Accuracy, bounds and compound measures
| English | 中文 | Pinyin |
|---|---|---|
| lower bound/ˈləʊə baʊnd/ | 下界 | xià jiè |
Is the printed measurement exact?
- A rectangular panel is labelled 8.0 cm by 5.0 cm, each to the nearest 0.1 cm. Its true area is not fixed at 40 cm².
- This lesson studies lower bound 下界: The smallest possible value consistent with a stated rounding rule.
Choose the mathematical structure
- A value rounded to the nearest unit u lies from stated value-u/2 up to, but usually not including, stated value+u/2. For positive quantities, combine extremes according to the operation.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines lower bound?
The smallest possible value consistent with a stated rounding rule.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
The lengths satisfy 7.95≤L<8.05 and 4.95≤W<5.05. Since A=LW, 39.3525≤A<40.6525. For speed d/t, the largest speed uses the largest distance and smallest positive time.
Accuracy, bounds and compound measures
A value rounded to the nearest unit u lies from stated value-u/2 up to, but usually not including, stated value+u/2
Explain the different endpoint inclusions for a measurement rounded to 8.0 cm.
Find the lower bound of 3.4 rounded to the nearest 0.1.
Half the rounding unit is 0.05. Lower bound=3.4-0.05=3.35.
Test a tempting shortcut
- An upper bound is not automatically achieved. Dividing upper distance by upper time does not give the largest speed. Keep enough digits in intermediate calculations.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The greatest value of positive d/t uses the greatest d and greatest t. This claim is false. Explain which definition or assumption it violates.
Find the upper bound of 250 rounded to the nearest 10.
Half of 10 is 5. Upper bound=250+5=255; the upper endpoint is excluded.
The greatest value of positive d/t uses the greatest d and greatest t.
An upper bound is not automatically achieved. Dividing upper distance by upper time does not give the largest speed. Keep enough digits in intermediate calculations.
Interpret a new situation
- Distinguish measurement uncertainty from arithmetic rounding. A sensible reported precision cannot be finer than the measurements justify.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
A distance is at most 105 m and time at least 20 s. Find the upper speed bound.
For positive d and t, the largest quotient uses the greatest d and least t: 105/20=5.25.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · P. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The smallest possible value consistent with a stated rounding rule. Choose the relationship, show the method, check its assumptions and interpret the result.