Bounds of Error

When we say 8 cm “to the nearest centimetre,” the true length could be anything that rounds to 8. The precision unit is 1 cm, and the maximum error is half of that — 0.5 cm — in either direction. So the error interval is \(7.5 \leq \text{length} < 8.5\).

A length of 7.2 cm would round to 7, not 8. The bounds must be tighter: half a unit, not a whole unit. As shown on the number line, 7.5 is the exact lower boundary and is included.

To round 3.55 to 1 decimal place, we look at the second decimal digit: 5. Following the “round half up” convention, 3.55 rounds up to 3.6, not down to 3.5. This is why the error interval for 3.5 (1 d.p.) is \(3.45 \leq x < 3.55\) — the upper bound of 3.55 is excluded using a strict inequality (<).

This matters because the upper bound marks the exact point where a value would tip into the next rounded value. If the upper bound were included, 3.55 would belong to two error intervals simultaneously (3.5’s and 3.6’s), which would make the system ambiguous.

When 400 is given to 1 significant figure, only the 4 is significant — it tells us “roughly 4 hundreds.” The precision unit is 100 (the place value of the significant figure), and the maximum error is half of that: 50. So the bounds are 400 ± 50, giving \(350 \leq x < 450\).

The interval \(399.5 \leq x < 400.5\) would be correct if 400 were given to the nearest integer (precision unit = 1). The key question is always “what was the degree of accuracy?” A value like 370 clearly rounds to 400 when given to 1 s.f., yet 370 falls outside 399.5 ≤ x < 400.5. It does fall within 350 ≤ x < 450, confirming the wider interval is correct.

Truncation means chopping off digits beyond a certain point without rounding. If a number is truncated to 1 decimal place and the result is 3.2, the original number must have started with 3.2… — for example, 3.21, 3.25, or 3.29 all truncate to 3.2. So the error interval for truncation is \(3.2 \leq x < 3.3\).

A value of 3.1 could never truncate to 3.2 — truncating 3.1 gives 3.1 (nothing to chop off), and truncating 3.19 gives 3.1 (not 3.2). This is different from rounding, where the error interval for 3.2 (1 d.p.) would be \(3.15 \leq x < 3.25\) and would not include 3.29. Truncation shifts the entire interval: it starts at the stated value and extends one precision unit above it.

Example: 5.1

Another: 4.8

Creative: 4.5 — exactly the lower bound. It IS included because the lower inequality uses ≤. Many students assume boundary values are excluded, but only the upper bound is.

Trap: 5.5 — it looks like it belongs because 5.5 appears in the interval notation, but the strict inequality (<) means 5.5 is excluded. A value of 5.5 would round to 6, not 5, confirming it shouldn’t be in the interval.

Example: 50 (to 1 s.f.) → \(45 \leq x < 55\), width = 10

Another: 200 (to 1 s.f.) → \(150 \leq x < 250\), width = 100

Creative: 3000 (to 1 s.f.) → \(2500 \leq x < 3500\), width = 1000. The larger the number, the wider the interval at 1 s.f.

Trap: 8 (to 1 s.f.) → \(7.5 \leq x < 8.5\), width = 1. A student might choose any single-digit number thinking “1 significant figure must mean a big error interval,” but for single-digit numbers the width is exactly 1, not greater than 1.

Example: 7.3 — within \(6.5 \leq x < 7.5\) ✓ but outside \(6.95 \leq x < 7.05\) ✓

Another: 6.6 — within \(6.5 \leq x < 7.5\) ✓ but outside \(6.95 \leq x < 7.05\) ✓

Creative: 7.4999 — barely inside the integer interval, as far from 7.0’s interval as possible while still qualifying.

Trap: 7.02 — a student might think “this isn’t 7.0 so it can’t be in the 7.0 interval.” But 7.02 IS within \(6.95 \leq x < 7.05\), so it lies in both intervals. It fails the “but NOT” condition.

Example: 3 − 3 → minimum = 2.5 − 3.5 = −1

Another: 1 − 1 → minimum = 0.5 − 1.5 = −1

Creative: 10 − 10 → minimum = 9.5 − 10.5 = −1. Use the formula: \(\text{Lower Bound} (A – B) = \text{Lower Bound}(A) – \text{Upper Bound}(B)\).

Trap: 10 − 5 → minimum = 9.5 − 5.5 = 4, which is still positive. For the minimum to be negative, the two rounded values must be equal (or the first must be less than the second).

Example: \(10 \div 2\). The Max answer is \(10.5 \div 1.5 = 7\).

Explanation: To make a fraction as large as possible, you want the numerator to be as big as it can be (Upper Bound) and the denominator to be as small as it can be (Lower Bound).

Trap: \(10 \times 2\). For multiplication, the Upper Bound is \(UB \times UB\), not \(UB \times LB\). Students often confuse the rules for division and multiplication bounds.

This is true for decimal places and “nearest” rounding: every number rounded to 1 d.p. has an error interval with width 0.1, every number rounded to the nearest 10 has width 10, and so on. The number doesn’t matter.

But it fails for significant figures. With s.f., the width depends on the column value of the last significant digit. 50 (to 1 s.f.) has width 10 because the significant digit is in the tens column. 500 (to 1 s.f.) has width 100 because the digit is in the hundreds column.

The error interval always uses a strict inequality (<) for the upper bound. If a value equalled the upper bound, it would round up to the next value, not to the stated one. For example, 5.45 rounds to 5.5 (1 d.p.), not to 5.4, so 5.45 is not part of the error interval for 5.4.

This is why the notation is always \(\text{lower} \leq x < \text{upper}\), not \(\text{lower} \leq x \leq \text{upper}\). The lower bound IS included (a value exactly at the lower bound rounds up to the stated value), but the upper bound is always excluded.

For rounding, the bounds are always symmetric about the rounded value: the value ± half the precision unit. So the rounded value sits exactly at the centre. For example: 5.3 (1 d.p.) has bounds 5.25 and 5.35, midpoint = (5.25 + 5.35) ÷ 2 = 5.3. Or 400 (nearest 100) has bounds 350 and 450, midpoint = (350 + 450) ÷ 2 = 400.

This is a useful property — it means you can always find the bounds by going half a precision unit either side. Note that this applies specifically to rounding; for truncation, the stated value is the lower bound, not the midpoint.

Truncation removes digits without rounding up, so the result is either smaller than or equal to the original — never larger. But it doesn’t always make it strictly smaller.

TRUE case: 3.74 truncated to 1 d.p. gives 3.7, and 3.7 < 3.74. FALSE case: 3.40 truncated to 1 d.p. gives 3.4, and 3.4 = 3.40. The value hasn’t become smaller — it’s stayed the same. Truncation only reduces the value when the removed digits contain non-zero values.

The student made two errors. First, the bounds should be half a unit either side, not a whole unit — the error interval should go from 2.5 to 3.5, not from 2 to 4. A parcel weighing 2.3 kg would round to 2, not 3, so 2.3 can’t be in the interval. This is the “whole-unit bounds” misconception.

Second, the upper bound should use a strict inequality (<), not ≤. A weight of exactly 3.5 kg would round up to 4, not stay at 3. The correct error interval is \(2.5 \leq \text{weight} < 3.5\).

The answer is correct — 3.65 is NOT within the error interval — but the reasoning is wrong. The number of decimal places doesn’t determine if a value fits. The value 3.65 is excluded because it sits exactly at the upper bound. Imagine zooming in on the number line:

Values like 3.649 are closer to 3.6. At 3.65 exactly, the ruler tips over to 3.7.

The student is applying a “±0.5 rule” regardless of the degree of accuracy. This only works when the precision unit is 1. For 500 to 2 significant figures, the last significant digit (the second 0) is in the tens column, so the precision unit is 10. The error is ±5.

The correct interval is \(495 \leq x < 505\). To see why the student’s answer must be wrong: a value like 497 clearly rounds to 500 when given to 2 significant figures, yet 497 falls outside the student’s interval of \(499.5 \leq x < 500.5\).

The student is confusing truncation with rounding. Truncation means chopping off digits beyond a certain point — it does not round up, ever. If you truncate 2.38 to 1 d.p., you get 2.3. If you truncate 2.34 to 1 d.p., you also get 2.3. But 2.29 truncated to 1 d.p. gives 2.2, not 2.3.

The correct error interval for truncation is \(2.3 \leq x < 2.4\). The lower bound is the stated value itself (not half a unit below it), and the upper bound is one full precision unit above. Notice this is an asymmetric interval — the error is entirely above the stated value.

The student has forgotten that people are discrete data (you can’t have half a person). While the upper bound of the continuous interval is indeed 850 (because anything less than 850 rounds to 800), you cannot have 849.5 people.

For discrete data, the maximum integer value is the largest whole number that rounds to 800, which is 849. If there were 850 people, it would round up to 900.