1.3 Percentage Error and Estimation

SL 1.6

In this section you will learn

how to calculate the percentage error of an approximate value
where errors come from: rounding, and the limits of a measuring instrument
how to find the maximum percentage error of a result calculated from a measurement
how to estimate an answer, and how to decide whether a calculated answer is reasonable

An approximate value is close to the exact value, but not equal to it. The percentage error tells us how big the difference is, compared with the exact value.

Percentage error

Percentage error In the formula booklet

\[ \varepsilon = \left| \frac{v_A - v_E}{v_E} \right| \times 100\% \]

$\varepsilon$: the percentage error (the Greek letter epsilon)
$v_A$: the approximate value of $v$ (an estimate, a rounded value or a measurement)
$v_E$: the exact value of $v$

The absolute value signs $|\;\;|$ make the error positive, so it does not matter whether the approximation is too big or too small. Always divide by the exact value $v_E$.

Example 1 Two common approximations for $\pi$ are $3.14$ and $\dfrac{22}{7}$. Find the percentage error of each one. Which is the better approximation?

Solution

Here $v_E = \pi$.

\[ \varepsilon = \left| \frac{3.14 - \pi}{\pi} \right| \times 100\% = 0.050\,69\ldots\% \approx 0.0507\% \]

\[ \varepsilon = \left| \frac{\frac{22}{7} - \pi}{\pi} \right| \times 100\% = 0.040\,24\ldots\% \approx 0.0402\% \]

$\dfrac{22}{7}$ has the smaller percentage error, so it is the better approximation.

Example 2 A shop’s income in one month is $\$48\,637$. In a report, it is rounded to 2 significant figures (s.f.). Find the percentage error caused by the rounding.

Solution

To 2 s.f., $v_A = \$49\,000$. The exact value is $v_E = \$48\,637$.

\[ \varepsilon = \left| \frac{49\,000 - 48\,637}{48\,637} \right| \times 100\% = 0.746\,3\ldots\% \approx 0.746\% \]

Example 3 Hana estimates that a tree is $12$ m tall. She then measures it accurately, and finds that it is $13.6$ m tall.

(a) Find the percentage error of her estimate.

(b) Interpret your answer.

Solution

(a) $v_A = 12$ and $v_E = 13.6$:

\[ \varepsilon = \left| \frac{12 - 13.6}{13.6} \right| \times 100\% = 11.76\ldots\% \approx 11.8\% \]

(b) Her estimate is about $12\%$ too small. This is a reasonable estimate by eye, but not accurate enough for building work.

Measurement errors and the maximum percentage error

Every measuring instrument has a limit. A ruler marked in millimetres cannot measure to $0.01$ mm. So a measurement is a rounded value, and the exact value lies between its bounds (see 1.2).

When we calculate with a measurement (for example an area from a length), the error in the measurement causes an error in the result. The exact result could be at either end of its range, so we check both.

Maximum percentage error of a calculated result Not in the formula booklet

Calculate the result $v_A$ from the measured value(s).
Find the lower and upper bounds of the measurement(s), and calculate the result at the lower bound and at the upper bound.
Use each of these as $v_E$ in the percentage error formula.
The larger of the two answers is the maximum percentage error.

Example 4 The radius of a circular pond is measured as $3.8$ m, correct to 1 decimal place (d.p.).

(a) Calculate the area of the pond, using the measured radius.

(b) Find the maximum percentage error in the area.

Solution

(a) $A = \pi \times 3.8^2 = 45.36\ldots \approx 45.4 \text{ m}^2$.

(b) The radius lies in $3.75 \le r < 3.85$.

At the lower bound: $\pi \times 3.75^2 = 44.17\ldots \text{ m}^2$. $\quad$ At the upper bound: $\pi \times 3.85^2 = 46.56\ldots \text{ m}^2$.

\[ \varepsilon_1 = \left| \frac{45.364\ldots - 44.178\ldots}{44.178\ldots} \right| \times 100\% = 2.684\ldots\% \]

\[ \varepsilon_2 = \left| \frac{45.364\ldots - 46.566\ldots}{46.566\ldots} \right| \times 100\% = 2.580\ldots\% \]

The maximum percentage error is $2.68\%$ (3 s.f.).

Example 5 A cube-shaped box has edges of length $12$ cm, correct to the nearest centimetre.

(a) Find the maximum percentage error in the length of an edge.

(b) Find the maximum percentage error in the volume of the box.

(c) Comment on your answers.

Solution

The edge lies in $11.5 \le x < 12.5$. The measured value is $v_A = 12$ cm.

(a) The larger error comes from the lower bound (we divide by the smaller number):

\[ \varepsilon = \left| \frac{12 - 11.5}{11.5} \right| \times 100\% = 4.347\ldots\% \approx 4.35\% \]

(With the upper bound: $\left| \frac{12 - 12.5}{12.5} \right| \times 100\% = 4\%$, which is smaller.)

(b) Measured volume: $12^3 = 1728 \text{ cm}^3$. At the bounds: $11.5^3 = 1520.875$ and $12.5^3 = 1953.125$.

\[ \varepsilon_1 = \left| \frac{1728 - 1520.875}{1520.875} \right| \times 100\% = 13.61\ldots\% \]

\[ \varepsilon_2 = \left| \frac{1728 - 1953.125}{1953.125} \right| \times 100\% = 11.52\ldots\% \]

The maximum percentage error in the volume is $13.6\%$ (3 s.f.).

(c) An error of about $4\%$ in the length becomes an error of about $14\%$ in the volume. Errors grow when a measurement is multiplied by itself, so lengths that are used for volumes should be measured more accurately.

Estimation

An estimate is a quick, rough answer. Estimates help you to check a graphic display calculator (GDC) answer: a typing mistake on the GDC often gives an answer that is far too big or far too small.

Estimating and checking

Round each number to 1 significant figure.
Work out the answer by hand with these simple numbers.
Compare with the GDC answer. They should be close (not equal).
Also ask: is the answer reasonable? Lengths and areas cannot be negative, a probability is between $0$ and $1$, a number of people is a whole number, and the size should make sense in the context.

Example 6 A school buys $38$ adult tickets at $\$21.95$ each and $12$ child tickets at $\$8.50$ each.

(a) Estimate the total cost by rounding each number to 1 s.f.

(b) Use your GDC to find the exact total cost.

(c) Find the percentage error of your estimate.

(d) A student says that the total cost is $\$9361.00$. Use your estimate to comment.

Solution

(a) $38 \approx 40$, $\;21.95 \approx 20$, $\;12 \approx 10$, $\;8.50 \approx 9$:

\[ 40 \times 20 + 10 \times 9 = 800 + 90 = \$890 . \]

(b)

\[ \begin{aligned} 38 \times 21.95 + 12 \times 8.50 &= 834.10 + 102 \\ &= \$936.10 \end{aligned} \]

(c) $\varepsilon = \left| \dfrac{890 - 936.10}{936.10} \right| \times 100\% = 4.92\%$ (3 s.f.).

(d) $\$9361.00$ is about ten times the estimate of $\$890$, so it cannot be right. The student probably made a typing error (the correct answer, $\$936.10$, has the same digits).

Common errors

✗ Wrong	✓ Right	Why
$\varepsilon = \left\lvert \dfrac{v_A - v_E}{v_A} \right\rvert \times 100\%$	Divide by $v_E$	The error is compared with the exact value.
$\varepsilon = 0.0268$	$\varepsilon = 2.68\%$	Multiply by $100\%$.
Using only the upper bound for the maximum error	Check both bounds	The lower bound usually gives the larger error.
Giving a 1 s.f. estimate as the final answer	Use the estimate only to check	The final answer comes from the full calculation.

Exercises

1 The acceleration due to gravity is $9.81 \text{ m s}^{-2}$. In a quick calculation it is approximated by $10 \text{ m s}^{-2}$. Find the percentage error.

Solution

\[ \begin{aligned} \varepsilon &= \left| \dfrac{10 - 9.81}{9.81} \right| \times 100\% \\ &= 1.936\ldots\% \\ &\approx 1.94\% \end{aligned} \]

2 The population of a town is $186\,420$. A newspaper gives it correct to 2 s.f. Find the percentage error.

Solution

$v_A = 190\,000$.

\[ \begin{aligned} \varepsilon &= \left| \dfrac{190\,000 - 186\,420}{186\,420} \right| \times 100\% \\ &= 1.920\ldots\% \\ &\approx 1.92\% \end{aligned} \]

3 A running race is exactly $10$ km long. A runner’s watch records the distance as $10.23$ km.

(a) Find the percentage error of the watch.

(b) The watch shows $4.6$ km at one point in the race. If the watch has the same percentage error, and it always reads too high, estimate the true distance run.

Solution

(a) $\varepsilon = \left| \dfrac{10.23 - 10}{10} \right| \times 100\% = 2.3\%$.

(b) The watch reading is $1.023$ times the true distance, so the true distance is $\dfrac{4.6}{1.023} = 4.496\ldots \approx 4.50$ km.

4 Find the percentage error when $\sqrt{2}$ is approximated by $1.41$.

Solution

\[ \begin{aligned} \varepsilon &= \left| \dfrac{1.41 - \sqrt{2}}{\sqrt{2}} \right| \times 100\% \\ &= 0.2979\ldots\% \\ &\approx 0.298\% \end{aligned} \]

5 A builder estimates that a job will take $45$ minutes. His estimate is too small, and its percentage error is $25\%$. Find the time that the job really takes.

Solution

The estimate is $25\%$ below the exact value, so $45 = 0.75\, v_E$ and $v_E = \dfrac{45}{0.75} = 60$ minutes.

(Check: $\left| \dfrac{45 - 60}{60} \right| \times 100\% = 25\%$ ✓.)

6 The side of a square is measured as $7.4$ cm, correct to 1 d.p. Find the maximum percentage error in the area of the square.

Solution

Measured area: $7.4^2 = 54.76$. Bounds of the side: $7.35 \le s < 7.45$, so the area is between $54.0225$ and $55.5025$.

\[ \begin{aligned} \varepsilon_1 &= \left| \dfrac{54.76 - 54.0225}{54.0225} \right| \times 100\% \\ &= 1.365\ldots\% \end{aligned} \]

\[ \begin{aligned} \varepsilon_2 &= \left| \dfrac{54.76 - 55.5025}{55.5025} \right| \times 100\% \\ &= 1.337\ldots\% \end{aligned} \]

Maximum percentage error: $1.37\%$.

7 A ball is a sphere of radius $5.0$ cm, correct to 1 d.p. The volume of a sphere is $V = \frac{4}{3}\pi r^3$.

(a) Find the volume, using the measured radius.

(b) Find the maximum percentage error in the volume.

Solution

(a) $V = \frac{4}{3}\pi \times 5^3 = 523.59\ldots \approx 524 \text{ cm}^3$.

(b) $4.95 \le r < 5.05$. The factor $\frac{4}{3}\pi$ cancels in the fraction, so

$\varepsilon_1 = \left| \dfrac{5^3 - 4.95^3}{4.95^3} \right| \times 100\% = 3.061\ldots\%$ $\quad$ $\varepsilon_2 = \left| \dfrac{5^3 - 5.05^3}{5.05^3} \right| \times 100\% = 2.940\ldots\%$

Maximum percentage error: $3.06\%$.

8 A car travels $386$ km and uses $29.4$ litres of petrol.

(a) Estimate the number of kilometres travelled per litre, by rounding to 1 s.f.

(b) Find the value with your GDC.

(c) Another student’s GDC answer is $0.0762$. Use your estimate to explain the mistake.

Solution

(a) $\dfrac{400}{30} \approx 13$ km per litre.

(b) $\dfrac{386}{29.4} = 13.12\ldots \approx 13.1$ km per litre.

(c) $0.0762$ is far from $13$. The student divided the wrong way: $\dfrac{29.4}{386} = 0.0762$ is litres per kilometre.

9 A room is $4.86$ m long and $3.12$ m wide. Carpet costs $\$38.75$ per square metre.

(a) Estimate the cost of carpet for the room, by rounding each number to 1 s.f.

(b) Find the exact cost.

(c) Find the percentage error of your estimate.

Solution

(a) $5 \times 3 \times 40 = \$600$.

(b) $4.86 \times 3.12 \times 38.75 = 587.574 \approx \$587.57$.

(c) $\varepsilon = \left| \dfrac{600 - 587.574}{587.574} \right| \times 100\% = 2.11\%$ (3 s.f.).

10 Comment on whether each answer is reasonable.

(a) The height of a classmate: $16.5$ m.

(b) The probability that it rains tomorrow: $1.3$.

(c) The width of a road: $-7.2$ m.

(d) The average speed of a car on a motorway: $108 \text{ km h}^{-1}$.

Solution

(a) Not reasonable: people are about $1.5$ m to $2$ m tall. Perhaps the answer should be $1.65$ m.

(b) Not reasonable: a probability cannot be greater than $1$.

(c) Not reasonable: a length cannot be negative.

(d) Reasonable: this is a normal motorway speed.