Accuracy of Measuring Instruments
Every measurement is approximate. The precision of an instrument is its smallest unit, the absolute error is half of that, and the percentage error compares the absolute error with the measurement. In NSW Year 10 Advanced maths you find the limits of accuracy of a measurement and follow how errors carry through to perimeters, areas and volumes (NESA outcome MA5‑MAG‑C‑01).
You will find upper and lower limits for lengths, masses, times and populations rounded to significant figures, and calculate absolute and percentage errors from scales on jugs and stopwatches. You will find the least and greatest possible area of a triangle, travel time for a trip and volume of water in a tank, decide how much edging or how many tiles to buy, and give answers to a sensible number of significant figures. It is part of Year 10 Advanced Measurement.
Theory
Every measurement is an approximation. The absolute error is half the smallest unit on the instrument, and it gives the limits of accuracy: the least and greatest possible true values. This Year 10 Advanced maths page covers percentage error, comparing accuracy, and finding the least and greatest results of calculations made with rounded measurements.
No instrument measures exactly. A ruler marked in millimetres can only give a length to the nearest millimetre. The smallest marked unit is the precision or scale interval.
The absolute error (greatest possible error) of a measurement is half the precision. A length of \(4.7\) cm to the nearest \(0.1\) cm has an absolute error of \(\pm 0.05\) cm.
The limits of accuracy are the measurement minus and plus the absolute error. The lower limit is included and the upper limit is not: \(4.65 \le L \lt 4.75\). For whole-number counts, such as people, the greatest value is one less than the upper limit.
The percentage error compares the absolute error with the measurement. The smaller the percentage error, the more accurate the measurement.
Least and greatest results of a calculation:
| Calculation | Least result | Greatest result |
|---|---|---|
| \(a + b\) | least \(+\) least | greatest \(+\) greatest |
| \(a \times b\) | least \(\times\) least | greatest \(\times\) greatest |
| \(a - b\) | least \(a\) \(-\) greatest \(b\) | greatest \(a\) \(-\) least \(b\) |
| \(a \div b\) | least \(a\) \(\div\) greatest \(b\) | greatest \(a\) \(\div\) least \(b\) |
How to find the limits of a measurement
- Identify the precision: the scale interval, the place value it was rounded to, or the stated “nearest” unit.
- Halve it to get the absolute error.
- Subtract and add to get the lower and upper limits. Write them as an inequality with \(\le\) and \(\lt\).
- Divide the absolute error by the measurement and multiply by \(100\%\) for the percentage error.
Calculations with measured values
- Find the limits of every measurement first, all in the same unit.
- Choose the limits that make the result smallest or largest (table in the Formulas tab). Dividing or subtracting reverses the choice for the second number; an overlap or a cut takes away length, so use its greatest value for the least total.
- Be sure: if you must have enough (edging, tiles, turf), plan for the greatest value and round up. If something must be certain to last, use the least value and round down.
- Do two measurements agree? Write both intervals in the same unit. If they do not overlap, they cannot both be correct.
- Rounding error in a method: compare an answer found with rounded values (such as \(\pi \approx 3.14\)) with the exact one: \(\dfrac{\text{difference}}{\text{exact}} \times 100\%\).
- Working backwards: if the percentage error and the scale are known, the measurement is absolute error \(\div\) percentage error. If a limit involves a pronumeral, write the limit as an expression and solve.
(i) the absolute error
(ii) the limits of the true mass
(iii) the percentage error, correct to \(2\) decimal places.
(i) The absolute error is half the scale interval.
| \(\text{Absolute error}\) | \(=\) | \(\dfrac{1}{2} \times 20\) |
| \(\) | \(=\) | \(\mathbf{\pm 10}\text{ g}\) |
(ii) Subtract and add the absolute error.
| \(\text{Lower limit}\) | \(=\) | \(640 - 10 = \mathbf{630}\text{ g}\) |
| \(\text{Upper limit}\) | \(=\) | \(640 + 10 = \mathbf{650}\text{ g}\) |
(iii) Compare the error with the reading.
| \(\text{Percentage error}\) | \(=\) | \(\dfrac{\text{absolute error}}{\text{measurement}} \times 100\%\) |
| \(\) | \(=\) | \(\dfrac{10}{640} \times 100\%\) |
| \(\) | \(=\) | \(1.5625\ldots\%\) |
| \(\) | \(\approx\) | \(\mathbf{1.56\%}\) |
The true mass \(m\) satisfies \(630 \le m \lt 650\) g.
(i) A crowd is reported as \(36\,000\), correct to \(2\) significant figures. Find the least and greatest possible crowd.
(ii) Which is the more accurate measurement: a run of \(2.4\) km, to the nearest \(0.1\) km, or a shelf of \(85\) cm, to the nearest centimetre?
(i) Two significant figures in \(36\,000\) means it was rounded to the nearest \(1000\), so the error is \(\pm 500\).
| \(36\,000 - 500\) | \(=\) | \(35\,500\) |
| \(36\,000 + 500\) | \(=\) | \(36\,500\) |
The crowd \(N\) satisfies \(35\,500 \le N \lt 36\,500\). People are counted in whole numbers, so the least is \(\mathbf{35\,500}\) and the greatest is \(\mathbf{36\,499}\).
(ii) Different units, so compare percentage errors.
| \(\text{Run}\) | \(:\) | \(\dfrac{0.05}{2.4} \times 100\% \approx 2.08\%\) |
| \(\text{Shelf}\) | \(:\) | \(\dfrac{0.5}{85} \times 100\% \approx 0.59\%\) |
The shelf has the smaller percentage error, so the shelf measurement is more accurate, even though the run was measured with a bigger unit.
(i) Find the least and greatest possible area.
(ii) Turf is sold in rolls that each cover exactly \(1\text{ m}^2\). How many rolls must be bought to be sure of covering the lawn?
(i) Each side is within \(\pm 0.05\) m. The least area uses both lower limits; the greatest uses both upper limits.
| \(\text{Least}\) | \(=\) | \(6.35 \times 2.45\) |
| \(\) | \(=\) | \(\mathbf{15.5575}\text{ m}^2\) |
| \(\text{Greatest}\) | \(=\) | \(6.45 \times 2.55\) |
| \(\) | \(=\) | \(\mathbf{16.4475}\text{ m}^2\) |
(ii) To be sure, plan for the greatest area and round up.
| \(16.4475 \div 1\) | \(=\) | \(16.4475 \to \mathbf{17}\text{ rolls}\) |
The stated measurements give \(6.4 \times 2.5 = 16\text{ m}^2\), but \(16\) rolls might not be enough.
Write the limits first. The time is to the nearest \(5\) min, so it is within \(\pm 2.5\) min.
| \(17.5 \le d\) | \(\lt\) | \(18.5\text{ km}\) |
| \(42.5 \le t\) | \(\lt\) | \(47.5\text{ min}\) |
Speed \(= d \div t\). The least speed uses the shortest distance and the longest time; the greatest uses the longest distance and the shortest time. Change minutes to hours by dividing by \(60\).
| \(\text{Least}\) | \(=\) | \(\dfrac{17.5}{47.5 \div 60}\) |
| \(\) | \(=\) | \(22.105\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{22.1}\text{ km/h}\) |
| \(\text{Greatest}\) | \(=\) | \(\dfrac{18.5}{42.5 \div 60}\) |
| \(\) | \(=\) | \(26.117\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{26.1}\text{ km/h}\) |
The stated values give \(24\) km/h, but the true speed could be anywhere from about \(22.1\) to \(26.1\) km/h.
Common pitfalls
Frequently asked questions
What is absolute error in measurement?
It is the greatest possible difference between the measured value and the true value, equal to half the smallest unit of the measuring instrument. A ruler marked in millimetres has an absolute error of \(\pm 0.5\) mm.
How do you calculate percentage error?
Divide the absolute error by the measurement and multiply by \(100\%\). A reading of \(250\) mL on a jug marked every \(10\) mL has a percentage error of \(\dfrac{5}{250} \times 100\% = 2\%\).
What are the upper and lower limits of a measurement?
They are the measurement minus and plus the absolute error. For \(6.3\) m to the nearest \(0.1\) m, the lower limit is \(6.25\) m and the upper limit is \(6.35\) m. The true value can equal the lower limit but is less than the upper limit.
Which measurement is more accurate?
The one with the smaller percentage error. Absolute errors alone cannot compare measurements of very different sizes.
How do you find the greatest possible area from rounded measurements?
Find the upper limit of each length, by adding half the precision, then multiply the upper limits together. For the least area multiply the lower limits.