rounded reading sits in the middle of its interval; a truncated reading sits at the bottom.
(i) the least and greatest possible area
(ii) the upper limit of the perimeter.
Each side is \(\pm 0.05\) m: length \(3.55\) to \(3.65\) m, width \(1.75\) to \(1.85\) m.
(i) Least: both lower limits. Greatest: both upper limits.
| \(A_{\min}\) | \(=\) | \(3.55 \times 1.75 = \mathbf{6.2125}~\text{m}^2\) |
| \(A_{\max}\) | \(=\) | \(3.65 \times 1.85 = \mathbf{6.7525}~\text{m}^2\) |
(ii) Use both upper limits in \(P = 2(l + w)\).
| \(P\) | \(<\) | \(2(3.65 + 1.85) = \mathbf{11}~\text{m}\) |
The recorded area is \(3.6 \times 1.8 = 6.48~\text{m}^2\), but the true area could be anywhere in the range above.
Limits: distance \(41.5\) to \(42.5\) km; time \(1.45\) to \(1.55\) h.
Speed \(=\) distance \(\div\) time. Dividing by a smaller time gives a bigger speed, so pair the limits in opposite directions.
| \(v_{\min}\) | \(=\) | \(\dfrac{41.5}{1.55} = 26.7741\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{26.77}~\text{km/h}\) |
| \(v_{\max}\) | \(=\) | \(\dfrac{42.5}{1.45} = 29.3103\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{29.31}~\text{km/h}\) |
The least possible speed is below \(27\) km/h, so no, we cannot be sure.
Common pitfalls
Frequently asked questions
What is the limit of accuracy of a measuring instrument?
It is half of the smallest division on the instrument. A ruler marked in millimetres has a limit of accuracy of \(\pm 0.5\) mm.
What is the difference between precision and absolute error?
Precision is the smallest unit the instrument or reading shows. Absolute error is half of that, and is the most the reading can be out by.
How do you find the upper and lower limits of a measurement?
Halve the precision to get the absolute error, then subtract it from the reading for the lower limit and add it for the upper limit.
Why is the upper limit not included?
A value exactly at the upper limit would round up to the next unit, so it would not have been recorded as this reading. The lower limit is included because it rounds up to the reading.
How do you find the greatest possible area?
Use the upper limit of every length. Multiply the upper limit of the length by the upper limit of the width.
How do you find the greatest possible speed from rounded measurements?
Divide the greatest possible distance by the least possible time. A smaller time gives a larger speed.