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# How to calculate percentage error

In this article, we will discuss how to calculate percentage error and this applies to measurements in both physics and chemistry. Both physics and chemistry are sciences that depend on measurements. While doing experiments it is expected that we get values that are accurate but our instruments have limitations. Since our instruments are not accurate we get errors in measurements. This error is the difference between true/accepted value and the measured/experimental value.

Percent errors mean how accurate our results are when we measure something. Smaller percentage errors mean that we are close to the true/accepted value. A 2% error, for instance, suggests that we got quite close to the true value, while 50% means our results are not accurate and nowhere close to the true value.
In measurement errors are often unavoidable due to certain reasons like lack of skill of the person doing the measurement, limitations of measuring instruments, or some external factors beyond our control.

## Definition

Percentage error definition:

Percent error is the difference between the measured value and the true value, as a percentage of true value.

We use Percentage Error to compare a measured or experimental value to a true or accepted value value. This helps you to see how far away the measured value or the value you estimated is in a comparison to the true value or the accepted value.

## Formula

Percentage error formula is $$\text{Percentage Error}= \frac{|\text{Measured Value}-\text{True Value}|}{|\text{True Value}|}\times 100$$
Or,
$\text{% Error}=\frac{|T-M|}{|T|}\times 100$
Where,
$T=$ true/accepted value
$M=$ measured/experimental value
Also, here the symbol “|” means absolute value where negative becomes positive.

## Steps for calculating percentage error

The following are the steps you can follow for calculating percentage error when you have information about the measured value and true/accepted value of a physical or chemical quantity.

1. First find the difference between the true value and measured value.
2. Now find the absolute value of the result of the first step.
3. Divide that answer by the true/accepted value and you would get a decimal number
4. Convert that decimal number into a percentage (multiply that answer by 100 and add the % symbol ). Now that we know the steps involved for calculation let us have a look at some examples to clarify the concept.

## Percentage error example

### Example 1

A student measures the radius of a circular sheet of paper and finds that it is 15 cm long. The label on the package indicates that the radius is is 17 cm. Calculate the percentage error in the measurement.

Solution:
Measured value = 15 cm
Accepted value = 17 cm
Step 1: Subtract the measured value from the accepted value.
17 cm – 15 cm = 2 cm
Step 2: Take the absolute value of step 1
|2 cm| = 2 cm
Step 3: Divide that answer by the accepted value.
$\frac{2 cm}{17 cm}=0.12$ (by rounding off to two significant figures)
Step 4: Multiply that answer by 100 and add the % symbol to express the answer as a percentage.
0.12 x 100 = 12%
The percent error of the measurement was 12%.

### Example 2

Suki weighed an object on her balance and recorded a mass of 13.62 grams. Her teacher told her that there was obviously something wrong with her balance because it was giving her a reading which was 22.22% too high. Find the actual mass of the object?
Solution:
Our previous question was a straightforward question where you just have to put in the values and calculate the percentage error.
In this question we are given the mass of the object and percentage error. Here we have to calculate the actual mass of the object.
So ,
Measured value = 13.62 grams
percentage = 22.22%
$\text{Percentage Error}= \frac{|\text{Measured Value}-\text{True Value}|}{|\text{True Value}|}\times 100$
Putting in the values we get
$22.22=\frac{M-T}{T}\times 100$
$\frac{22.22T}{100}=M-T$
$1.2222T=M$
Given that $M=13.62 gm$ and putting in the value of $M$ in above formula we get
$T=\frac{M}{1.2222}=\frac{13.62}{1.2222}=11.143\ gm$
Which is the true/actual mass.
So here we have to rearrange the formula to calculate the desired quantity.

## Can percentage error be negative

We generally take percentage error to be positive because it is the amount of error involved that is important not its sign. However some sciences like chemistry requires it to specify the sign of the error.
In this case you do not take the absolute of the difference between two values. Here order of subtraction also matters where you subtract measured value from the true value.
So, if your percentage error is negative , then measured value is higher than the true/accepted value.
However until and unless it is desired you should always take the absolute value and report percentage error as a positive number.
For example Suppose you estimated the length of the board and made a measure of 22 cm. A peer measures the length of board as well and obtains a measurement of 18 cm. You need to know the "true" or "accepted value" of the length of the board to determine the percentage error of each of the measurements. After checking the manufacturer’s documentation, you found the length of the board is 20 cm in length. This would be used as the accepted value of our estimate.
The percentage error estimates for the two measures as seen below. Both measurements indicate an error of 1%. Your peer has made a mistake that is on the low side of the accepted value, so it is negative . You, on the other hand, have made an error on the high side of the accepted value and it is positive.
$\text{Percentage Error}= \frac{|\text{Measured Value}-\text{True Value}|}{|\text{True Value}|}\times 100$
$\text{Percentage Error}= \frac{22-20}{20}\times 100 = 10%$
$\text{Percentage Error}=+10%$
$\text{Percentage Error}= \frac{18-20}{20}\times 100 = -10%$
$\text{Percentage Error}=-10%$

## What is a good percent error

The amount of error that is appropriate or good, varies with the case under consideration. Suppose you calculate the distance between New Delhi and Gurugram on a map and you estimate it to be around 25 Km. If the actual distance is 30 Km, the chances are that the trip between the two cities would have little impact on this mistake. But if the calculations used to send satellites to designated orbit were to have the same degree of error, then whole mission would be in trouble.