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An absolute difference says how far a result is from a reference. Percentage error expresses the size of that difference relative to the reference. The reference must be identified, and its magnitude must be non-zero.

This is an application of the terminology introduced in Error, Accuracy and Precision. Here the focus is comparing results, recovering tolerance limits, and avoiding misleading normalisations.

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Define the reference and the sign convention

Write xx for the measured or estimated value and rr for the reference value. The signed difference is

e=x−r.e=x-r.

A negative value means the result is below the reference. The absolute error magnitude is |x−r||x-r|. This corresponds to the metrology definition of measurement error, which allows a reference quantity value rather than requiring an exactly known “true” value.

In this lesson, percentage error means the non-negative magnitude

E=|x−r||r|×100%,r≠0.E=\frac{|x-r|}{|r|}\times100\%,\qquad r\ne0.

The quotient must compare values expressed in the same unit. A signed normalised error can instead be reported as (x−r)/|r|(x-r)/|r| times 100%, but label it explicitly because sign conventions vary.

Calculate the difference before dividing

A length is reported as 19.6 cm against a 20.0 cm reference. The signed error is −0.4-0.4 cm; its magnitude is 0.4 cm. Thus

E=0.4 cm20.0 cm×100%=2%.E=\frac{0.4\text{ cm}}{20.0\text{ cm}}\times100\%=2\%.

An alternative reading of 20.4 cm has the same percentage error but the opposite signed error. The percentage magnitude alone does not identify the direction.

Converting both readings to millimetres leaves 4/200=0.024/200=0.02, so the relative error is unchanged. Multiplicative changes of unit cancel between numerator and denominator.

Recover permitted values from a relative limit

Suppose a mathematical specification allows an error magnitude of at most 3% relative to a positive reference of 80 mm. Write xx for the numerical reading in millimetres. Then

|x−80|≤0.03×80=2.4.|x-80|\le0.03\times80=2.4.

Solving the absolute-value inequality gives

77.6≤x≤82.4.77.6\le x\le82.4.

Thus the permitted measurements run from 77.6 mm to 82.4 mm.

The endpoints are included because the phrase was “at most”. If the requirement were “less than 3%”, both endpoints would be excluded. This is a stipulated tolerance calculation, not a general rule for deciding whether a physical product is acceptable.

Conversely, a bound of 0.2 mm relative to 50 mm represents 0.2/50×100%=0.4%0.2/50\times100\%=0.4\%. A bound is a permitted or stated range; it is not the same thing as an observed error.

A percentage tolerance becomes an absolute interval

An inclusive three percent tolerance around eightyA number line shows the permitted interval from 77.6 to 82.4 around a reference of 80 millimetres; filled endpoints mean equality is allowed.A 3% tolerance around a reference of 80 mm76777879808182838477.682.4Reference: 803% of 80 = 2.477.6 mm ≤ reading ≤ 82.4 mm
The example uses a positive reference and an inclusive maximum relative-error condition.
Diagram explanation and exact data

Three percent of 80 is 2.4. Subtracting and adding it yields both endpoints. The criterion says at most three percent, so the boundary values qualify.

A percentage tolerance becomes an absolute interval: exact reference data
Quantity Value
Reference 80
Maximum absolute error 2.4
Lower permitted value 77.6
Upper permitted value 82.4

Distinguish error from percentage difference

Sometimes two results are compared without choosing one as the reference. For positive values, a commonly used symmetric percentage difference divides the absolute difference by their arithmetic mean:

D=|x−y|(x+y)/2×100%.D=\frac{|x-y|}{(x+y)/2}\times100\%.

For 48 and 52, this gives 4/50×100%=8%4/50\times100\%=8\%. Treating 52 as the reference instead gives 4/52×100%≈7.69%4/52\times100\%\approx7.69\%. These are not conflicting calculations. They use different denominators to answer different questions.

State the formula because “percentage difference” is not used consistently in every field. Do not apply this positive-value formula unchanged to arbitrary signed quantities or to a pair whose mean is zero.

Recognise when a percentage is a poor description

At a zero reference, percentage error is undefined. Near a very small reference, a tiny absolute difference can produce a huge percentage. Report the absolute difference and its units when that is more meaningful.

Relative comparisons can also depend on the origin of a scale. Readings of 21 °C and 20 °C differ by 1 °C, producing a numerical ratio of 1/20=5%1/20=5\%. The corresponding Kelvin readings differ by 1 K but give 1/293.15≈0.0034111/293.15\approx0.003411, or about 0.341%.

The physical difference did not change. The denominator’s origin did. For thermodynamic temperature ratios, use an absolute scale; for other applications, explain the chosen reference convention. Percentage arithmetic does not automatically make every scale suitable for ratio interpretation.

Propagate stated bounds through the actual calculation

Suppose positive lengths are bounded by 9.9≤L≤10.19.9\le L\le10.1 cm and 4.9≤W≤5.14.9\le W\le5.1 cm. For A=LWA=LW, multiplication is increasing in both positive inputs, so

48.51 cm2≤A≤51.51 cm2.48.51\text{ cm}^2\le A\le51.51\text{ cm}^2.

Relative to the central product 10×5=5010\times5=50 cm², these limits correspond to a decrease of 2.98% and an increase of 3.02%. Simply adding the input percentages, 1% and 2%, gives a useful small-error approximation of 3%, but not both exact endpoints.

The calculation assumes every combination of the stated bounds is permissible. If the quantities are constrained together, the attainable range may be narrower. Statistical uncertainty propagation is another method with further assumptions; it should not be silently substituted for deterministic bounds.

Keep error separate from uncertainty and bias

An observed difference from a reference is one comparison. Measurement uncertainty describes the dispersion of values attributed to the measurand under the available information. JCGM’s uncertainty definition makes that distinction explicit.

A small percentage error in one trial does not demonstrate that an instrument will remain accurate, and opposite errors can cancel in a sample mean. Repeated observations and a suitable reference are needed to investigate systematic and random effects. Keep the interpretation proportional to the evidence.

Practice questions

  1. A value is 48 against a reference of 50 in the same unit. Find the signed error and percentage error.
  2. What interval is within 2% of a positive 150-unit reference, including the boundaries?
  3. Two positive results are 95 and 105. Calculate their symmetric percentage difference.
Check your answers
  1. The signed error is 48−50=−248-50=-2 units. Percentage error is 2/50×100%=4%2/50\times100\%=4\%.
  2. Two percent of 150 is 3, so the interval is [147,153][147,153] units.
  3. Their mean is 100 and their absolute difference is 10. The stated symmetric difference is 10/100×100%=10%10/100\times100\%=10\%.

Related topics

Use Estimation and Approximation for numerical bounds, then Mathematical Modelling to distinguish calculation error from an unsuitable model.

Sources and further reading