Errors in Measurement
- With measurements such as length and weight, these are continuous variables, so it is hard to find an exact value.
- It is possible to measures someone's height to the 100th decimal, but it is often not sensible, nor is it necessary, for you to do so.
- Instead, we can approximate the height to the nearest centimetre - this is often precise enough for us to use.
- e.g. instead of saying someone's height is 168.93163862973284cm, we typically round to the nearest centimeter and say this person's height is 169cm.
However, when we approximate these quantities, there will always the the potential for error - this is especially true for measured quantities. When we are taking measurements, we are reading the value off a scale.
- Say we are measuring the length of a square tile. You would typically place the tile against a ruler and find the length of the tile by seeing which value on the ruler it is closest to.
- For example, say the tile was measured using a 30cm ruler, with the smallest unit at millimeters. To measure the length of the tile, we would place the edge against the ruler, with one corner at 0cm, and the length would be determined by the millimeter mark which the second corner is closests to.
- However, when we do something like this, we are approximating the length. It is possible for the length of the tile to fall between two millimeter marks, but we are approximating it based on which millimeter mark it is closest to.
- e.g. Say we found the length of the tile to be 37mm. However, the actual edge of the tile lay in between the 36mm and 37mm mark, but it was closer to the 37mm mark than it was to the 36mm, so we round the length to 37mm.
- By rounding, we can say that our answer was inaccurate by up to half a millimeter - the actual value lies within ±0.5mm of the measured value.
This applies to all measurements - A measurement is accurate to ±1/2 of the smallest division on the scale.
- If you are measuring the weight of a pebble on a scale that gives you the weight to the nearest gram (i.e. gram is the smallest division), then you will get a value that is accurate to ±0.5g. This means that the actual weight of the pebble will lie between ±0.5g of the measured value.
- If you were asked to give a range for which the actual weight of the pebble should lie in, it would be x - 0.5g < w < x + 0.5g (where x = the measured weight and w = the actual weight).
This way of writing out the potential errors in measurements are extremely useful, especially in writing lab reports (for your IB science IAs), so it is an important skill to have.
Absolute and Percentage Error
As mentioned above, we approximate a lot of quantities in the real world, simply because there is no need for us to be that precise. However, this means that there will be a difference between the approximated value and the actual, exact value. This difference between the approximated and exact value is known as error.
The size (or magnitude) of the error, regardless of whether the approximated value is larger or smaller than the exact value, can be determined using absolute error. The absolute error can be found using the following formula:
Error is also often expressed using a percentage. The percentage error can be found using the following formula:
Here is a sample question where you would have to find the absolute and percentage error for an approximated value, given the exact value.
- Samuel, who works in construction, is measuring the weight of a brick, and he records down the weight as 56 grams. It turns out that the scale that Samuel uses was faulty, and his co-worker, Michael, finds that the actual weight of the brick is 56.4 grams. Find the absolute and percentage error.
- In this question, we are given the approximated value (VA), which is 56 grams. The exact value (VE) is also given - it is 56.4 grams.
- Substituting it into the absolute error equation, we get absolute error = | 56 - 56.4 | = | -0.4 | = 0.4
- Susbstituting these values into the percentage error equation, we get percentage error = ( |56 - 56.4 |/56.4) x 100% = (0.4/56.4) x 100% = 0.00709 x 100% = 0.709%
In the IBDP Mathematics exam, you may be expected to find the exact or approximated values with information given in the question, before being asked to determine the absolute/percentage error. However, don't worry about having to memorise these formulas. They will be given in the formula booklet, and questions related to absolute and percentage error should be fairly straight forward.
This is the end of this topic.