Year 10AC9M10M04

Measurement Error and Accuracy

Identify the impact of measurement errors on the accuracy of results in practical contexts.

Every measurement is wrong. The question is by how much, and what happens to that error once you start calculating with it — because error does not stay the same size.

Builds onYr 10 · Surface Area and Volume of Composite Solids

Method

The measurement is a range, not a number

\text{measured } 24 \text{ cm to nearest cm} \ \Rightarrow\ 23{,}5 \leq \ell < 24{,}5

Half the smallest unit either side. That range is the only honest statement about the length.

Absolute error is half the smallest unit

\text{error} = \tfrac{1}{2} \times \text{smallest unit}

Measuring to the nearest millimetre gives an absolute error of $0{,}5$ mm.

Percentage error compares error to size

\%\text{ error} = \frac{\text{absolute error}}{\text{measured value}} \times 100

A $0{,}5$ cm error on $2$ cm is serious. The same error on $200$ cm is not.

Error grows when you calculate

\text{area} \approx 2\times \qquad \text{volume} \approx 3\times

Percentage error roughly doubles for an area and triples for a volume, because the measurement is used twice or three times.

Worked example

e.g.

\text{A square tile is measured as } 20 \text{ cm to the nearest centimetre. Find the range of its area.}

Write the bounds of the measurement. Half a unit either side. $19{,}5 \leq \ell < 20{,}5$
Find the smallest possible area using the lower bound. $19{,}5^2 = 380{,}25 \text{ cm}^2$
Find the largest possible area using the upper bound. $20{,}5^2 = 420{,}25 \text{ cm}^2$
State the range. Never give a single value for a calculated result. $380{,}25 \leq A < 420{,}25$
Compare the percentage errors to see what the calculation did. $\text{length: } \tfrac{0{,}5}{20} = 2{,}5\% \qquad \text{area: } \tfrac{20}{400} = 5\%$
Note the pattern. Squaring the measurement roughly doubled the percentage error. $2{,}5\% \ \rightarrow\ 5\%$

Practice

Bounds first, then calculate with the bounds — never with the rounded value.

\text{A length reads } 8{,}4 \text{ cm to the nearest millimetre. Give the bounds.}

Answer

8{,}35 \leq \ell < 8{,}45

\text{Find the percentage error in a measurement of } 2 \text{ cm to the nearest centimetre.}

Answer

\tfrac{0{,}5}{2}\times 100 = 25\%

\text{Find the percentage error in } 200 \text{ cm to the nearest centimetre.}

Answer

\tfrac{0{,}5}{200}\times 100 = 0{,}25\%

\text{A cube edge is } 5 \text{ cm to the nearest cm. Find the largest possible volume.}

Answer

5{,}5^3 = 166{,}375 \text{ cm}^3

\text{Roughly what happens to a } 3\% \text{ length error when you calculate a volume?}

Answer

\text{It roughly triples, to about } 9\%.

\text{Why is it wrong to give a calculated area as } 400 \text{ cm}^2 \text{ exactly?}

Answer

\text{The measurement was a range, so the area is a range. A single value claims accuracy you never had.}