Errors — absolute, relative, percentage; systematic vs random: Worked examples

Before anything, three plain-word reminders so no symbol is unearned:

Recall The three quantities in one breath

Absolute error $Δ a$ — how far off, in the same units (cm, g, s).
Relative error $δ_{r} = Δ a / a$ — the miss as a fraction of the thing, no units.
Percentage error $δ_{%} = δ_{r} \times 100%$ — the same fraction spoken as a percent.

" $\pm$ " (read "plus or minus") means "the true value lives somewhere in this band." $\sum$ (a big Greek S) just means "add all of these up." $\overset{a}{ˉ}$ or $a_{mean}$ is the average.

The scenario matrix

Every error problem you will meet falls into one of these cells. Each row is a distinct kind of question; the examples below are labelled with the cell they belong to.

Cell	The situation	The trap it tests
C1	Full pipeline: mean → abs → rel → %	Rounding the error to the value's decimals
C2	Systematic (zero) error present	Averaging can't remove it; you subtract it
C3	Both flavours in one dataset	Separating scatter from bias
C4	Degenerate input: all readings identical	Zero error ≠ "no uncertainty" (least count)
C5	Tiny quantity vs huge quantity	Same absolute error, wildly different %
C6	Limiting case: value $\to$ 0	Relative error blows up
C7	Real-world word problem	Translate words → numbers
C8	Exam twist: given %, work backwards	Reverse the pipeline

We will hit all eight cells with 8 worked examples. Two of them get figures because geometry makes the idea click.

C1 — The full pipeline

Worked example C1 · A length measured five times

A rod is measured (cm): $12.34, 12.30, 12.36, 12.31, 12.29$ . Report $L = L_{mean} \pm Δ L$ , and the percentage error.

Forecast: the readings spread over about $0.07$ cm. Guess: will the % error be near $0.2%$ or near $2%$ ? (Write your guess before reading on.)

Step 1 — Mean. Why this step? The average is our single best estimate; random highs and lows partly cancel. $L_{mean} = \frac{12.34 + 12.30 + 12.36 + 12.31 + 12.29}{5} = \frac{61.60}{5} = 12.32 cm .$

Step 2 — Absolute error of each reading. Why? Each $Δ L_{i} = ∣ L_{mean} - L_{i} ∣$ is how far that reading missed. $0.02, 0.02, 0.04, 0.01, 0.03 cm .$

Step 3 — Mean absolute error. Why? The typical miss is what we quote as uncertainty. $Δ L = \frac{0.02 + 0.02 + 0.04 + 0.01 + 0.03}{5} = \frac{0.12}{5} = 0.024 cm .$

Step 4 — Round the error, then match the value. Why? The error is itself fuzzy, so keep it to 1 significant figure: $Δ L \approx 0.02$ cm. Round the value to the same decimal place. $L = 12.32 \pm 0.02 cm$

Step 5 — Percentage. $δ_{%} = \frac{0.024}{12.32} \times 100% = 0.195% \approx 0.2%$ .

Verify: $12.32$ sits between the min ( $12.29$ ) and max ( $12.36$ ) — a mean must. And $0.2%$ matches the forecast band (readings agree to two decimals, so a fraction-of-a-percent error is expected). ✓

C2 — Systematic zero error (a figure earns its place)

Worked example C2 · A vernier caliper that never read zero

A vernier caliper shows $0.05$ cm when its jaws are fully closed (nothing between them). You then measure a ball bearing four times: $2.55, 2.56, 2.55, 2.54$ cm. Give the true diameter.

Forecast: if you just averaged and reported, would you be too big or too small — and by how much?

Step 1 — Identify the flavour. Why? A fixed reading with nothing in the jaws is a zero error: the same $+ 0.05$ cm is baked into every reading. That is a systematic error — averaging can never remove it.

Step 2 — Mean of the raw readings. Why? First get the best estimate of what the instrument reported. $d_{raw} = \frac{2.55 + 2.56 + 2.55 + 2.54}{4} = \frac{10.20}{4} = 2.55 cm .$

Step 3 — Subtract the known offset. Why? Once a systematic error is identified and measured, you correct it exactly. Look at the figure: the blue "true" scale is shifted from the red "instrument" scale by a fixed $0.05$ cm everywhere. $d_{true} = 2.55 - 0.05 = 2.50 cm .$

Step 4 — Random error survives, and we still report it. Absolute errors from the mean: $0.00, 0.01, 0.00, 0.01$ ; mean abs $= 0.005 \approx 0.01$ cm. $d = 2.50 \pm 0.01 cm$

Verify: the correction shifts every reading down by exactly the same amount, so the scatter (and thus the $\pm 0.01$ ) is unchanged — correcting a bias never changes random spread. ✓

C3 — Both flavours living in one dataset

Worked example C3 · A stopwatch that lags and a shaky thumb

A stopwatch is known to start $0.10$ s late every time (bias). Five timings of a fixed interval: $8.4, 8.6, 8.3, 8.7, 8.5$ s. Find the corrected best value and its random uncertainty.

Forecast: which number here is systematic and which is the random $\pm$ ? Name them before computing.

Step 1 — Split the two effects. Why? They are fought differently: subtract the bias, average away the scatter.

The $- 0.10$ s always-late start is systematic (every reading reads $0.10$ s too short).
The spread across $8.3$ – $8.7$ is random.

Step 2 — Mean of raw readings. $t_{raw} = \frac{8.4 + 8.6 + 8.3 + 8.7 + 8.5}{5} = \frac{42.5}{5} = 8.50 s .$

Step 3 — Correct the bias. Why? The clock starts late, so it undercounts; the true interval is longer. Add $0.10$ s. $t_{true} = 8.50 + 0.10 = 8.60 s .$

Step 4 — Random uncertainty. Abs errors from $8.50$ : $0.10, 0.10, 0.20, 0.20, 0.00$ ; mean abs $= 0.60/5 = 0.12$ s. $t = 8.60 \pm 0.12 s$

Verify: the bias moved the centre ( $8.50 \to 8.60$ ) but left the width ( $\pm 0.12$ ) untouched — exactly what a systematic shift does. Units are seconds throughout. ✓

C4 — Degenerate input: all readings identical

Worked example C4 · When your readings agree perfectly

A length is read three times with a ruler of least count $0.1$ cm and comes out $7.5, 7.5, 7.5$ cm every time. Is the uncertainty zero?

Forecast: the mean-absolute-error formula will give $0$ . Is the honest error therefore $0$ ? Guess yes/no.

Step 1 — Apply the formula. Why? To see what it says on its own. Mean $= 7.5$ ; every $∣7.5 - 7.5∣ = 0$ , so $Δ a_{mean} = 0$ .

Step 2 — Reality check. Why? Zero scatter does not mean zero uncertainty. You still cannot read between the ruler's smallest marks. The instrument's least count sets a floor.

Step 3 — Use the least-count rule. Why? When statistical scatter under-reports the truth, the uncertainty is at least the least count (often taken as half of it, but the safe exam convention is the full least count). $L = 7.5 \pm 0.1 cm$

Verify: the reported $\pm 0.1$ equals the least count — it can never legitimately fall below what the instrument can resolve. ✓ (See Least Count and Vernier Calipers.)

C5 — Same absolute error, opposite worlds (a figure)

Worked example C5 · One millimetre on a hair and on a road

A caliper misses by the same absolute error $Δ = 0.001$ m in two jobs: (a) a hair of diameter $0.002$ m; (b) a road of length $2.000$ m. Compare the percentage errors.

Forecast: the same $0.001$ m — will the two % errors be equal, or wildly different?

Step 1 — Percentage for the hair. Why divide by the value? Relative error asks "how big is the miss compared to the thing?" $δ_{%}^{hair} = \frac{0.001}{0.002} \times 100% = 50%.$

Step 2 — Percentage for the road. $δ_{%}^{road} = \frac{0.001}{2.000} \times 100% = 0.05%.$

Step 3 — Read the picture. Why? The green bar (road) barely notices a $1$ mm error; the red bar (hair) is half wrong. Identical absolute error → $1000 \times$ difference in percentage.

Verify: ratio of percentages $= 50/0.05 = 1000$ , which equals the ratio of the two lengths $2.000/0.002 = 1000$ . Consistent. ✓

C6 — Limiting case: the value heads toward zero

Worked example C6 · Weighing something almost weightless

A balance has a fixed random uncertainty of $Δ m = 0.01$ g. Compute the percentage error as the measured mass shrinks: $m = 10.00$ g, then $0.10$ g, then $0.02$ g.

Forecast: as $m \to 0$ with $Δ m$ fixed, does $δ_{%}$ settle down, or run away to infinity?

Step 1 — Three percentages. Why? To watch the trend. $δ_{%} = \frac{0.01}{m} \times 100%$ . $m = 10.00 : 0.1%, m = 0.10 : 10%, m = 0.02 : 50%.$

Step 2 — Interpret the limit. Why? Because $m$ sits in the denominator, shrinking $m$ with $Δ m$ fixed sends $δ_{%} \to \infty$ . Small quantities are fractionally hard to measure even when the absolute error is tiny.

Step 3 — The practical lesson. Why? To pin down a small quantity to good percentage, you need a finer instrument (smaller $Δ m$ ) — or measure many together and divide (measure 100 grains, get one grain's mass with $1/100$ the error).

Verify: $0.01/0.02 = 0.5 = 50%$ ; the sequence $0.1%, 10%, 50%$ strictly increases as $m$ falls — matches " $δ_{%} \to \infty$ as $m \to 0$ ." ✓

C7 — Real-world word problem

Worked example C7 · Fuel per 100 km

A car's trip meter reads a distance $d = 240 \pm 2$ km and the fuel gauge $V = 16.0 \pm 0.5$ L. What is the percentage error in the fuel-per-distance figure $V / d$ ?

Forecast: which of the two inputs, $d$ or $V$ , contributes more error — the big number or the small one?

Step 1 — Percentage error of each input. Why? For a quotient, percentage errors add (see Combination of Errors). $δ_{%} (d) = \frac{2}{240} \times 100% = 0.83%, δ_{%} (V) = \frac{0.5}{16.0} \times 100% = 3.13%.$

Step 2 — Combine for the ratio $V / d$ . Why add? In a product or quotient, fractional errors combine by addition; that is the rule for how uncertainties propagate. $δ_{%} (V / d) = 0.83% + 3.13% = 3.96% \approx 4.0%.$

Verify: the fuel measurement dominates ( $3.13%$ vs $0.83%$ ) — matches the forecast that the smaller-precision input (fuel) rules. The final $4.0%$ is bigger than either input alone, as adding positives requires. ✓

C8 — Exam twist: work backwards from a percentage

Recall Self-test before you leave

Which cell is each of these?

"All five readings came out $3.20$ mm." ::: C4 (degenerate → least count sets the floor).
"The ammeter reads $0.2$ A with the circuit open." ::: C2 (systematic zero error → subtract).
"Error in area $=$ sum of % errors in two sides." ::: C7-style (combination, percentages add).
"Same $\pm 0.1$ mm on a $0.5$ mm wire vs a $500$ mm rod." ::: C5 (fractional error differs enormously).