---
title: "Error propagation and effect of averaging"
subject: "Metrology"
chapter: 2
key: "semester_3/metrology:error-propagation-and-averaging"
source: https://www.bctengineer.com/q/semester_3/metrology/error-propagation-and-averaging
site: BCT Engineering
---

# Error propagation and effect of averaging

Metrology · Chapter 2: Errors in measurement

- Asked in: year not given, 4+4 marks
- Video: [Lesson 10 Propagation of Errors](https://www.youtube.com/watch?v=WMLVy38mPV0) — Analytical Chemistry I
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/metrology/error-propagation-and-averaging
- Whole chapter: https://www.bctengineer.com/md/semester_3/metrology/chapter-2

## Question

(a) A cylindrical slug has diameter $d = 25.00 \pm 0.02$ mm and height $h = 60.00 \pm 0.05$ mm. Calculate its volume and the maximum possible error and the probable (root-sum-square) error in the volume. (b) Eight readings of the diameter of a shaft (mm) are: 25.02, 24.98, 25.01, 25.03, 24.99, 25.00, 25.02, 25.01. Find the mean, the sample standard deviation and the standard error of the mean. How many readings are needed to halve the standard error?

## Short answer

Volume V = πd²h/4 = 29,452 mm³. Maximum error = 72 mm³, probable (RSS) error = 53 mm³. Mean diameter = 25.0075 mm, sample standard deviation s = 0.0167 mm, standard error = 0.0059 mm. To halve the standard error, 32 readings are needed (n quadrupled).

## Answer

#### (a) Error propagation
For $V = \dfrac{\pi d^2 h}{4}$ the relative error depends on each variable as

$$
\frac{\delta V}{V} = 2\frac{\delta d}{d} + \frac{\delta h}{h} \quad (\text{maximum}), \qquad \frac{\delta V}{V} = \sqrt{\left(2\frac{\delta d}{d}\right)^2 + \left(\frac{\delta h}{h}\right)^2} \quad (\text{probable})
$$

$$
V = \frac{\pi (25)^2 (60)}{4} = 29\,452\ \text{mm}^3
$$

- $2\,\delta d/d = 2(0.02/25) = 0.0016$
- $\delta h/h = 0.05/60 = 0.000833$

Maximum error: $\delta V = 29452 \times (0.0016 + 0.000833) = 71.7\ \text{mm}^3$ (0.243 %).

Probable error: $\delta V = 29452\sqrt{0.0016^2 + 0.000833^2} = 29452 \times 0.001804 = 53.1\ \text{mm}^3$ (0.180 %).

The diameter contributes most of the error because it is squared.

**Answer: $V = 29\,452 \pm 72$ mm³ (maximum), $\pm 53$ mm³ (probable).**

#### (b) Averaging
Number of readings $n = 8$.

$$
\bar d = \frac{25.02+24.98+25.01+25.03+24.99+25.00+25.02+25.01}{8} = \frac{200.06}{8} = 25.0075\ \text{mm}
$$

| Reading | $x - \bar d$ (mm) | $(x-\bar d)^2$ ($\times10^{-6}$) |
|---|---|---|
| 25.02 | +0.0125 | 156.25 |
| 24.98 | -0.0275 | 756.25 |
| 25.01 | +0.0025 | 6.25 |
| 25.03 | +0.0225 | 506.25 |
| 24.99 | -0.0175 | 306.25 |
| 25.00 | -0.0075 | 56.25 |
| 25.02 | +0.0125 | 156.25 |
| 25.01 | +0.0025 | 6.25 |

Sum of squares $= 1950\times10^{-6}$ mm².

$$
s = \sqrt{\frac{1950\times10^{-6}}{n-1}} = \sqrt{278.6\times10^{-6}} = 0.0167\ \text{mm}
$$

$$
\sigma_{\bar d} = \frac{s}{\sqrt{n}} = \frac{0.0167}{\sqrt 8} = 0.0059\ \text{mm}
$$

Since the standard error varies as $1/\sqrt n$, halving it needs $n$ four times larger: $4 \times 8 = 32$ readings. Averaging reduces random error but not systematic error.

**Answer: mean = 25.0075 mm, $s$ = 0.0167 mm, standard error = 0.0059 mm; 32 readings for half the standard error.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For V = pi d^2 h / 4, the maximum relative error in V is:
   A. delta d/d + delta h/h
   B. sqrt((delta d/d)^2 + (delta h/h)^2)
   C. 2 delta d/d + delta h/h
   D. delta d/d x delta h/h
   Answer: C. Since d is squared, its relative error counts twice; for the maximum error the terms are added directly. (Key point: Maximum error formula)

2. How is the probable error in V obtained from the individual contributions?
   A. By adding the contributions directly
   B. By taking the root of the sum of squares of the contributions
   C. By multiplying the contributions
   D. By using only the largest contribution
   Answer: B. The probable error uses root-sum-square: sqrt((2 delta d/d)^2 + (delta h/h)^2), which is smaller than the maximum error. (Key point: Probable error)

3. For d = 25.00 ± 0.02 mm and h = 60.00 ± 0.05 mm, V = 29 452 mm^3. What is the maximum error?
   A. About ±53 mm^3
   B. About ±36 mm^3
   C. About ±143 mm^3
   D. About ±72 mm^3
   Answer: D. Relative max error = 0.0016 + 0.000833 = 0.00243, giving 29452 x 0.00243 = about 72 mm^3. (53 mm^3 is the probable error.) (Key point: Result of part (a))

4. For the eight shaft readings, the mean is 25.0075 mm and the sum of squared deviations is 1950 x 10^-6 mm^2. What divisor gives the sample standard deviation?
   A. n - 1 = 7
   B. n = 8
   C. sqrt(8)
   D. n + 1 = 9
   Answer: A. Sample standard deviation divides by n - 1, giving s = sqrt(278.6 x 10^-6) = 0.0167 mm. (Key point: Standard deviation and mean)

5. How many readings are needed to halve the standard error of the mean when 8 were taken?
   A. 16
   B. 32
   C. 64
   D. 12
   Answer: B. Standard error varies as 1/sqrt(n), so halving it needs four times as many readings: 32. (Key point: Standard error of the mean)

## Related in Metrology

- [Thread errors and effect of pitch error](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Ascrew-thread-errors-and-pitch-effect) — Metrology, chapter 12
- [Vernier caliper and micrometer readings](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Avernier-and-micrometer-readings) — Metrology, chapter 4
- [Fringe measurement of flatness error](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Ainterference-fringes-flatness-numerical) — Metrology, chapter 8
- [Limits of a 40 H7/g6 hole-shaft fit](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Afit-40-h7-g6-calculation) — Metrology, chapter 9
- [Sine bar principle, setting and error](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Asine-bar-setting-and-error) — Metrology, chapter 5
- [Effective diameter by three-wire method](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Athree-wire-method-effective-diameter) — Metrology, chapter 12

## Related in other subjects

- [Settling tank 30 m x 3 m, 0.02 mm at 25°C; 0.01 mm at 10°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-as70) — Water Supply Engineering, chapter 6, asked 3×
- [Cylindrical shell with spherical ends: volume change](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acyl-shell-spherical-ends-volume-change) — Strength of Materials, chapter 5
- [Cylindrical vessel 3 m long: change in volume](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acyl-vessel-3m-volume-change) — Strength of Materials, chapter 5
- [Cantilever 5 m: maximum and 25 mm bending stress](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acantilever-5m-8kn-15kn-bending-stresses) — Strength of Materials, chapter 7