Relative Error Formula - Definition, Steps, Examples

Key Points

• The relative error formula is $\frac{\lvert m - t \rvert}{\lvert t \rvert}$, the absolute error divided by the true or accepted value.

• Absolute error carries the units of the measurement, while relative error is a dimensionless ratio.

• Multiplying the relative error by 100 gives the percent error, which is the usual reporting form.

• The denominator is the true value, not the measured one, because the true value is the authoritative reference.

• Modern metrology reports measurement uncertainty rather than error, since the true value is normally unknown. NIST represents each uncertainty component as an estimated standard deviation.

What Is the Relative Error Formula?

The relative error formula scales an error against the quantity being measured:

$$\text{relative error} = \frac{\lvert m - t \rvert}{\lvert t \rvert}$$

where $m$ is the measured value and $t$ the true or accepted value. The numerator $\lvert m - t \rvert$ is the absolute error, and dividing by $\lvert t \rvert$ turns it into a proportion.

Why an Error Needs Scaling

An absolute error of 1 metre means nothing by itself. On a 2-metre doorway it is a serious mistake; on the distance from London to Paris it is beyond excellent.

Dividing by the true value supplies the missing context, and the units cancel in the process. A relative error is therefore comparable across quantities measured in completely different units, which an absolute error never is.

That is the same reasoning behind the coefficient of variation, which scales a standard deviation by a mean. Both turn an absolute spread into a relative one so that different scales can be compared.

Variable Key

Examples of the Relative Error Formula

Example 1

A length is measured as 9.8 cm when the true value is 10.0 cm. Find the absolute, relative and percent error.

$$\lvert m - t \rvert = \lvert 9.8 - 10.0 \rvert = 0.2 \text{ cm}$$ $$\text{relative error} = \frac{0.2}{10.0} = 0.02$$

Final answer: absolute error 0.2 cm, relative error 0.02, percent error 2 per cent.

Example 2

A mass is measured as 48 g against a true value of 50 g. Find the relative error by dividing by the measured value.

The measured figure is the one in hand, so it looks like the natural denominator:

$$\frac{2}{48} \approx 0.0417$$

That gives 4.17 per cent. But the true value is the standard the measurement is being judged against, and letting the measurement set its own denominator means a worse measurement is judged against a base it has itself distorted.

Dividing by the true value instead:

$$\frac{2}{50} = 0.04$$

Final answer: relative error 0.04, or 4 per cent exactly.

Example 3

A calculation gives 3.15 for a quantity whose accepted value is 3.14159. Find the percent error to three decimal places.

$$\frac{\lvert 3.15 - 3.14159 \rvert}{3.14159} \times 100 = \frac{0.00841}{3.14159} \times 100 \approx 0.268$$

Final answer: about 0.268 per cent

Example 4

Two instruments report 199 and 205 against a true value of 200. Which is more accurate?

$$\frac{1}{200} = 0.005, \qquad \frac{5}{200} = 0.025$$

Final answer: The first instrument, with a relative error of 0.5 per cent against 2.5 per cent.

Both errors are judged against the same true value, so here the comparison could have been made on absolute errors alone. Relative error becomes necessary only when the quantities measured differ.

Example 5

A thermometer reads 0.5 degrees Celsius when the true temperature is 0 degrees. Find the relative error.

The absolute error is 0.5 degrees, but the denominator is zero:

$$\frac{0.5}{0} \text{ is undefined}$$

Final answer: The relative error does not exist. With a true value of zero the formula fails, so report the absolute error of 0.5 degrees instead.

This is a genuine limitation rather than an edge case. Near-zero true values make relative error unstable long before it becomes undefined.

Don't Confuse These Four Measures

Error and uncertainty are not interchangeable. An error needs a known true value, whereas uncertainty quantifies the doubt about a result when no true value is available, which is the usual situation in real measurement.

Where Do Students Trip Up on Relative Error?

Mistake 1: Dividing by the measured value

Where it slips in: The measured value is the number produced by the experiment, so it feels like the subject of the calculation.

Don't do this: $\frac{\lvert m - t \rvert}{\lvert m \rvert}$

The correct way: Divide by $\lvert t \rvert$. The true value is the reference, and using the measurement as its own base lets a poor result flatter itself.

Students who label the two numbers as "reference" and "observation" before dividing pick the right denominator reliably. The habit is the same one that distinguishes percent error from percent difference.

Mistake 2: Dropping the absolute value

Where it slips in: The subtraction is done in whichever order the numbers appear.

Don't do this: Report a relative error of minus 0.04.

The correct way: Take the absolute value. Relative error measures the size of a discrepancy, so it is never negative. If the direction matters, report the signed deviation separately and call it bias.

Mistake 3: Treating error as uncertainty

Where it slips in: Both describe imperfect measurement and both are often quoted as plus or minus a figure.

Don't do this: Report a relative error when no true value is known.

The correct way: Use uncertainty when the true value is unavailable. NIST expresses each uncertainty component as an estimated standard deviation, which is a statement about doubt rather than about a known discrepancy.

The Short Version

• The relative error formula is $\frac{\lvert m - t \rvert}{\lvert t \rvert}$, with percent error being the same figure times 100.

• Scaling by the true value makes errors comparable across quantities measured in different units.

• The denominator is always the true value, never the measurement being judged.

• Relative error fails at a true value of zero, and modern metrology reports uncertainty instead when no true value is known.

• To work through measurement and accuracy with a teacher, explore a high school math tutor or math tuition.

Quick Self-Check, Try These

A reading of 7.6 is taken where the accepted value is 8.0. Find the absolute, relative and percent error. Then decide which is more accurate: 99 against a true 100, or 495 against a true 500.

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Read More

• Percent Difference Formula

• Coefficient of Variation Formula

• Sample Standard Deviation Formula

• What Is a Percent

• What Is Ratio