Percentage Error

The gap, measured against the exact value.

The error

A metal rod is exactly 200 cm long. Measured with a tape that has stretched, it reads 194 cm. The error is the size of the gap between the measured value and the exact value: 200 − 194 = 6 cm.

An error is a size, so it is never negative. A reading of 206 cm would also have an error of 6 cm.

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The reading, 194, is 6 below the exact value, 200.

Compare the error with the exact value

Is 6 cm a large error? That depends on what was measured. To compare the error with the length itself, divide the error by the exact value: 6 ÷ 200 = 0.03. Then multiply by 100 to write it as a percentage: 0.03 × 100 = 3%.

The percentage error is 3%: the error is 3 hundredths of the length that was being measured.

of 200 cm6 cm0%25%50%75%100%3%

The exact 200 cm is the whole bar, 100%. Against it, the 6 cm error is 3%.

The same error on a shorter rod

Now a rod that is exactly 20 cm long is measured as 14 cm. The error is 6 cm again, but 6 ÷ 20 = 0.3, which is 30%.

The same 6 cm is ten times the percentage error, because this rod is ten times shorter. That is what a percentage error is for: it compares errors in measurements of very different sizes, such as an error on a bridge with an error on a bolt.

of 20 cm6 cm0%25%50%75%100%30%

Against an exact 20 cm, the same 6 cm error is 30%.

The rule

As one rule: percentage error = |measured − exact| ÷ exact × 100. The bars around measured − exact take its absolute value, the size of the difference whatever its sign.

For the first rod, |194 − 200| ÷ 200 × 100 = 6 ÷ 200 × 100 = 3%.

The usual mistakes

Divide by the exact value, not by the measured value. 6 ÷ 194 × 100 is about 3.1%, which is not the percentage error: the error is judged against the true length the measurement was trying to find.

Subtract before dividing. 194 ÷ 200 × 100 = 97% is the reading as a percentage of the exact value, not the error as a percentage of it.

Worked example: Two Measurements Compared by Their Percentage Errors

Question In a science lesson a student measures the length of a desk as 120 cm and the length of a corridor as 19.8 m. The true lengths are 125 cm and 20 m. (a) Find the percentage error in the measurement of the desk. (b) Find the percentage error in the measurement of the corridor, and say which measurement is more accurate.

  1. 1.The error in the desk measurement is 125 − 120 = 5 cm.

    Deskmeasured 120 cmtrue length 125 cmerror 5 cm
    Deskmeasured 120 cmtrue length 125 cmerror 5 cm
    The error in the desk measurement is 125 − 120 = 5 cm.

  2. 2.Percentage error = errortrue value × 100%. (a) For the desk it is 5125 × 100% = 4%.

    Deskmeasured 120 cm4% errortrue length 125 cmerror 5 cm
    Deskmeasured 120 cm4% errortrue length 125 cmerror 5 cm
    (a) The percentage error is 5125 × 100% = 4%.

  3. 3.The error in the corridor measurement is 20 − 19.8 = 0.2 m, which is 20 cm.

    Deskmeasured 120 cm4% errortrue length 125 cmerror 5 cmCorridormeasured 19.8 mtrue length 20 merror 0.2 m
    Deskmeasured 120 cm4% errortrue length 125 cmerror 5 cmCorridormeasured 19.8 mtrue length 20 merror 0.2 m
    The error in the corridor measurement is 20 − 19.8 = 0.2 m.

  4. 4.(b) For the corridor the percentage error is 0.220 × 100% = 1%.

    Deskmeasured 120 cm4% errortrue length 125 cmerror 5 cmCorridormeasured 19.8 m1% errortrue length 20 merror 0.2 m
    Deskmeasured 120 cm4% errortrue length 125 cmerror 5 cmCorridormeasured 19.8 m1% errortrue length 20 merror 0.2 m
    (b) The percentage error is 0.220 × 100% = 1%.

  5. 5.The corridor measurement is more accurate, because 1% is less than 4%. Its error of 20 cm is four times the error for the desk, but the corridor is 16 times as long as the desk. Check: 1% of 20 m is 0.2 m, and 4% of 125 cm is 5 cm.

    Deskmeasured 120 cm4% errortrue length 125 cmerror 5 cmCorridormeasured 19.8 m1% errortrue length 20 merror 0.2 mThe corridor has the larger error and the smaller percentageerror.
    Deskmeasured 120 cm4% errortrue length 125 cmerror 5 cmCorridormeasured 19.8 m1% errortrue length 20 merror 0.2 mThe corridor has the larger error and the smallerpercentage error.
    The corridor measurement is more accurate, because 1% is less than 4%.

Answer: (a) 4%; (b) 1%, so the corridor measurement is more accurate, although its error of 20 cm is larger

Common mistakes

  • Dividing the error by the measured value, 5120, which gives about 4.2%. The error is compared with the true value, 125 cm, because that is the value the measurement was meant to find.
  • Saying that the desk was measured more accurately because 5 cm is less than 20 cm. An error of 20 cm in 20 m is a small part of the length, and an error of 5 cm in 125 cm is a larger part.

More significant figures and estimation problems, worked step by step →

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