0580 · 1.10Limits of accuracy

Finding bounds from rounded values

0:00A length recorded as eighteen metres might not be exactly eighteen metres. It could be slightly shorter or slightly longer, because the measurement has been rounded. That uncertainty also affects any calculation using the length. In this video, we will find the limits of rounded measurements and explain how those limits affect calculated results. The starting idea is ordinary rounding to the nearest specified unit. Rounding replaces a more precise value with a nearby value at the requested accuracy.

0:27Several different original measurements can therefore produce the same recorded result. Bounds describe the edges of that range, rather than guessing which original measurement was actually taken. Suppose a length is eighteen metres, correct to the nearest metre. A length of seventeen point six metres rounds to eighteen metres. So does eighteen point four metres. The recorded value tells us which rounding interval contains the true length, but it does not identify one exact value.

0:53The boundaries lie halfway between eighteen and its neighbouring whole numbers. Halfway between seventeen and eighteen is seventeen point five. Halfway between eighteen and nineteen is eighteen point five. These are called the lower bound and the upper bound. They mark the edges of the possible values. Using the usual rounding convention, seventeen point five rounds up to eighteen, so the lower endpoint is included. However, eighteen point five rounds up to nineteen, so the upper endpoint is excluded.

1:19The length is greater than or equal to seventeen point five and less than eighteen point five. We still call eighteen point five the upper bound, even though that endpoint is not a possible length. So here the upper bound is not a value that the length can actually take. Here, a length can approach eighteen point five as closely as we like without reaching it. Do not replace the upper bound with eighteen point four, nine, nine.

1:41That would wrongly exclude possible lengths lying even closer to the boundary. For data rounded to the nearest specified unit, each boundary is half that unit away from the recorded value. Subtract half the rounding unit to find the lower bound, and add it to find the upper bound. This works because the boundaries are the halfway points between neighbouring rounded values. The rounding unit matters more than the number of digits displayed.

2:04For the nearest metre, the unit is one metre, so the half-unit is zero point five metres. For the nearest ten metres, the half-unit is five metres. For the nearest hundredth, it is zero point zero zero five.

Bounds with decimals and significant figures

2:16Our first exam-style example gives a mass of three point six kilograms, correct to one decimal place. The rounding unit is zero point one kilograms. Dividing zero point one by two gives a half-unit of zero point zero five kilograms. Subtracting zero point zero five from three point six gives a lower bound of three point five five kilograms. Adding zero point zero five to three point six gives an upper bound of three point six five kilograms.

2:41The mass is greater than or equal to three point five five kilograms. It is also less than three point six five kilograms. These endpoints are exact boundaries, so do not round them back to one decimal place. Significant figures use the same halfway reasoning, but you must identify the place value of the last significant digit. Suppose a distance is two thousand four hundred metres, correct to two significant figures.

3:04The second significant digit is in the hundreds column, so the rounding unit is one hundred metres. Half that unit is fifty metres. The lower bound is two thousand three hundred and fifty metres, and the upper bound is two thousand four hundred and fifty metres.

Bounds in perimeter and area

3:18Calculations with rounded data inherit the uncertainty in their inputs. Using the recorded values alone gives one result, but it does not give the full range. To find a bound, choose the permitted input values that push the result in the required direction. Think about what makes the result larger or smaller before substituting numbers. Our second example is a rectangle with recorded length twelve centimetres and width eight centimetres, both measured to the nearest centimetre.

3:40The half-unit is zero point five centimetres. The length bounds are eleven point five and twelve point five centimetres. The width bounds are seven point five and eight point five centimetres. Perimeter is twice the sum of the length and width. A longer or wider rectangle has a larger sum, so the upper bound of the perimeter uses the largest possible length and width. Decreasing either dimension makes the perimeter smaller.

4:04Therefore, the lower bound uses the smallest possible length and width. For the upper bound, 12.5 + 8.5 = 21 centimetres. Multiplying twenty-one by two gives forty-two centimetres. For the lower bound, 11.5 + 7.5 = 19 centimetres. Multiplying nineteen by two gives thirty-eight centimetres. Area is length multiplied by width, and both dimensions are positive. A longer or wider rectangle has a larger area, so its upper bound uses the largest possible length and width.

4:3612.5 × 8.5 = 106.25 square centimetres. 11.5 × 7.5 gives the lower bound of eighty-six point two five square centimetres. The positive quantities in these examples make addition and multiplication straightforward. Subtraction needs a different choice because subtracting a larger amount makes the result smaller. For the upper bound of a difference, take the largest possible first value and subtract the smallest possible second value. For the lower bound, take the smallest first value and subtract the largest second value.

5:08Division also requires careful reasoning rather than automatically choosing two matching bounds. For positive distance and time, speed equals distance divided by time. A larger distance increases speed, but a larger time decreases it. Therefore, the lower bound of the speed uses the shortest distance divided by the longest time.

Speed bounds and exam technique

5:27Our third exam-style example asks us to find the lower bound of a speed. The distance is one hundred and fifty metres, rounded to the nearest ten metres. The time is twenty seconds, rounded to the nearest second. We first find the bounds of each measurement separately. Half of ten metres is five metres. Subtracting five gives a distance lower bound of one hundred and forty-five metres. Adding five gives an upper bound of one hundred and fifty-five metres.

5:51For time, the half-unit is zero point five seconds, giving bounds of nineteen point five and twenty point five seconds. To make the speed as small as possible, combine the shortest distance boundary with the longest time boundary. The lower speed bound is 145 ÷ 20.5. Multiplying numerator and denominator by two gives the exact fraction 290/41. Its decimal value begins seven point zero seven three one seven metres per second.

6:18If the question requests three significant figures, the answer is approximately seven point zero seven metres per second. Keep the unrounded calculation until that final step. For an upper speed bound, reverse the choices: divide the upper distance bound by the lower time bound. This represents more distance covered in less time. Common mistakes include using half of the recorded value instead of half of the rounding unit. Another is dividing the shortest distance by the shortest time when the lowest speed is wanted.

6:45Those changes act in opposite directions, so that choice does not reliably minimise speed. Avoid both mistakes by identifying the rounding unit and explaining how each input affects the result. In examinations, “write down the upper bound” usually calls for the boundary value, not a long calculation. For “calculate” or “find” questions, show the input bounds and the expression used. These methods must work without a calculator on Paper Two.

7:09Paper Four allows a calculator, but it does not replace the reasoning needed to select the correct bounds. Unless instructed otherwise, answers that are not exact are normally given to three significant figures. Exact bounds should be kept exact, and intermediate values should not be rounded. The central idea is that a rounded measurement represents an interval. Find its halfway boundaries, then choose the combination that makes the requested calculation as large or as small as possible.