1.6 Approximation and errorIB Maths: Applications and Interpretation HL: Revision notes

To round to decimal places (d.p.), count digits after the decimal point. To round to significant figures (s.f.), count from the first non-zero digit. Look at the next digit: $5$ or more rounds up. $0.004567=0.0046$ to 2 s.f. and $0.005$ to 3 d.p. Zeros at the start do not count as significant figures. Choosing accuracy: an answer cannot be more accurate than the data it comes from. If a calculation uses $41.8$ (3 s.f.) and $2.3$ (2 s.f.), give the result to 2 s.f.: $41.8\div2.3=18.17\ldots\approx18$. In IB exams, unless told otherwise, give exact values or 3 s.f. (money to 2 d.p.), and keep full calculator values until the final answer.

A rounded number can come from a range of values. A value rounded to a given accuracy lies within half a unit of the last digit either side. If $x=4.1$ to one decimal place, then $4.05\leq x<4.15$. The lower bound is included and the upper bound is not, because $4.15$ would round up to $4.2$. Other examples: $85$ m to the nearest metre gives $84.5\leq l<85.5$; $0.86$ to 2 d.p. gives $0.855\leq w<0.865$. To find bounds of a calculation, combine bounds sensibly. For an area, the lower bound uses both lower bounds: $84.5\times41.5=3506.75$; the upper bound uses both upper bounds: $85.5\times42.5=3633.75$. Bounds also show how accurately a result can be stated: if both bounds round to the same value to 2 s.f., you may quote the answer to 2 s.f.

The percentage error compares an approximate value $v_A$ with the exact value $v_E$: $\varepsilon=\left|\frac{v_A-v_E}{v_E}\right|\times100\%.$ Worked example: the radius of a circle is measured as $2.5$ cm but is really $2.53$ cm. Then $v_A=\pi(2.5)^{2}=19.635\ldots$ and $v_E=\pi(2.53)^{2}=20.109\ldots$, so $\varepsilon=\frac{|19.635-20.109|}{20.109}\times100=2.36\%$. Errors arise from rounding (a value to 1 d.p. can be wrong by up to $0.05$) and from measurement limits (a ruler marked in mm). The greatest possible error in a rounded value is half a unit of the last digit: for $4.1$ cm it is $0.05$ cm, which is $\frac{0.05}{4.1}\times100=1.22\%$ of the measurement. Errors grow in calculations: a radius of $2.5$ cm (1 d.p.) can give an area from $\pi(2.45)^{2}$ to $\pi(2.55)^{2}$, so the greatest possible error in the area is $\pi(2.55)^{2}-\pi(2.5)^{2}=0.793$ cm$^{2}$, about $4.04\%$ of the calculated area.

Estimating means rounding each number to 1 significant figure to check that an answer is of the right size. For $41.8\div2.3$: $40\div2=20$, so an answer of $18$ is reasonable but $1.8$ or $180$ is not. Check that answers make sense in context: a length cannot be negative, a probability lies between $0$ and $1$, a person's age is not $250$, and a percentage of a quantity cannot exceed the whole unless a rise is stated. If an answer is unreasonable, look for a misplaced decimal point, a wrong unit or a sign error.