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Upper and lower bounds - Higher

Upper and lower bounds - Higher

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Summary

Upper and lower bounds

​​In a nutshell

An upper and lower bound for a rounded measurement gives a range of values within which the true values of the measurement could lie. The range of numbers is called an error interval.



Upper and lower bounds

When a measurement is given correct to a certain number of decimal places or significant figures, find the upper and lower bound by halving the number the unit is rounded to, and then adding or subtracting this number from the measurement to find the upper and lower bounds.


Example 1

The mass of a man is 93kg93kg​ to the nearest kgkg​. Find the upper and lower bounds for the mass of the man.


The mass is rounded to the nearest 1kg1kg. Half of this is 0.5kg0.5kg. Add 0.5kg0.5kg​ onto 93kg93kg to find the upper bound and subtract 0.5kg0.5kg from 93kg93kg to find the lower bound.


The lower bound is 92.5 kg\underline{92.5 \ kg} and the upper bound is 93.5 kg\underline{93.5 \ kg}.

 


Upper and lower bounds within a calculation

You may have to find the value of a quantity to an appropriate degree of accuracy from a formula, where the quantities in the formula have been rounded. Find the upper and lower bounds of the quantities in the formula to work out all possible answers for the quantity being calculated. Then it is possible to give the answer to an appropriate degree of accuracy.


Example 2

A plot of land is 2.52.5m wide and 6.06.0m long, correct to 22 significant figures. Find the area of the plot of land to an appropriate degree of accuracy.


Find the upper and lower bounds for the width.

Upper bound of the width =2.55= 2.55m

Lower bound of the width =2.45= 2.45m

2.45mwidth<2.55m2.45m\le width \lt 2.55m​​


Find the upper and lower bounds for the length.

Upper bound of length =6.05= 6.05m

Lower bound of length =5.95= 5.95m

5.95mlength<6.05m5.95m \le length \lt 6.05m


Multiply the lower bound for each measurement to find the minimum area and multiply the higher bound for each length to find the maximum area.

Minimum area =5.95×2.45=14.5775= 5.95 \times 2.45 = 14.5775

Maximum area =6.05×2.55=15.4275= 6.05 \times 2.55 = 15.4275


The answer lies between 14.5775m214.5775m^{2} and 15.4275m215.4275m^{2}. Rounding both of these numbers to 22  significant figures gives the same answer.


Area=15m2 (2 s.f.)\underline{Area = 15m^{2} \ (2 \ s.f.)}


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FAQs - Frequently Asked Questions

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