Measurement errors - Year 9 Mathematics

What you will learn

calculate absolute error, relative error, and percentage error,
determine upper and lower bounds for rounded or measured values,
understand the concept of maximum possible error for a given precision,
apply error propagation to sums, differences, products, and quotients,
evaluate whether a measurement or result is acceptably precise for a given context.

Worked example 0 Real-world example: is the dosage safe?

A pharmacist measures 250 mg of a drug, but the actual required dose is 245 mg. Is the measurement within the acceptable 5% tolerance?

Absolute error $= |250 - 245| = 5$ mg.
Relative error $= \dfrac{5}{245} \approx 0.0204$ .
Percentage error $= 0.0204 \times 100\% \approx 2.0\%$ .
Since $2.0\% < 5\%$ , the dosage is within acceptable tolerance.

Key idea: percentage error gives a standardised way to judge whether an error is large or small relative to the quantity being measured.

1. Types of error

Error formulas

Absolute error

\text{Absolute error} = |\text{measured value} - \text{actual value}|

Relative error

\text{Relative error} = \frac{\text{absolute error}}{\text{actual value}}

Percentage error

\text{Percentage error} = \text{relative error} \times 100\%

Worked example 1 Calculating all three error types

A student measures the length of a table as 1.53 m. The actual length is 1.50 m.

Absolute error $= |1.53 - 1.50| = 0.03$ m.
Relative error $= \dfrac{0.03}{1.50} = 0.02$ .
Percentage error $= 0.02 \times 100\% = 2\%$ .

Worked example 2 Comparing errors

Measurement A: a 5 cm length measured as 5.2 cm. Measurement B: a 200 cm length measured as 201.5 cm. Which is more accurate?

A: absolute error $= 0.2$ cm, percentage error $= \dfrac{0.2}{5} \times 100\% = 4\%$ .
B: absolute error $= 1.5$ cm, percentage error $= \dfrac{1.5}{200} \times 100\% = 0.75\%$ .
Measurement B has a larger absolute error but a much smaller percentage error, so it is more accurate in relative terms.

2. Upper and lower bounds

When a measurement is rounded to a certain precision, the true value lies within a range.

Bounds for a rounded measurement

\text{Lower bound} = \text{stated value} - \frac{1}{2} \times \text{precision}

\text{Upper bound} = \text{stated value} + \frac{1}{2} \times \text{precision}

The maximum possible error of a rounded measurement is half the precision (the smallest unit).

Worked example 3 Bounds of a single measurement

A length is given as 4.7 cm, measured to the nearest 0.1 cm.

Precision $= 0.1$ cm.
Lower bound $= 4.7 - 0.05 = 4.65$ cm.
Upper bound $= 4.7 + 0.05 = 4.75$ cm.
The actual length $l$ satisfies $4.65 \leq l < 4.75$ .

Worked example 4 Bounds for a calculated area

A rectangle is measured as 12 cm by 8 cm, each to the nearest centimetre. Find the upper and lower bounds of the area.

Length bounds: $11.5 \leq l < 12.5$ . Width bounds: $7.5 \leq w < 8.5$ .
Lower bound of area $= 11.5 \times 7.5 = 86.25$ cm $^2$ .
Upper bound of area $= 12.5 \times 8.5 = 106.25$ cm $^2$ .
The stated area is $12 \times 8 = 96$ cm $^2$ , but the true area could be anywhere from $86.25$ to $106.25$ cm $^2$ .

Upper and lower bounds for a length measured to the nearest 0.1 cm.

3. Propagation of errors

When measurements are combined in calculations, errors accumulate.

Error propagation rules

Addition or subtraction

\text{Max error of } (A \pm B) = \text{max error of } A + \text{max error of } B

Multiplication (percentage errors)

\text{Max \% error of } (A \times B) \approx \text{\% error of } A + \text{\% error of } B

Division (percentage errors)

\text{Max \% error of } \frac{A}{B} \approx \text{\% error of } A + \text{\% error of } B

Worked example 5 Error propagation in addition

Two planks are measured as 2.4 m and 1.8 m, each to the nearest 0.1 m. Find the upper and lower bounds for their combined length.

Max error for each plank $= 0.05$ m.
Combined max error $= 0.05 + 0.05 = 0.10$ m.
Stated total $= 2.4 + 1.8 = 4.2$ m.
Lower bound $= 4.2 - 0.10 = 4.10$ m. Upper bound $= 4.2 + 0.10 = 4.30$ m.

Worked example 6 Error propagation in a product

A speed is calculated as distance $\div$ time. Distance $= 100$ m (to the nearest metre) and time $= 12.5$ s (to the nearest 0.1 s).

Distance max error $= 0.5$ m, so percentage error $= \dfrac{0.5}{100} \times 100\% = 0.5\%$ .
Time max error $= 0.05$ s, so percentage error $= \dfrac{0.05}{12.5} \times 100\% = 0.4\%$ .
Speed $= \dfrac{100}{12.5} = 8$ m/s. Maximum percentage error $\approx 0.5\% + 0.4\% = 0.9\%$ .
Maximum absolute error in speed $\approx 8 \times 0.009 = 0.072$ m/s. So speed $= 8.0 \pm 0.1$ m/s (rounded sensibly).

Practice

Fluency

Tier 1: basic skills

The actual mass of a parcel is 3.2 kg. A scale reads 3.35 kg. Find the absolute error.
Find the relative error for the measurement in Q1.
Find the percentage error for the measurement in Q1.
A length is 15.8 cm, measured to the nearest 0.1 cm. State the upper and lower bounds.
A time is recorded as 24 seconds, to the nearest second. State the upper and lower bounds.
A mass is given as 500 g, to the nearest 10 g. What is the maximum possible error?
A stick is measured as 1.2 m to the nearest 0.1 m, and another as 0.8 m to the nearest 0.1 m. Find the bounds of their combined length.
A student estimates $\pi$ as 3.14. Find the percentage error (use $\pi = 3.14159\ldots$ ).
The temperature is recorded as $22°$ C to the nearest degree. What are the upper and lower bounds?
A car’s odometer reads 45 230 km, rounded to the nearest km. State the bounds.

Reasoning

Tier 2: mixed practice

Two students measure the same length. Student A gets 12.4 cm (actual 12.0 cm) and Student B gets 47.2 cm (actual 46.0 cm). Who has the smaller percentage error?
A rectangle is measured as 20 cm $\times$ 15 cm, each to the nearest cm. Find the upper and lower bounds of the area.
A runner completes 400 m (to the nearest metre) in 52.3 s (to the nearest 0.1 s). Calculate the speed and find the maximum percentage error in the speed.
A square has a side measured as 8.0 cm to the nearest mm. Find the upper and lower bounds of its area.
Explain why percentage error is more useful than absolute error when comparing measurements of different magnitudes.
A cylindrical tank has radius 1.2 m and height 3.0 m, each to the nearest 0.1 m. Calculate the volume and find the upper and lower bounds.
A thermometer has a precision of $0.5°$ C. If it reads $37.0°$ C, state the error bounds and express the maximum percentage error.

Reasoning

Tier 3: explain and apply

A surveyor measures the distance between two points as 84.3 m. The true distance is 85.0 m. She then uses this measurement to calculate the area of a square plot. Find the percentage error in the distance and in the area.
The density of an object is calculated using $\rho = \dfrac{m}{V}$ . The mass is $120 \pm 0.5$ g and the volume is $50 \pm 1$ cm $^3$ . Find the density and the maximum percentage error.
Explain why the maximum error of a difference $A - B$ uses the sum of the individual errors, not the difference.
A GPS device is accurate to $\pm 3$ m. Two points are recorded 200 m apart. What is the maximum percentage error in this distance? If the points were only 10 m apart, how would the percentage error change?

Challenge

Reasoning

Harder reasoning

The area of a circle is calculated from a measured diameter of $14.0 \pm 0.1$ cm. Find the percentage error in the area. (Hint: area depends on $d^2$ .)
A student measures the acceleration due to gravity using $g = \dfrac{2s}{t^2}$ , where $s = 1.200 \pm 0.005$ m and $t = 0.495 \pm 0.005$ s. Calculate $g$ and the maximum percentage error. Comment on whether this is an acceptable result given that the accepted value is $9.81$ m/s $^2$ .
Two measurements are subtracted: $A = 50.0 \pm 0.5$ and $B = 49.0 \pm 0.5$ . Find the result and its percentage error. Explain why subtraction of nearly equal quantities leads to a large relative error (this is called catastrophic cancellation).
A map has a scale of 1 : 25 000. A distance on the map is measured as $4.2 \pm 0.1$ cm. Find the actual distance and the bounds (in metres) of the actual distance.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Tier 1

Absolute error $= |3.35 - 3.2| = 0.15$ kg.
Relative error $= \dfrac{0.15}{3.2} = 0.046875$ .
Percentage error $= 0.046875 \times 100\% \approx 4.7\%$ .
Lower bound $= 15.75$ cm, upper bound $= 15.85$ cm.
Lower bound $= 23.5$ s, upper bound $= 24.5$ s.
Maximum possible error $= \dfrac{10}{2} = 5$ g.
Combined stated length $= 2.0$ m. Max error $= 0.05 + 0.05 = 0.10$ m. Lower bound $= 1.90$ m, upper bound $= 2.10$ m.
Absolute error $= |3.14 - 3.14159| = 0.00159$ . Percentage error $= \dfrac{0.00159}{3.14159} \times 100\% \approx 0.051\%$ .
Lower bound $= 21.5°$ C, upper bound $= 22.5°$ C.
Lower bound $= 45\,229.5$ km, upper bound $= 45\,230.5$ km.

Tier 2

Student A: $\dfrac{0.4}{12.0} \times 100\% = 3.33\%$ . Student B: $\dfrac{1.2}{46.0} \times 100\% = 2.61\%$ . Student B has the smaller percentage error.
Length bounds: $19.5$ to $20.5$ cm. Width bounds: $14.5$ to $15.5$ cm. Lower area $= 19.5 \times 14.5 = 282.75$ cm $^2$ . Upper area $= 20.5 \times 15.5 = 317.75$ cm $^2$ .
Speed $= \dfrac{400}{52.3} \approx 7.65$ m/s. Distance % error $= \dfrac{0.5}{400} \times 100\% = 0.125\%$ . Time % error $= \dfrac{0.05}{52.3} \times 100\% \approx 0.096\%$ . Max % error in speed $\approx 0.125\% + 0.096\% = 0.22\%$ .
Side bounds: $7.95$ cm to $8.05$ cm. Lower area $= 7.95^2 = 63.2025$ cm $^2$ . Upper area $= 8.05^2 = 64.8025$ cm $^2$ .
Percentage error scales the error to the size of the quantity, enabling fair comparison. A 1 cm error on a 10 cm measurement (10%) is far more significant than a 1 cm error on a 1000 cm measurement (0.1%), but the absolute errors are identical.
Volume $= \pi r^2 h = \pi \times 1.2^2 \times 3.0 \approx 13.57$ m $^3$ . Lower: $\pi \times 1.15^2 \times 2.95 \approx 12.26$ m $^3$ . Upper: $\pi \times 1.25^2 \times 3.05 \approx 14.97$ m $^3$ .
Bounds: $36.75°$ C to $37.25°$ C. Max error $= 0.25°$ C. Percentage error $= \dfrac{0.25}{37.0} \times 100\% \approx 0.68\%$ .

Tier 3

Distance percentage error $= \dfrac{|84.3 - 85.0|}{85.0} \times 100\% = \dfrac{0.7}{85.0} \times 100\% \approx 0.82\%$ . Area using measured distance $= 84.3^2 = 7106.49$ m $^2$ . Actual area $= 85.0^2 = 7225$ m $^2$ . Area percentage error $= \dfrac{|7106.49 - 7225|}{7225} \times 100\% \approx 1.64\%$ , which is approximately double the distance error (since area $\propto d^2$ ).
Density $= \dfrac{120}{50} = 2.4$ g/cm $^3$ . Mass % error $= \dfrac{0.5}{120} \times 100\% \approx 0.42\%$ . Volume % error $= \dfrac{1}{50} \times 100\% = 2.0\%$ . Max % error in density $\approx 0.42\% + 2.0\% = 2.42\%$ . Density $= 2.4 \pm 0.06$ g/cm $^3$ .
The error in $A - B$ is maximised when $A$ is at its upper bound and $B$ is at its lower bound (or vice versa). This gives the largest possible spread: $(A + \delta_A) - (B - \delta_B) = (A - B) + (\delta_A + \delta_B)$ . The individual errors add regardless of whether we add or subtract the measurements.
Max error $= 3 + 3 = 6$ m. For 200 m: % error $= \dfrac{6}{200} \times 100\% = 3\%$ . For 10 m: % error $= \dfrac{6}{10} \times 100\% = 60\%$ . The GPS is unreliable for short distances.

Challenge

Radius $= 7.0$ cm, so area $= \pi \times 7.0^2 = 49\pi \approx 153.94$ cm $^2$ . Diameter % error $= \dfrac{0.1}{14.0} \times 100\% \approx 0.714\%$ . Since area $\propto d^2$ , area % error $\approx 2 \times 0.714\% = 1.43\%$ .
$g = \dfrac{2 \times 1.200}{0.495^2} = \dfrac{2.400}{0.245025} \approx 9.795$ m/s $^2$ . Distance % error $= \dfrac{0.005}{1.200} \times 100\% \approx 0.42\%$ . Time % error $= \dfrac{0.005}{0.495} \times 100\% \approx 1.01\%$ . Since $g \propto t^{-2}$ , time contributes $2 \times 1.01\% = 2.02\%$ . Total % error $\approx 0.42\% + 2.02\% = 2.44\%$ . This gives $g = 9.80 \pm 0.24$ m/s $^2$ . The accepted value $9.81$ falls within this range, so the result is acceptable.
Result $= 50.0 - 49.0 = 1.0$ . Max error $= 0.5 + 0.5 = 1.0$ . Percentage error $= \dfrac{1.0}{1.0} \times 100\% = 100\%$ . When two nearly equal quantities are subtracted, the result is small but the absolute error remains the sum of the original errors, leading to a very large relative error. This catastrophic cancellation should be avoided in practice by redesigning the measurement approach.
Map distance $= 4.2$ cm, so actual distance $= 4.2 \times 25\,000 = 105\,000$ cm $= 1050$ m. Lower bound: $(4.2 - 0.1) \times 25\,000 = 4.1 \times 25\,000 = 102\,500$ cm $= 1025$ m. Upper bound: $(4.2 + 0.1) \times 25\,000 = 4.3 \times 25\,000 = 107\,500$ cm $= 1075$ m. Actual distance is between 1025 m and 1075 m.

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