Calculating absolute and Relative error

Description:

Refractive index of water was measured as 1.29, 1.33, 1.34, 1.35, 1.32, 1.36, 1.30 and 1.33, calculate

• Mean value
• Mean absolute error
• Fractional error
• Percentage error
• Value with error

Solution

Mean value

Mean = ∑a/n

a_mean = 10.62/8 = 1.328 = 1.33

Mean absolute value

Errors in the individual measurement values are

Δa₁ = a₁ - a_mean,

Δa₂ = a₂ - a_mean,

.............

Δa_n = a_n - a_mean

So,

Δa₁ = 1.29 - 1.33 = -0.04

Δa₂ = 1.33 - 1.33 = 0.00

Δa₃ = 1.34 - 1.33 = 0.01

Δa₄ = 1.35 - 1.33 = 0.02

Δa₅ = 1.32 - 1.33 = -0.01

Δa₆ = 1.36 - 1.33 = 0.04

Δa₇ = 1.30 - 1.33 = -0.03

Δa₈ = 1.33 - 1.33 = 0.00

Δa_mean = ∑a/n

Δa_mean = 0.15/8 = 0.019 ≈ 0.02

Fractional error

Fractional error = Δa_mean/a_mean

Fractional error = 0.019/1.33 = 0.01428 ≈ 0.014

Percentage error

Percentage error = Fractional error × 100

Percentage error = 0.01428 × 100 = 1.428% = 1.4%

Value with error

Value with error = 1.33 ± 0.2

or

Value with error = 1.33 ± 1.4%