Errors — absolute, relative, percentage; systematic vs random: Visual walkthrough

We will use one running story: you timed a swinging pendulum five times and wrote down five numbers of seconds.

$2.63, 2.56, 2.42, 2.71, 2.80$

Step 1 — Lay the raw readings on a number line

WHY do this first? Because error analysis is entirely about how spread out these dots are and where their centre sits. You cannot reason about spread until you can see it. The picture below is the whole problem, drawn honestly.

PICTURE: the five red dots are scattered. Some sit low (the $2.42$ ), some high (the $2.80$ ). None of them is "the truth" — the true period is a hidden point we are trying to guess.

Figure — Errors — absolute, relative, percentage; systematic vs random

Step 2 — Find the balance point: the mean

WHY the average and not, say, the biggest or the middle one? Random wobbles push readings above and below the true value roughly equally. If highs and lows are balanced, the point that sits in the middle of the tug-of-war is the least-biased single guess. Summing all readings and sharing equally ("divide by how many") is exactly that balance point.

Look at each symbol as it appears:

$a_{mean} = share equally among n \frac{1}{n} add up all readings a_{1} \dots a_{n} i = 1 \sum n a_{i}$

$n$ ::: how many readings you took. Here $n = 5$ .
$i = 1 \sum n a_{i}$ ::: the big "S" (Greek sigma) is a shorthand for "add them all": start at $i = 1$ , stop at $i = n$ , and total every $a_{i}$ . Here it is $2.63 + 2.56 + 2.42 + 2.71 + 2.80 = 13.12$ .
$\frac{1}{n}$ ::: split that total into $n$ equal shares — that is what "average" means.

$a_{mean} = \frac{13.12}{5} = 2.624 s$

PICTURE: the seesaw balances exactly under $2.624$ . That black triangle is our best single guess.

Recall Why is the balance point the average? (peek)

Question ::: Why does the seesaw balance at the arithmetic mean and not somewhere else? Answer ::: Because the mean is the one point where the total "pull" of dots on its left exactly equals the total pull of dots on its right — the sum of signed distances to it is zero. That is the definition of a balance point.

Step 3 — Measure how far each dot missed the balance point

WHY the word "absolute"? A reading can sit to the left (too small, a negative gap) or to the right (too big, a positive gap). We only care how far off, not which side — a miss of 6 thousandths is a miss of 6 thousandths whether it landed left or right. So we throw away the sign. The two vertical bars $∣ ∣$ mean exactly that: "keep the size, drop the minus."

$Δ a_{i} = balance point a_{mean} - this reading a_{i}$

$Δ$ ::: the Greek letter delta, used everywhere in physics to mean "a small change / gap / amount of." Read $Δ a_{i}$ as "the gap of reading $i$ ."
$a_{mean} - a_{i}$ ::: the raw gap, which may be $+$ or $-$ .
$∣ \dots ∣$ ::: strip the sign, keep the length.

Computing all five (using the un-rounded $2.624$ ):

$∣2.624 - 2.63∣ = 0.006, 0.064, 0.204, 0.086, 0.176$

PICTURE: each red bracket spans from a dot to the black balance line. The longest bracket ( $0.204$ , the $2.42$ reading) is our worst miss; the shortest ( $0.006$ ) is nearly a bullseye.

Step 4 — Average the misses: the mean absolute error

WHY average the misses? One miss could be a fluke. The typical miss tells you honestly how much any single reading tends to wander. That typical wander is precisely what "our uncertainty" should mean.

$Δ a_{mean} = share equally \frac{1}{n} add up all the misses i = 1 \sum n Δ a_{i}$

Same machinery as Step 2 — sigma adds, $\frac{1}{n}$ shares — but now it eats the misses, not the readings.

$Δ a_{mean} = \frac{0.006 + 0.064 + 0.204 + 0.086 + 0.176}{5} = \frac{0.536}{5} = 0.1072 \approx 0.11 s$

Intuition Why this spread measure and not the standard deviation?

There are three common ways to say "how spread out are the readings," and it is worth knowing you are choosing one:

Mean absolute deviation $Δ a_{mean} = \frac{1}{n} \sum ∣Δ a_{i} ∣$ — average the misses directly. Simplest to compute by hand and to explain; this is the school/JEE convention and the one the parent note uses.
Standard deviation $σ = \frac{1}{n} \sum (Δ a_{i})^{2}$ — average the squared misses, then square-root. It weights big misses more heavily and is what statistics prefers. See Mean and Standard Deviation.
Standard error of the mean $σ / n$ — how uncertain the mean itself is (it shrinks as you take more readings).

We deliberately use the MAD here because this walkthrough matches the parent note and needs no squares or square-roots — every step stays a picture a 12-year-old can draw. For the same data $σ \approx 0.13$ s, a little larger; the conclusion ("about a tenth of a second of wobble") is the same either way.

PICTURE: all five bracket-lengths are stacked as bars; the dashed red line is their average height — the one number $0.1072$ that stands in for "how far off, typically."

Step 5 — Round once, then draw the honesty fence

WHY round only here, and to matching decimals? Because rounding is lossy — every round throws away a sliver of information. Round early and the slivers pile up through Steps 3–4 (this is rounding-error propagation). Round once, at the last moment, and there is nothing downstream to corrupt. Matching decimals stops you from claiming a value more precise than its own error bar allows — writing $2.624 \pm 0.11$ pretends you know the third decimal that the $\pm 0.11$ says you do not.

WHY a band, not a single number? Because claiming one exact number would be a lie — you don't know the true value to infinite precision. The band is the honest statement: "somewhere in here." Notice every raw dot except the far-out $2.42$ falls inside; the band's job is to typically contain readings, not all of them.

PICTURE: the red band brackets the balance point. This picture is the whole point of the chapter made visible.

Step 6 — Make it comparable: relative and percentage error

WHY divide? Division cancels the units (seconds ÷ seconds = a pure number) and rescales every measurement to a common ruler: fraction of itself. Now a timing error and a mass error can be compared head-to-head — something absolute errors, stuck in different units, can never do.

$δ_{r} = \frac{Δ a _{mean} the miss (s)}{the size (s) a _{mean}} = \frac{0.11}{2.62} = 0.042 (a pure number, no units)$

$δ$ ::: Greek delta (lower-case), a conventional label for "error of." The subscript is a label, not a variable — it names which kind of error.
subscript $r$ in $δ_{r}$ ::: stands for "relative" (fractional).
subscript $%$ in $δ_{%}$ (below) ::: a mnemonic label reading literally "the error expressed as a percentage." Some books write $δ_{p}$ or just append " $%$ " to the number; the symbol is only a name, so read whichever you meet as "percentage error." We use $δ_{%}$ to match the parent note.

Dress it as a percent by multiplying by 100 — because humans read " $4.2%$ " faster than " $0.042$ ":

$δ_{%} = 0.042 \times 100% = 4.2%$

PICTURE: a bar showing the tiny red $\approx 4%$ slice against the whole $100%$ of the measured value — a fraction you can see.

Step 7 — The edge cases you must never trip on

Case A — all readings identical. Say you measured $2.62, 2.62, 2.62$ . Every gap $Δ a_{i} = 0$ , so $Δ a_{mean} = 0$ . Does this mean zero error? No. It means your random scatter is below your instrument's resolution. The real floor is then the least count of your device — see Least Count and Vernier Calipers. A stopwatch reading "0 spread" still can't beat its own tick size.

Case B — the balance point sits at zero. Relative error $δ_{r} = Δ a_{mean} / a_{mean}$ blows up if $a_{mean} = 0$ : you would be dividing by nothing. Physically this warns you that "relative error" is meaningless for a quantity whose best estimate is zero — a $\pm 0.1$ around $0$ is not " $\infty%$ off," it just means the true value could be either sign.

Case C — a systematic offset hides inside the mean. If your stopwatch always starts $0.3$ s late, every reading is $+ 0.3$ too big. Averaging (Step 2) keeps that $+ 0.3$ — the balance point is shifted, the band is centred on the wrong place. The pictures above assumed pure random scatter. Systematic error is not in these formulas at all; it must be found and subtracted separately. This is exactly why precise ≠ accurate.

PICTURE: two number lines. Top = random-only (band straddles the true value, good). Bottom = systematic shift (tight band, but the whole band is slid off the true value — precisely wrong).

The one-picture summary

Recall Feynman retelling — explain the whole walk to a 12-year-old

You timed a swing five times and got five slightly different numbers, because your thumb on the stopwatch isn't magic. First you find the middle of those five — the seesaw balance point — and call it your best guess ( $2.624$ s, and you keep that extra digit for now so you don't spoil the next steps). Then for each try you ask "how far did I miss the middle?" and ignore whether you were early or late — just the size of the miss. You average those misses to get "how far off I usually am" ( $0.107$ s). Only at the end do you tidy up: the wobble rounds to $0.11$ , so the answer rounds to match — "about 2.62 seconds, give or take 0.11." That "give or take," compared to the 2.62 itself, is about 4 out of 100 — a 4% wobble, small. Two warnings: don't round in the middle or the little chops pile up; and if your stopwatch always lags by the same amount, taking more tries won't rescue you — every number is shifted the same way, so the middle is shifted too. Averaging fixes shaky-hand mistakes, never the-ruler-is-broken mistakes.

Recall Quick self-test

The whole pipeline in order ::: mean (keep guard digits) → absolute error of each → mean absolute deviation → round once → write $a_{mean} \pm Δ a_{mean}$ → divide for relative → ×100 for percent. Why strip the sign in $Δ a_{i}$ ? ::: We want the size of the miss, not its direction; left-misses and right-misses are equally bad. Why round only at the very end? ::: Rounding is lossy; rounding early lets small chops propagate through every later step and corrupt the final band. Why use mean absolute deviation instead of standard deviation here? ::: It needs no squares or square-roots, so every step stays a simple picture; it matches the parent-note convention. Standard deviation is the statistically preferred alternative. If all readings are identical, is the error zero? ::: No — random scatter is zero, but the least count and any systematic offset still limit you. Why can't relative error be computed when $a_{mean} = 0$ ? ::: You would divide by zero; "fraction of the value" is undefined for a value of zero.

See also: Mean and Standard Deviation · Combination of Errors · Significant Figures and Rounding · Dimensional Analysis.