Error intervals and bounds
Error intervals for rounding
A rounded number could have been up to half a unit either side. The gives every value it could have been.
- Find the unit it was rounded to (for example 0.1 or 100).
- Halve it.
- = rounded value − half a unit. = rounded value + half a unit.
- Write lower bound ≤ x < upper bound.
Example: a length l is 7.4 cm to 1 decimal place. Write its error interval.
- The unit is 0.1, and half of 0.1 is 0.05.
- 7.4 − 0.05 = 7.35 and 7.4 + 0.05 = 7.45
- 7.35 ≤ l < 7.45
Example: n = 3200 to the nearest 100.
- Half of 100 is 50.
- 3150 ≤ n < 3250
Common mistake
Use < for the upper bound, not ≤. 7.45 itself would round up to 7.5.
Key terms
Error intervals for truncation
To a number, you chop off the digits after a certain place without rounding. 7.48 truncated to 1 decimal place is 7.4.
A truncated number could have been anything from itself up to (but not including) one whole unit more.
- Lower bound = the truncated value.
- Upper bound = the truncated value + 1 unit.
- Write lower bound ≤ x < upper bound.
Example: x is 12 when truncated to a whole number. Write its error interval.
- Lower bound: 12
- Upper bound: 12 + 1 = 13
- 12 ≤ x < 13
Compare rounding: if x were 12 to the nearest whole number, the interval would be 11.5 ≤ x < 12.5.
Key terms
Bounds in calculations (Higher tier only)
When you calculate with rounded measurements, choose the bounds that make the answer as big or as small as possible.
| Calculation | Maximum | Minimum |
|---|---|---|
| a + b | upper + upper | lower + lower |
| a − b | upper − lower | lower − upper |
| a × b | upper × upper | lower × lower |
| a ÷ b | upper ÷ lower | lower ÷ upper |
Example: a rectangle is 8.4 cm by 5.2 cm, both to 1 decimal place. Find the upper bound of its area.
- Upper bounds: 8.45 cm and 5.25 cm
- Maximum area = 8.45 × 5.25
- 44.3625 cm²
Exam tip
For the biggest answer to a subtraction or division, use the lower bound of the number you take away or divide by.
Bounds to a suitable accuracy (Higher tier only)
A question may ask for an answer "to a suitable degree of accuracy". Work out both bounds, then round them until they agree.
- Work out the maximum and minimum values.
- Round both to the same accuracy, trying fewer and fewer significant figures.
- Give the most accurate value they both round to.
Example: a car travels 245.6 km (to 1 decimal place) in 3.84 hours (to 2 decimal places). Find its average speed to a suitable degree of accuracy.
- Distance: 245.55 ≤ d < 245.65. Time: 3.835 ≤ t < 3.845
- Maximum speed = 245.65 ÷ 3.835 = 64.05… km/h
- Minimum speed = 245.55 ÷ 3.845 = 63.86… km/h
- To 3 s.f. they are 64.1 and 63.9, which differ.
- To 2 s.f. both are 64, so the speed is 64 km/h