SME
Grade 10 Science · Problem type
Accuracy, precision and percent error
Use this to judge a set of trials against the accepted value and find the percent error.
Diagram
Mean = 2.69 g/cm³; percent error = 0.3704%.
Accuracy: the mean is within the first ring (1%), so the trials are accurate.
Precision: they spread over 0.02 g/cm³ (0.7407%), so they are precise.
This is an example. Type your own values to start a new problem.
x₁
Trial 1 density
example · g/cm³
x₂
Trial 2 density
example · g/cm³
x₃
Trial 3 density
example · g/cm³
x̄
Mean density
calculated · g/cm³
A
Accepted density
example · g/cm³
E
Percent error
calculated · %
Formulas
x̄
=
(
x₁
+
x₂
+
x₃
)/3
2.69 = (2.68 + 2.7 + 2.69)/3
E
=
|
x̄
−
A
|/
A
×
100
0.37 = |2.69 − 2.7|/2.7 × 100
Assumptions
•
Three students measured the density of aluminum; its accepted value is 2.70 g/cm³.
•
Accurate means the mean is close to the accepted value.
•
Precise means the trials are close to each other.
Step-by-step
WE KNOW
x₁ = 2.68 g/cm³, x₂ = 2.7 g/cm³, x₃ = 2.69 g/cm³, A = 2.7 g/cm³
FIND
mean density (x̄), percent error (E)
Step 1 · Find mean density (x̄)
Use x̄ = (x₁ + x₂ + x₃)/3
Add the three trials and divide by 3.
x̄
=
(2.68
+
2.7
+
2.69)/3
x̄
=
8.07/3
x̄
=
2.69
g/cm³
Step 2 · Find percent error (E)
Use E = |x̄ − A|/A × 100
The error is how far the mean is from the accepted value, as a percent of it.
E
=
|2.69
−
2.7|/2.7
×
100
E
=
|−0.01|/2.7
×
100
E
=
0.01/2.7
×
100
E
=
0.003704
×
100
E
=
0.37%
Check
Put the numbers back in. Both sides should match.
2.69 = (2.68 + 2.7 + 2.69)/3 ✓
0.37 ≈ |2.69 − 2.7|/2.7 × 100 ✓
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