# Accuracy in measurement (percentage error)

Lesson

Measurement errors probably caused this to happen

Measurement errors can easily be made. When each measurement taken is slightly out, this can be seen as a small problem. But it can easily grow into a bigger problem.

If I wanted to produce packets of tea bags, I would need to know how large the packet needs to be. If someone told me my tea bags were $3$3 cm by $3.5$3.5 cm by $0.5$0.5 cm I would go off and make a packet that was slightly larger than those measurements. Easy, right? Well what if each of those measurements given were slightly off? I may end up with a packet that my tea bag can't fit into at all. Or I may find the packet looks too big. Those small measurement errors mean that I would end up with a packet I wouldn't be able to use - a waste of time and money.

To calculate the absolute error made, you simply need to find the difference between the true measurement and the incorrect measurement.

When considering errors in measurement, it is useful to calculate the difference in terms of a percentage, to see how much of an error has been made. When calculating the percentage error you need use this formula:
$\frac{\text{absolute error}}{\text{true value}}\times100%=\text{percentage error}$absolute errortrue value×100%=percentage error

Sometimes the error made will be for an area or volume, and you will be asked to find the maximum area/volume of an object or minimum area/volume.

Remember!

Area:

$\text{maximum}=\text{upper}\times\text{upper}$maximum=upper×upper

$\text{minimum}=\text{lower}\times\text{lower}$minimum=lower×lower

Volume:

$\text{maximum}=\text{upper}\times\text{upper}\times\text{upper}$maximum=upper×upper×upper

$\text{minimum}=\text{lower}\times\text{lower}\times\text{lower}$minimum=lower×lower×lower

#### Worked Examples

##### Question 1

A dishwasher weighs exactly $30.3$30.3 kg, but was measured by Luke to be $35$35 kg.

1. Determine the absolute difference of this measurement.

2. Now determine the relative error, correct to one decimal place.

When considering measurement errors it's important to know how to find the maximum and minimum values for areas or volumes.

##### question 2

The dimensions in the figure have been measured to the nearest centimeter.

1. Calculate its volume.

2. What is the smallest possible volume?