Bounds and error intervals

Lesson overview

Number — accuracy and bounds

When a number is rounded, the true value lies within an **error interval** (also called **bounds**). If a length is given as $5.3\text{ cm}$ to $1$ decimal place, the true length $L$ satisfies: $5.25 \leq L < 5.35$ The **lower bound** is $5.25$ and the **upper bound** is $5.35$ (but not including $5.35$ ). General rule: if rounded to $n$ decimal places (or $n$ significant figures), the maximum error is **half a unit** at the last significant digit. For calculations with bounds: - **Addition**: upper bound $=$ UB $+$ UB, lower bound $=$ LB $+$ LB - **Subtraction**: upper bound $=$ UB $-$ LB, lower bound $=$ LB $-$ UB - **Multiplication**: upper bound $=$ UB $\times$ UB, lower bound $=$ LB $\times$ LB - **Division**: upper bound $=$ UB $\div$ LB, lower bound $=$ LB $\div$ UB

Finding bounds for a rounded number

A mass is given as $12.4\text{ kg}$ to $1$ decimal place. Find the error interval.

Half of $0.1$ is $0.05$ . Subtract and add to $12.4$ .

Bounds for calculations

Rectangle: length $= 8.3\text{ cm}$ (1 d.p.), width $= 4.7\text{ cm}$ (1 d.p.). Find the bounds for the perimeter.

Length LB $8.25$

UB: $8.35$

Width LB $4.65$

UB: $4.75$

For addition: LB $+$ LB gives the lower bound, UB $+$ UB gives the upper bound.

Bounds for division — finding maximum and minimum values

Speed $= \dfrac{\text{distance}}{\text{time}}$ . Distance $= 120\text{ m}$ (nearest metre), time $= 9.6\text{ s}$ (1 d.p.). Find bounds for the speed.

Max speed $\dfrac{120.5}{9.55}$

$\approx 12.6\text{ m/s}$

Min speed $\dfrac{119.5}{9.65}$

$\approx 12.4\text{ m/s}$

For division: max $=$ UB $\div$ LB, min $=$ LB $\div$ UB.

Worked example

A rectangle has area $54\text{ cm}^2$ to $2$ significant figures. The width is $6.0\text{ cm}$ to $1$ decimal place. Find the bounds for the length.

Area bounds: $54$ to 2 s.f. means $53.5 \leq A < 54.5$ .
Width bounds: $6.0$ to 1 d.p. means $5.95 \leq W < 6.05$ .
Length $= \dfrac{\text{Area}}{\text{Width}}$ . Maximum length $= \dfrac{54.5}{5.95} \approx 9.16\text{ cm}$ .
Minimum length $= \dfrac{53.5}{6.05} \approx 8.84\text{ cm}$ . So $8.84 \leq L < 9.16\text{ cm}$ .

Try it

Find the bounds for each value first, then combine them carefully for calculations.

Question 1

A length is $7.2\text{ cm}$ to $1$ decimal place. What is the lower bound?

Question 2

A number is $350$ to $2$ significant figures. What is the upper bound?

Question 3

If $a = 12.3$ (1 d.p.) and $b = 4.6$ (1 d.p.), what is the upper bound of $a - b$ ?

Question 4

For division $\dfrac{a}{b}$ , which combination gives the maximum possible value?

Question 5

A square has side $3.4\text{ cm}$ (1 d.p.). What is the lower bound for its area?

Common mistakes

Watch for these when working through the lesson.

For subtraction: using LB $-$ LB instead of LB $-$ UB for the lower bound.
For division: using UB $\div$ UB instead of UB $\div$ LB for the upper bound.
Writing the upper bound as inclusive ( $\leq$ ) instead of exclusive ( $<$ ).

Bounds and error intervals

Lesson overview

Worked example

Try it

Common mistakes

Related topics

Independent practice will plug in here