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Measurement Error

Probability and Statistics · FE Reference Handbook section

Probability and Statistics
1 formulas
10 exam-style examples
~47 min
All Probability and Statistics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • Measurement error is defined as: Measured quantity value minus a reference quantity value. [Source: ISO JCGM 200:2012
  • Sources of errors in measurements arise from imperfections and disturbances in the measurement process, and added noise. One
  • where x is the measurand (value being measured), xref is the reference value, dsystematic is a disturbance from the measurement
  • process such as a drift or bias, and drandom is a disturbance such as random noise.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Measurement error, residual and combined uncertainty — Measurement Error

A distance with an accepted value of 98.00 ft is observed as 100.3 ft. The instrument contributes a standard uncertainty of ±0.12 ft and the target centring ±0.10 ft. Report the measurement error, the residual, and the combined standard uncertainty.

Given

  • Truevalue=98.00ftTrue value = 98.00 ft
  • Observation=100.3ftObservation = 100.3 ft
  • σ1=0.12ft,σ2=0.10ft\sigma_{1} = 0.12 ft, \sigma_{2} = 0.10 ft

Find

Error, residual and combined uncertainty

Start with the thinking

  • Error is observation minus true value; the residual is the same magnitude with the sign convention of the adjustment.
  • Independent uncertainties combine in quadrature, never by simple addition.

Step-by-step solution

  1. Formula

    e=observed−truee = observed - true
  2. Substituting

    e=100.3−98.00=2.300fte = 100.3 - 98.00 = 2.300 ft
  3. Relative error — 2.3469%

  4. Formula

    uc=(σ12+σ22)u_c = \sqrt(\sigma_{1}^{2} + \sigma_{2}^{2})
  5. Substituting

    uc=(0.122+0.102)=0.1562ftu_c = \sqrt(0.12^{2} + 0.10^{2}) = 0.1562 ft
  6. 95% expanded uncertainty

    U=2uc=0.3124ftU = 2u_c = 0.3124 ft
Answer:
Error=2.300ft,combineduncertainty=±0.156ftError = 2.300 ft, combined uncertainty = \pm0.156 ft

Why the other options are there

  • ±0.220 ft (uncertainties added directly)
  • -2.300 ft (sign reversed)

Reference: FE Reference Handbook — Probability and Statistics → Measurement Error

Example 2
Measurement error, residual and combined uncertainty — Measurement Error (2)

A distance with an accepted value of 106.0 ft is observed as 106.9 ft. The instrument contributes a standard uncertainty of ±0.09 ft and the target centring ±0.13 ft. Report the measurement error, the residual, and the combined standard uncertainty.

Given

  • Truevalue=106.0ftTrue value = 106.0 ft
  • Observation=106.9ftObservation = 106.9 ft
  • σ1=0.09ft,σ2=0.13ft\sigma_{1} = 0.09 ft, \sigma_{2} = 0.13 ft

Find

Error, residual and combined uncertainty

Start with the thinking

  • Error is observation minus true value; the residual is the same magnitude with the sign convention of the adjustment.
  • Independent uncertainties combine in quadrature, never by simple addition.

Step-by-step solution

  1. Formula

    e=observed−truee = observed - true
  2. Substituting

    e=106.9−106.0=0.900fte = 106.9 - 106.0 = 0.900 ft
  3. Relative error — 0.8491%

  4. Formula

    uc=(σ12+σ22)u_c = \sqrt(\sigma_{1}^{2} + \sigma_{2}^{2})
  5. Substituting

    uc=(0.092+0.132)=0.1581ftu_c = \sqrt(0.09^{2} + 0.13^{2}) = 0.1581 ft
  6. 95% expanded uncertainty

    U=2uc=0.3162ftU = 2u_c = 0.3162 ft
Answer:
Error=0.900ft,combineduncertainty=±0.158ftError = 0.900 ft, combined uncertainty = \pm0.158 ft

Why the other options are there

  • ±0.220 ft (uncertainties added directly)
  • -0.900 ft (sign reversed)

Reference: FE Reference Handbook — Probability and Statistics → Measurement Error

Example 3
Measurement error, residual and combined uncertainty — Measurement Error (3)

A distance with an accepted value of 198.5 ft is observed as 198.5 ft. The instrument contributes a standard uncertainty of ±0.14 ft and the target centring ±0.05 ft. Report the measurement error, the residual, and the combined standard uncertainty.

Given

  • Truevalue=198.5ftTrue value = 198.5 ft
  • Observation=198.5ftObservation = 198.5 ft
  • σ1=0.14ft,σ2=0.05ft\sigma_{1} = 0.14 ft, \sigma_{2} = 0.05 ft

Find

Error, residual and combined uncertainty

Start with the thinking

  • Error is observation minus true value; the residual is the same magnitude with the sign convention of the adjustment.
  • Independent uncertainties combine in quadrature, never by simple addition.

Step-by-step solution

  1. Formula

    e=observed−truee = observed - true
  2. Substituting

    e=198.5−198.5=0.000fte = 198.5 - 198.5 = 0.000 ft
  3. Relative error — 0.0000%

  4. Formula

    uc=(σ12+σ22)u_c = \sqrt(\sigma_{1}^{2} + \sigma_{2}^{2})
  5. Substituting

    uc=(0.142+0.052)=0.1487ftu_c = \sqrt(0.14^{2} + 0.05^{2}) = 0.1487 ft
  6. 95% expanded uncertainty

    U=2uc=0.2973ftU = 2u_c = 0.2973 ft
Answer:
Error=0.000ft,combineduncertainty=±0.149ftError = 0.000 ft, combined uncertainty = \pm0.149 ft

Why the other options are there

  • ±0.190 ft (uncertainties added directly)
  • 0.000 ft (sign reversed)

Reference: FE Reference Handbook — Probability and Statistics → Measurement Error

Example 4
Measurement error, residual and combined uncertainty — Measurement Error (4)

A distance with an accepted value of 139.0 ft is observed as 140.0 ft. The instrument contributes a standard uncertainty of ±0.09 ft and the target centring ±0.09 ft. Report the measurement error, the residual, and the combined standard uncertainty.

Given

  • Truevalue=139.0ftTrue value = 139.0 ft
  • Observation=140.0ftObservation = 140.0 ft
  • σ1=0.09ft,σ2=0.09ft\sigma_{1} = 0.09 ft, \sigma_{2} = 0.09 ft

Find

Error, residual and combined uncertainty

Start with the thinking

  • Error is observation minus true value; the residual is the same magnitude with the sign convention of the adjustment.
  • Independent uncertainties combine in quadrature, never by simple addition.

Step-by-step solution

  1. Formula

    e=observed−truee = observed - true
  2. Substituting

    e=140.0−139.0=1.000fte = 140.0 - 139.0 = 1.000 ft
  3. Relative error — 0.7194%

  4. Formula

    uc=(σ12+σ22)u_c = \sqrt(\sigma_{1}^{2} + \sigma_{2}^{2})
  5. Substituting

    uc=(0.092+0.092)=0.1273ftu_c = \sqrt(0.09^{2} + 0.09^{2}) = 0.1273 ft
  6. 95% expanded uncertainty

    U=2uc=0.2546ftU = 2u_c = 0.2546 ft
Answer:
Error=1.000ft,combineduncertainty=±0.127ftError = 1.000 ft, combined uncertainty = \pm0.127 ft

Why the other options are there

  • ±0.180 ft (uncertainties added directly)
  • -1.000 ft (sign reversed)

Reference: FE Reference Handbook — Probability and Statistics → Measurement Error

Example 5
Measurement error, residual and combined uncertainty — Measurement Error (5)

A distance with an accepted value of 172.5 ft is observed as 173.7 ft. The instrument contributes a standard uncertainty of ±0.04 ft and the target centring ±0.04 ft. Report the measurement error, the residual, and the combined standard uncertainty.

Given

  • Truevalue=172.5ftTrue value = 172.5 ft
  • Observation=173.7ftObservation = 173.7 ft
  • σ1=0.04ft,σ2=0.04ft\sigma_{1} = 0.04 ft, \sigma_{2} = 0.04 ft

Find

Error, residual and combined uncertainty

Start with the thinking

  • Error is observation minus true value; the residual is the same magnitude with the sign convention of the adjustment.
  • Independent uncertainties combine in quadrature, never by simple addition.

Step-by-step solution

  1. Formula

    e=observed−truee = observed - true
  2. Substituting

    e=173.7−172.5=1.200fte = 173.7 - 172.5 = 1.200 ft
  3. Relative error — 0.6957%

  4. Formula

    uc=(σ12+σ22)u_c = \sqrt(\sigma_{1}^{2} + \sigma_{2}^{2})
  5. Substituting

    uc=(0.042+0.042)=0.0566ftu_c = \sqrt(0.04^{2} + 0.04^{2}) = 0.0566 ft
  6. 95% expanded uncertainty

    U=2uc=0.1131ftU = 2u_c = 0.1131 ft
Answer:
Error=1.200ft,combineduncertainty=±0.057ftError = 1.200 ft, combined uncertainty = \pm0.057 ft

Why the other options are there

  • ±0.080 ft (uncertainties added directly)
  • -1.200 ft (sign reversed)

Reference: FE Reference Handbook — Probability and Statistics → Measurement Error

Example 6
Measurement error, residual and combined uncertainty — Measurement Error (6)

A distance with an accepted value of 72.00 ft is observed as 71.80 ft. The instrument contributes a standard uncertainty of ±0.18 ft and the target centring ±0.16 ft. Report the measurement error, the residual, and the combined standard uncertainty.

Given

  • Truevalue=72.00ftTrue value = 72.00 ft
  • Observation=71.80ftObservation = 71.80 ft
  • σ1=0.18ft,σ2=0.16ft\sigma_{1} = 0.18 ft, \sigma_{2} = 0.16 ft

Find

Error, residual and combined uncertainty

Start with the thinking

  • Error is observation minus true value; the residual is the same magnitude with the sign convention of the adjustment.
  • Independent uncertainties combine in quadrature, never by simple addition.

Step-by-step solution

  1. Formula

    e=observed−truee = observed - true
  2. Substituting

    e=71.80−72.00=−0.200fte = 71.80 - 72.00 = -0.200 ft
  3. Relative error — 0.2778%

  4. Formula

    uc=(σ12+σ22)u_c = \sqrt(\sigma_{1}^{2} + \sigma_{2}^{2})
  5. Substituting

    uc=(0.182+0.162)=0.2408ftu_c = \sqrt(0.18^{2} + 0.16^{2}) = 0.2408 ft
  6. 95% expanded uncertainty

    U=2uc=0.4817ftU = 2u_c = 0.4817 ft
Answer:
Error=−0.200ft,combineduncertainty=±0.241ftError = -0.200 ft, combined uncertainty = \pm0.241 ft

Why the other options are there

  • ±0.340 ft (uncertainties added directly)
  • 0.200 ft (sign reversed)

Reference: FE Reference Handbook — Probability and Statistics → Measurement Error

Example 7
Measurement error, residual and combined uncertainty — Measurement Error (7)

A distance with an accepted value of 150.5 ft is observed as 152.8 ft. The instrument contributes a standard uncertainty of ±0.13 ft and the target centring ±0.02 ft. Report the measurement error, the residual, and the combined standard uncertainty.

Given

  • Truevalue=150.5ftTrue value = 150.5 ft
  • Observation=152.8ftObservation = 152.8 ft
  • σ1=0.13ft,σ2=0.02ft\sigma_{1} = 0.13 ft, \sigma_{2} = 0.02 ft

Find

Error, residual and combined uncertainty

Start with the thinking

  • Error is observation minus true value; the residual is the same magnitude with the sign convention of the adjustment.
  • Independent uncertainties combine in quadrature, never by simple addition.

Step-by-step solution

  1. Formula

    e=observed−truee = observed - true
  2. Substituting

    e=152.8−150.5=2.300fte = 152.8 - 150.5 = 2.300 ft
  3. Relative error — 1.5282%

  4. Formula

    uc=(σ12+σ22)u_c = \sqrt(\sigma_{1}^{2} + \sigma_{2}^{2})
  5. Substituting

    uc=(0.132+0.022)=0.1315ftu_c = \sqrt(0.13^{2} + 0.02^{2}) = 0.1315 ft
  6. 95% expanded uncertainty

    U=2uc=0.2631ftU = 2u_c = 0.2631 ft
Answer:
Error=2.300ft,combineduncertainty=±0.132ftError = 2.300 ft, combined uncertainty = \pm0.132 ft

Why the other options are there

  • ±0.150 ft (uncertainties added directly)
  • -2.300 ft (sign reversed)

Reference: FE Reference Handbook — Probability and Statistics → Measurement Error

Example 8
Measurement error, residual and combined uncertainty — Measurement Error (8)

A distance with an accepted value of 127.0 ft is observed as 126.1 ft. The instrument contributes a standard uncertainty of ±0.11 ft and the target centring ±0.09 ft. Report the measurement error, the residual, and the combined standard uncertainty.

Given

  • Truevalue=127.0ftTrue value = 127.0 ft
  • Observation=126.1ftObservation = 126.1 ft
  • σ1=0.11ft,σ2=0.09ft\sigma_{1} = 0.11 ft, \sigma_{2} = 0.09 ft

Find

Error, residual and combined uncertainty

Start with the thinking

  • Error is observation minus true value; the residual is the same magnitude with the sign convention of the adjustment.
  • Independent uncertainties combine in quadrature, never by simple addition.

Step-by-step solution

  1. Formula

    e=observed−truee = observed - true
  2. Substituting

    e=126.1−127.0=−0.900fte = 126.1 - 127.0 = -0.900 ft
  3. Relative error — 0.7087%

  4. Formula

    uc=(σ12+σ22)u_c = \sqrt(\sigma_{1}^{2} + \sigma_{2}^{2})
  5. Substituting

    uc=(0.112+0.092)=0.1421ftu_c = \sqrt(0.11^{2} + 0.09^{2}) = 0.1421 ft
  6. 95% expanded uncertainty

    U=2uc=0.2843ftU = 2u_c = 0.2843 ft
Answer:
Error=−0.900ft,combineduncertainty=±0.142ftError = -0.900 ft, combined uncertainty = \pm0.142 ft

Why the other options are there

  • ±0.200 ft (uncertainties added directly)
  • 0.900 ft (sign reversed)

Reference: FE Reference Handbook — Probability and Statistics → Measurement Error

Example 9
Measurement error, residual and combined uncertainty — Measurement Error (9)

A distance with an accepted value of 131.0 ft is observed as 132.5 ft. The instrument contributes a standard uncertainty of ±0.15 ft and the target centring ±0.18 ft. Report the measurement error, the residual, and the combined standard uncertainty.

Given

  • Truevalue=131.0ftTrue value = 131.0 ft
  • Observation=132.5ftObservation = 132.5 ft
  • σ1=0.15ft,σ2=0.18ft\sigma_{1} = 0.15 ft, \sigma_{2} = 0.18 ft

Find

Error, residual and combined uncertainty

Start with the thinking

  • Error is observation minus true value; the residual is the same magnitude with the sign convention of the adjustment.
  • Independent uncertainties combine in quadrature, never by simple addition.

Step-by-step solution

  1. Formula

    e=observed−truee = observed - true
  2. Substituting

    e=132.5−131.0=1.500fte = 132.5 - 131.0 = 1.500 ft
  3. Relative error — 1.1450%

  4. Formula

    uc=(σ12+σ22)u_c = \sqrt(\sigma_{1}^{2} + \sigma_{2}^{2})
  5. Substituting

    uc=(0.152+0.182)=0.2343ftu_c = \sqrt(0.15^{2} + 0.18^{2}) = 0.2343 ft
  6. 95% expanded uncertainty

    U=2uc=0.4686ftU = 2u_c = 0.4686 ft
Answer:
Error=1.500ft,combineduncertainty=±0.234ftError = 1.500 ft, combined uncertainty = \pm0.234 ft

Why the other options are there

  • ±0.330 ft (uncertainties added directly)
  • -1.500 ft (sign reversed)

Reference: FE Reference Handbook — Probability and Statistics → Measurement Error

Example 10
Measurement error, residual and combined uncertainty — Measurement Error (10)

A distance with an accepted value of 198.5 ft is observed as 200.7 ft. The instrument contributes a standard uncertainty of ±0.06 ft and the target centring ±0.19 ft. Report the measurement error, the residual, and the combined standard uncertainty.

Given

  • Truevalue=198.5ftTrue value = 198.5 ft
  • Observation=200.7ftObservation = 200.7 ft
  • σ1=0.06ft,σ2=0.19ft\sigma_{1} = 0.06 ft, \sigma_{2} = 0.19 ft

Find

Error, residual and combined uncertainty

Start with the thinking

  • Error is observation minus true value; the residual is the same magnitude with the sign convention of the adjustment.
  • Independent uncertainties combine in quadrature, never by simple addition.

Step-by-step solution

  1. Formula

    e=observed−truee = observed - true
  2. Substituting

    e=200.7−198.5=2.200fte = 200.7 - 198.5 = 2.200 ft
  3. Relative error — 1.1083%

  4. Formula

    uc=(σ12+σ22)u_c = \sqrt(\sigma_{1}^{2} + \sigma_{2}^{2})
  5. Substituting

    uc=(0.062+0.192)=0.1992ftu_c = \sqrt(0.06^{2} + 0.19^{2}) = 0.1992 ft
  6. 95% expanded uncertainty

    U=2uc=0.3985ftU = 2u_c = 0.3985 ft
Answer:
Error=2.200ft,combineduncertainty=±0.199ftError = 2.200 ft, combined uncertainty = \pm0.199 ft

Why the other options are there

  • ±0.250 ft (uncertainties added directly)
  • -2.200 ft (sign reversed)

Reference: FE Reference Handbook — Probability and Statistics → Measurement Error

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