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ESO Y4A · Mathematics
Numbers · Real numbers

Absolute and relative error

Being out by 1 cm when measuring the wall of your bedroom is nothing. Being out by 1 cm when measuring the thickness of a coin is a disaster. The error is the same number, but it does not weigh the same. That is why science and engineering use two types of error, not one.

Before you start, make sure you know: Approximation and errors (3rd ESO)
Before you start, make sure you know: Percentages (1st ESO)

Absolute error and relative error

Visual comparison of absolute and relative error when measuring a 150 cm corridor and a 20 cm strip. Corridor: true length 150 cm, you measure 148 cm Eₐ = 2 cm  ·  Eᵣ = 2/150 ≈ 0,013 = 1,3 % → good precision Strip: true length 20 cm, you measure 20,5 cm Eₐ = 0,5 cm  ·  Eᵣ = 0,5/20 = 0,025 = 2,5 % → worse, even though the absolute error is smaller
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Calculate the absolute error and the relative error made when approximating the number $3,456$ by $3,46$. Express the relative error as a percentage.

Calculate the absolute error and the relative error made when approximating the number $3,456$ by $3,46$. Express the relative error as a percentage.

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Two measurements were taken in a technology class: in the first, an absolute error of $0,5$ cm was made when measuring a strip of wood of $20$ cm, and in the second, an…

Two measurements were taken in a technology class: in the first, an absolute error of $0,5$ cm was made when measuring a strip of wood of $20$ cm, and in the second, an absolute error of $2$ cm when measuring a plank of $100$ cm. Which of the two measurements is more precise? Justify your answer by calculating the relative errors.

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Comparing precision

Relative error helps us compare precision, as long as the true value is not zero. Whether a measurement is good enough also depends on its use and on the tolerance required.

Visual comparison of precision and relative error between two instruments or measurements of different magnitude. Stopwatch A: Eₐ = 0,02 s Eᵣ = 0,02/9,81 ≈ 0,20 % Stopwatch B: Eₐ = 0,01 s Eᵣ = 0,01/9,81 ≈ 0,10 % → more precise
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Given that the relative error of a weighing in the laboratory is $5\ \%$ and that the true value of the mass is $250$ g, determine the absolute error made.

Given that the relative error of a weighing in the laboratory is $5\ \%$ and that the true value of the mass is $250$ g, determine the absolute error made.

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A space telescope measures the distance to a distant star, obtaining $4,25 \cdot 10^8$ km. Given that the value considered exact by the astronomical community after…

A space telescope measures the distance to a distant star, obtaining $4,25 \cdot 10^8$ km. Given that the value considered exact by the astronomical community after multiple calibrations is $4,23 \cdot 10^8$ km, calculate the absolute error and the percentage relative error of the telescope's measurement.

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Error bound when you do not know the exact value

The resolution of an instrument is the smallest increment its reading can distinguish. It does not guarantee its accuracy: there may be calibration, reading or reaction errors.

Number line showing the uncertainty interval and the absolute error bound of 0,005 s for a measurement of 9,81 s. ideal reading rounded to 0,01 s: 9,81 s 9,8059,819,815 0,0050,005 true value in [9,805 ; 9,815) · Eₐ ≤ 0,005 s
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When measuring the length of a corridor, an approximate value of $150$ cm was obtained with a maximum absolute error of $2$ cm. Write down the interval in which the…

When measuring the length of a corridor, an approximate value of $150$ cm was obtained with a maximum absolute error of $2$ cm. Write down the interval in which the actual value of the length definitely lies.

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In a 100-metre sprint, the official timer has a precision of one hundredth of a second (that is, the absolute error bound is $0,01$ s). If the time displayed on the…

In a 100-metre sprint, the official timer has a precision of one hundredth of a second (that is, the absolute error bound is $0,01$ s). If the time displayed on the scoreboard for the winning athlete is $9,81$ s, calculate the relative error bound affecting this measurement and express it as a percentage.

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Propagating errors: when you calculate with approximate measurements

If you measure $L$ and $A$ with some error, and then calculate $L \cdot A$, the error is also carried over into the result. The most direct way to bound it is to work with the possible extreme values.

Propagation of the side's margin of error in calculating the minimum, approximate and maximum area of a square. side: 40 m ± 0,8 m (given margin) side 39,2 mminimum side 40,8 mmaximum minimum area1536,64 m² maximum area1664,64 m²
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The radius of a circular metal part has been measured using callipers, obtaining an approximate value of $r = 15$ mm, given that the instrument has a maximum margin of…
r = 15 ±0,5

The radius of a circular metal part has been measured using callipers, obtaining an approximate value of $r = 15$ mm, given that the instrument has a maximum margin of error of $\pm 0,5$ mm. Calculate the interval containing the actual area of the part. Then, using the measured value to find an approximate area, determine the maximum absolute error bound that could be made for this area.

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A surveyor measures the side of a square plot of land, obtaining $L = 40$ m. They know that the relative error of their measuring instrument is $2\ \%$ (that is,…

A surveyor measures the side of a square plot of land, obtaining $L = 40$ m. They know that the relative error of their measuring instrument is $2\ \%$ (that is, $E_r(L) = 0,02$). Show mathematically that the relative error bound made when calculating the area of the land from this measurement is approximately $4\ \%$.

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In a density experiment, the mass of an object is measured, obtaining $m = 120$ g with an absolute error bound of $2$ g, and its volume $V = 30$ cm$^3$ with an absolute…

In a density experiment, the mass of an object is measured, obtaining $m = 120$ g with an absolute error bound of $2$ g, and its volume $V = 30$ cm$^3$ with an absolute error bound of $1$ cm$^3$. Calculate the maximum and minimum possible actual values of the density ($d = \frac{m}{V}$). Using the central approximate value ($d_{aprox} = \frac{120}{30}$), what is the absolute error bound for the density?

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Back to the beginning: If the wall measures $300\text{ cm}$, being out by $1\text{ cm}$ means $E_r = 1/300 \approx 0{,}33\,\%$; for a coin $0{,}2\text{ cm}$ thick, that same centimetre means $E_r = 1/0{,}2 = 500\,\%$, an unacceptable deviation.

Summary

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