Estimating Values and Error Intervals

Welcome back. In the previous lesson, you evaluated exact numerical expressions by following the order of operations. That skill remains important today: estimation still preserves the structure of a calculation, even though the numbers are replaced by easier nearby values.

This lesson develops two connected ideas. First, you will make sensible estimates to judge the likely size of an answer. Second, you will turn a rounded value into an error interval, stating precisely which original values could have produced it. These are useful throughout O Level Mathematics and later when Physics and Chemistry questions involve measured data.

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Estimation is controlled approximation, not guessing

An estimate is a deliberately approximate answer. You round awkward values to nearby values that are easy to calculate mentally, then retain the original operations.

Use the approximation symbol $\approx$, rather than the equals sign, because an estimate is not exact.

For example, estimate

$$\frac{48.7\times 3.92}{0.198}$$

Choose convenient values:

$$48.7\approx 50,3.92\approx 4,0.198\approx 0.2$$

Then retain the multiplication and division:

$$\frac{48.7\times 3.92}{0.198}\approx \frac{50\times 4}{0.2}=\frac{200}{0.2}=1000$$

So the estimated value is

$$1000$$

The estimate does not claim that the exact answer is $1000$. It tells you the answer should be in that general region. If a calculator later gives $96.4$, for example, you immediately know that a keying or method error is likely.

Choosing sensible rounded values

In many exam questions, rounding each number to one significant figure is a reliable starting point. A significant figure begins at the first non-zero digit.

For instance:

$$73.6\approx 70,0.0864\approx 0.09,418\approx 400$$

However, the real aim is not a mechanical “one significant figure” rule. It is to make the arithmetic manageable without making the estimate absurdly rough.

For example:

$$19.8+3.07+0.994$$

can be estimated as

$$20+3+1=24$$

For a product such as

$$12.1\times 4.98$$

the convenient estimate is

$$12\times 5=60$$

When a question includes units, include them in the estimate:

$$14.8\text{ m}\times 6.1\text{ m}\approx 15\text{ m}\times 6\text{ m}=90{\text{ m}}^2$$

The squared unit matters because an area is being estimated.

A short estimation routine

When asked to estimate a calculation:

1. Identify the operations, brackets, and fractions first.
2. Round each value to a nearby convenient number.
3. Carry out the simplified calculation using the same order of operations.
4. State the answer with $\approx$ and the correct unit where needed.
5. Check whether the final size is plausible.

Do not round only some numbers and then treat the result as exact. Also, do not round every value upwards: that tends to create a systematic overestimate. In a sensible estimate, some values may round up and others down.

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From a rounded number to an interval

Estimation asks, “What is a sensible approximate answer?” An error interval asks a more precise question:

If a reported value has been rounded, what original values could it represent?

Suppose a distance is recorded as $30\text{ m}$, correct to the nearest $10\text{ m}$.

The neighbouring multiples of ten are $20$, $30$, and $40$. The halfway points are $25$ and $35$.

• $25$ rounds to $30$, so it is included.
• Any value below $35$, such as $34.999$, rounds to $30$.
• $35$ rounds to $40$, so it is not included.

Therefore, if the true distance is $d$,

$$25\text{ m}\le d<35\text{ m}$$

The lower bound is $25\text{ m}$: the smallest possible value.
The upper bound is $35\text{ m}$: the boundary above which the rounded value changes.

This is why ordinary rounding intervals nearly always have the form

$$\text{lower bound}\le x<\text{upper bound}$$

The left inequality includes equality; the right inequality does not.

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The half-the-rounding-unit rule

There is a general method behind every ordinary rounding interval.

If a value $r$ is rounded to the nearest unit $u$, the uncertainty on either side is half that unit:

$$\frac{u}{2}$$

So the interval is

$$r-\frac{u}{2}\le x<r+\frac{u}{2}$$

The important step is identifying $u$, the rounding unit.

Stated accuracy	Rounding unit $u$	Half-unit
nearest integer	$1$	$0.5$
nearest $10$	$10$	$5$
nearest $0.1$	$0.1$	$0.05$
nearest $0.01$	$0.01$	$0.005$

Consider a mass of $8.4\text{ kg}$, correct to one decimal place. One decimal place means the nearest $0.1\text{ kg}$, so the half-unit is $0.05\text{ kg}$.

$$8.4-0.05=8.35$$ $$8.4+0.05=8.45$$

Thus, for true mass $m$,

$$8.35\text{ kg}\le m<8.45\text{ kg}$$

Notice that the bounds have more decimal places than the stated measurement. This is normal: the boundaries lie halfway between the recorded values.

A time of $15.89\text{ s}$, correct to the nearest $0.01\text{ s}$, similarly has half-unit $0.005\text{ s}$:

$$15.885\text{ s}\le t<15.895\text{ s}$$

Keep trailing zeroes when they communicate the stated precision. For example, $7.40\text{ cm}$ is different in meaning from $7.4\text{ cm}$ if the question says the former was recorded to two decimal places.

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Significant figures: identify the place value first

Bounds involving significant figures can look less familiar because the rounding unit depends on the size of the number.

Suppose a length is $5300\text{ m}$, correct to two significant figures. The two significant digits are $5$ and $3$. The second significant digit, $3$, is in the hundreds place. Therefore, the value has been rounded to the nearest $100\text{ m}$.

The half-unit is

$$\frac{100}{2}=50$$

So:

$$5300-50=5250$$ $$5300+50=5350$$

If $l$ is the true length,

$$5250\text{ m}\le l<5350\text{ m}$$

Here is another useful example. A value is $0.047$, correct to two significant figures. The significant digits are $4$ and $7$, and the $7$ is in the thousandths place. Therefore the rounding unit is $0.001$, not $0.01$.

Its half-unit is

$$0.0005$$

The error interval is therefore

$$0.0465\le x<0.0475$$

A dependable method for significant-figure bounds is:

1. Locate the last stated significant digit.
2. Find its place value.
3. Halve that place value.
4. Subtract it for the lower bound and add it for the upper bound.
5. Use $\le$ at the lower boundary and $<$ at the upper boundary.

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Common errors and how to prevent them

Using the full rounding unit instead of half

If $62\text{ kg}$ is correct to the nearest kilogram, the true value is not between $61$ and $63$. The boundaries are halfway to those neighbouring whole-number values:

$$61.5\text{ kg}\le m<62.5\text{ kg}$$

Including the upper boundary

For a value rounded by the usual rule, the upper boundary belongs to the next rounded result. Thus:

$$34.5\le x<35.5$$

is correct for $35$ to the nearest integer, but

$$34.5\le x\le 35.5$$

is not.

Ignoring the stated degree of accuracy

The values $4.8$, $4.80$, and $4.800$ may represent different precisions when a question explicitly states decimal places. Always read the phrase after the number: “nearest centimetre,” “one decimal place,” “two significant figures,” and so on.

Mixing up an estimate and an error interval

These ideas both involve approximation, but they answer different questions.

Idea	Given	What you produce
Estimate	A calculation with inconvenient values	A manageable approximate answer
Error interval	A rounded measurement or number	The full range of possible original values

For example, estimating a journey time might give approximately $50\text{ min}$. If a stopwatch reading is $50\text{ min}$ correct to the nearest minute, its interval is instead

$$49.5\text{ min}\le t<50.5\text{ min}$$

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Exam presentation

For an estimate, show the substitutions clearly:

$$\frac{48.7\times 3.92}{0.198}\approx \frac{50\times 4}{0.2}=1000$$

For bounds, make the accuracy and half-unit visible:

$$\text{nearest }0.1\Rightarrow \text{half-unit }0.05$$ $$8.35\le x<8.45$$

This makes your reasoning easy to check and reduces sign errors in the inequality.

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Key takeaways

Estimation replaces awkward values with sensible nearby numbers while preserving the calculation’s structure and units. It gives a reasonableness check, not an exact result.

For a normally rounded value, find half the rounding unit. Subtract it to obtain the included lower bound and add it to obtain the excluded upper bound:

$$r-\frac{u}{2}\le x<r+\frac{u}{2}$$

For significant figures, first identify the place value of the final significant digit. In the next lesson, you will shift from numerical approximation to Chemistry and use kinetic particle theory to explain diffusion and changes of state.