1.03 Percentage errors, upper and lower bounds, and estimation
Measurements are never exact. This can be for a variety of reasons, including the following:
- Measurement devices have varying degrees of accuracy or precision (i.e. a device measuring mass, may only be capable of measuring to the nearest kilogram, whereas a more accurate device could measure to the nearest gram)
- Human errors in reading the measuring device
- Calibration errors in the device (i.e. a scale not set to zero before making a measurement)
Calibration errors can usually be avoided by calibrating the device before measurements are made. Human error can be reduced by taking multiple readings of the same measurement, and averaging the results.
If we can avoid human and calibration errors then the main source of error comes from the precision of the measuring device. The term 'error', in this case, can seem misleading because it does not mean a mistake has been made. The error comes from limitations in the measuring device itself.
In this section, we look at how we can account for errors in measurement that arise due to the precision of the device.
Absolute error
The absolute error of a measurement is half of the smallest unit on the measuring device. The smallest unit is called the precision of the device.
For example, the ruler below a precision of $0.5$0.5 cm, because the smallest scale markings are $0.5$0.5 cm apart. The absolute error, of any measurement made with the ruler, will be $0.25$0.25 cm (i.e. half of $0.5$0.5 cm).
If we have already taken a measurement, we can also find the absolute error by finding the difference between the actual or exact measurement value, $V_E$VE, and the measured (approximate) measurement value, $V_A$VA.
For a measuring device:
| Absolute error | $=$= | $\frac{1}{2}\times\text{precision }$12×precision |
If we know the exact measurement, $V_E$VE, and have an approximate measurement, $V_A$VA, then:
| Absolute error | $=$= | $|V_A-V_E|$|VA−VE| |
The limits of accuracy
Because every measurement is prone to error, we can use the absolute error of the measuring device to determine the interval, within which the true measurement will lie.
This interval is defined by two values, a lower bound and an upper bound, that form what is known as the limits of accuracy. The lower bound is the smallest possible value that the true measurement could be. The upper bound is the largest possible value of the true measurement.
The limits of accuracy for a measurement are the possible upper and lower bounds of the measurement.
| Upper bound | $=$= | $\text{measurement }+\text{absolute error }$measurement +absolute error |
| Lower bound | $=$= | $\text{measurement }-\text{absolute error }$measurement −absolute error |
Sometimes, the limits of accuracy of a measurement will be expressed in the form:
$\text{measurement }\pm\text{absolute error }$measurement ±absolute error
Worked example
Example 1
A person's height is measured to be $1.68$1.68 metres rounded to the nearest centimetre.
Calculate,
- The absolute error of the measuring device
- The upper bound of their height
- The lower bound of their height
Solution
Because the measurement has been made to the nearest centimetre, the precision of the measuring device is $0.01$0.01 m (i.e. $1$1 cm).
- The absolute error is half the precision,
Absolute error $=$= $\frac{1}{2}\times\text{precision }$12×precision $=$= $\frac{0.01}{2}$0.012 $=$= $0.005$0.005 m
- The upper bound is the largest possible value of their height.
Upper bound $=$= $\text{measurement }+\text{absolute error }$measurement +absolute error $=$= $1.68+0.005$1.68+0.005 $=$= $1.685$1.685 m
- The lower bound is the smallest possible value of their height.
Lower bound $=$= $\text{measurement }-\text{absolute error }$measurement −absolute error $=$= $1.68-0.005$1.68−0.005 $=$= $1.675$1.675 m
This means that the person's true height lies somewhere between $1.675$1.675 and $1.685$1.685 metres.
Example 2
On an architectural drawing, the height of a door is indicated to be $2040\pm5$2040±5 mm.
Write down the maximum and minimum possible heights of the door in metres.
Solution
The maximum height is the upper bound:
| Maximum height | $=$= | $2040+5$2040+5 mm |
| $=$= | $2045$2045 mm | |
| $=$= | $2.045$2.045 m |
The minimum height is the lower bound:
| Maximum height | $=$= | $2040-5$2040−5 mm |
| $=$= | $2035$2035 mm | |
| $=$= | $2.035$2.035 m |
Example 3
A city has a population of $148572$148572 people. A newspaper reports the city to have approximately $150000$150000 people. Find the absolute error of this approximation.
| Absolute error | $=$= | $|V_A-V_E|$|VA−VE| |
| $=$= | $|150000-148572|$|150000−148572| | |
| $=$= | $1428$1428 |
Practice questions
Question 1
Between what limits does the cost of a CD lie if it is known to be $\$50$$50 correct to the nearest $\$5$$5?
Upper bound = $\$$$$\editable{}$
Lower bound = $\$$$$\editable{}$
Question 2
State the limits of accuracy for a distance measured to be $13.45$13.45 km.
Upper bound = $\editable{}$ km
Lower bound = $\editable{}$ km
question 3
The length of a piece of rope is measured to be $19.99$19.99 m using a ruler. What is the upper bound of the largest possible length of this rope?
Percentage error
To compare the sizes of errors across different measurements and situations, it is more useful to use the percentage error, rather than the absolute error.
The percentage error of a measurement is the absolute error expressed as a percentage of the measurement.
| Percentage error | $=$= | $\frac{\text{absolute error }}{\text{exact measurement }}\times100$absolute error exact measurement ×100% |
| $=$= | $\frac{|V_A-V_E|}{V_E}\times100%$|VA−VE|VE×100% |
Note that the absolute error and the measurement must be in the same units before calculating the percentage error.
Worked example
Example 4
The capacity of a bottle is measured as $1.25$1.25 litres, correct to the nearest $10$10 millilitres.
Calculate the percentage error for this measurement.
Solution
The precision of the measuring device is $10$10 mL, so the absolute error is half of this value, which is $5$5 mL.
Note that the measurement is in litres and the absolute error is in millilitres. To convert the measurement into millilitres we multiply by $1000$1000:
| $1.25$1.25 L | $=$= | $1.25\times1000$1.25×1000 mL |
| $=$= | $1250$1250 mL |
Now, we can calculate the percentage error:
| Percentage error | $=$= | $\frac{\text{absolute error }}{\text{exact measurement }}\times100$absolute error exact measurement ×100 % |
| $=$= | $\frac{5}{1250}\times100$51250×100 % | |
| $=$= | $0.4$0.4% |
EXAMPLE 5
A room is 3.8 metres by 4.1 metres in size.
(a) Find the exact area of the room.
(b) Approximate the area of the room by rounding each dimension to the nearest metre.
(c) Find the percentage area of your approximation.
(a)
| Exact Area, $V_E$VE | $=$= | $3.8\times4.1$3.8×4.1 |
| $=$= | $15.58\text{ m}^2$15.58 m2 |
(b)
| Approximate Area, $V_A$VA | $=$= | $4\times4$4×4 |
| $=$= | $16\text{ m}^2$16 m2 |
(c)
| Percentage error | $=$= | $\frac{|V_A-V_E|}{V_E}\times100%$|VA−VE|VE×100% |
| $=$= | $\frac{|16-15.58|}{15.58}\times100%$|16−15.58|15.58×100% | |
| $=$= | $2.70%$2.70% |
Practice question
Question 4
Find the percentage error for a measurement of $55$55 kg, given that it is measured to the nearest kg. Express your answer to the nearest $0.01%$0.01%.
Area and volume calculations
In calculations involving areas and volumes, each measured side length will have an upper and lower bound.
- To find the maximum possible area or volume we multiply the upper bound values for each side.
- To find the minimum possible area of volume we multiply the lower bound values of each side.
Practice question
Question 5
A field has dimensions $15.4$15.4 $\text{m }\times17.6$m ×17.6 $\text{m }$m , to the nearest $10$10 cm.
What is the upper bound of the area of the field?
What is the lower bound of the area of the field?
What is the upper bound of the perimeter of the field?
What is the lower bound of the perimeter of the field?