Question

In an experiment with a common balance the mass of ring is 2.52, 2.53, 2.51, 2.44, 2.54.find a)absolute error, mean absolute error, fractional error, percentage error, relative error

Answer 1

Given :

a₁ = 2.52

a₂ = 2.53

a₃ = 2.51

a₄ = 2.44

a₅ = 2.54

To find :

a) Absolute error

b) Mean absolute error

c) Fractional error

d) Percentage error

e) Relative error

Solution :

a) Absolute Error

$Mean = \frac{2.52 + 2.53 + 2.51 + 2.44 + 2.54}{5}$

$Mean = \frac{12.54}{5}$

Mean = 2.508

Hence, absolute errors

Δa₁ = 2.52 - 2.508 = 0.012

Δa₂ = 2.53 - 2.508 = 0.022

Δa₃ = 2.51 - 2.508 = 0.002

Δa₄ = 2.44 - 2.508 = 0.064

Δa₅ = 2.54 - 2.508 = 0.032

b) Mean Absolute Error

$MeanAbsolute = \frac{0.012 + 0.022+0.002+0.064+0.032}{5}$

$MeanAbsolute = \frac{0.132}{5}$

Mean absolute error = 0.0264

c) Fractional Error

$\delta a = \frac{Mean Absolute Error}{Mean Value }$

$\delta a = \frac{0.0264}{2.508}$

δa = 0.01053

Hence, Fractional error = 0.01053

d) Percentage Error

$PercentageError = \frac{Mean Absolute Error}{Mean Value }*100$

$PercentageError = \frac{0.0264}{2.508}*100$

Percentage Error = 1.053%

e) Relative Error

Since, fractional error and relative error are same.

Therefore, relative error = 0.01053