FN15–FN16, HN15–HN16 - Accuracy and error intervals

Rounding intervals

Two lengths are reported as $12.4\,\mathrm{cm}$ and $12.40\,\mathrm{cm}$. They have the same numerical value, but do they give the same information about accuracy? The answer depends on the place to which each was rounded.

Suppose a positive length rounds to $12.4\,\mathrm{cm}$ to the nearest tenth of a centimetre. Values closer to $12.4$ than to $12.3$ or $12.5$ produce this result. The halfway boundaries are $12.35$ and $12.45$.

Use this convention throughout the positive rounding examples: a value exactly halfway rounds to the larger neighbouring value. Thus $12.35$ rounds to $12.4$, while $12.45$ rounds to $12.5$.

If $x$ is the length in centimetres, write $12.35\le x<12.45$. Read the chain as two statements: $x$ is at least $12.35$, and $x$ is less than $12.45$. The interval concerns uncertainty from rounding, rather than a mistake in the calculation.

The rounding step is the gap between neighbouring possible rounded answers. To find the boundaries, subtract and add half the step, not half the reported measurement.

Now change only the stated precision. The centre stays the same, but the interval becomes narrower.

The number lines use different scales to make both intervals visible. Their inequalities show the actual widths.

[DIAGRAM: asset_name: Rounded measurement intervals; asset_slug: rounded-measure-intervals; caption: The extra decimal place gives a narrower range of possible lengths.; description: Two number lines show 12.35 less than or equal to x less than 12.45 for 12.4 to one decimal place, and 12.395 less than or equal to x less than 12.405 for 12.40 to two decimal places. Filled circles mark included lower endpoints and open circles mark excluded upper endpoints. The scales differ.]

For significant figures, identify the last retained place value first. A positive mass of $480\,\mathrm g$, correct to two significant figures, is rounded to the nearest $10\,\mathrm g$. Its interval is $475\le m<485$, where $m$ is in grams.

The lower bound is the lower limiting value and the upper bound is the upper limiting value. Here $485$ is the upper bound, but it is not a possible mass in the interval. There is no greatest possible real value below $485$: however close you choose, another allowed value can be closer.

Find the place value before writing both inequality signs.

Truncation intervals

Truncation discards digits instead of choosing the nearest value. Knowing which process was used changes the error interval.

For example, $3.68$ truncated to one decimal place is $3.6$, but rounded to one decimal place it is $3.7$. The corresponding negative value $-3.68$ truncates to $-3.6$, moving to the right on a number line. “Always round down” would therefore be a misleading description.

Now change only the sign. The discarded digits still move the value towards zero, so the included endpoint changes sides.

Zero needs its own check. If a signed number truncates to $0.0$ to one decimal place, it may have come from either side: $0.08$ and $-0.08$ both give zero. Hence the interval is $-0.1<x<0.1$, including $0$ but excluding both outer endpoints. If the original quantity is stated to be non-negative, its allowed interval is instead $0\le x<0.1$.

[DIAGRAM: asset_name: Signed truncation intervals; asset_slug: signed-truncation-intervals; caption: Truncation towards zero gives different endpoint patterns for positive, negative and zero results.; description: Three number lines show 3.6 less than or equal to x less than 3.7, minus 3.7 less than x less than or equal to minus 3.6, and minus 0.1 less than x less than 0.1. Filled circles mean included endpoints and open circles mean excluded endpoints.]

Apply the endpoint check with two decimal places.

Interpreting accuracy

The way a number was obtained tells you whether it has a rounding interval. A count, a measured length and a price set by a shop can have very different meanings even if they contain the same digits.

A complete count of $240$ tickets means exactly $240$, if every ticket was counted. An attendance reported as $240$ to the nearest ten could instead mean any whole-number count from $235$ to $244$. The accuracy statement matters: the final zero alone does not establish how the number was obtained.

Likewise, a fixed price of £$2.50$ is exactly $250$ pence. A length reported as $2.50\,\mathrm m$ to two decimal places is a rounded measurement. Do not automatically attach an interval to every decimal.

Choose reported accuracy that suits the data and the question. Use the stated unit and requested decimal places or significant figures. Avoid adding digits that suggest more precision than the measurements justify.

For example, a ruler reading reported as $6.2\,\mathrm{cm}$ to the nearest $0.1\,\mathrm{cm}$ has interval $6.15\le r<6.25$. Rewriting it as $6.200\,\mathrm{cm}$ correct to three decimal places would claim a much narrower interval without taking a better measurement. Adding zeros can preserve a numerical value; it cannot create new information about the actual length.

Change only how the number was obtained. There is now no rounding step to reverse.

This resolves the opening question. If $12.4\,\mathrm{cm}$ means correct to one decimal place and $12.40\,\mathrm{cm}$ means correct to two, their centres match but the second interval is ten times narrower. A smaller rounding step gives more precise information; the two actual lengths need not be identical.

Compare a known count with a measured value.

Quick Recap

• A positive rounded measurement lies between two halfway boundaries, with the lower included and the upper excluded under the stated convention.
• Truncation removes digits towards zero. Test signed and zero endpoints rather than copying the positive rounding rule.
• Exact counts do not acquire an interval just because they end in zero.
• The stated precision distinguishes the information carried by $12.4$ and $12.40$.