Question

The diameter of a wire as measured by a screw guage was found to be 0.026cm,0.028cm,0.029cm,0.027cm,0.024cm,0.027cm.Calculate 1)mean value of the diameter.2)mean absolute error.3)relative error.4)percentage error.Also express the result in terms of absolute and percentage error.

Answer 1

Given observations are
---0.026,0.028, 0.027, 0.029, 0.024, 0.027

No. of Observations = 6
∵ Mean = Sum of all observations/No. of observations.
= (0.026 + 0.028 + 0.027 + 0.029 + 0.024 + 0.027)/6
= 0.161/6 = 0.0268 cm

.For the Approximate Calculations,
Taking Mean = 0.027 cm.

By taking in approx, answer can be easily calculated

.For the Relative Error,

x₁ = |0.026 - 0.027| = 0.001

x₂ = |0.028 - 0.027| = 0.001

x₃ = |0.027 - 0.027| = 0.000

x₄ = |0.029 - 0.027| = 0.002

x₅ = |0.024 - 0.027| = |-0.003| = 0.003

x₆ = |0.027 - 0.027| = 0.000

∵ Absolute Error(Δx) = (x₁ + x₂ + x₃ + x₄ + x₅ + x₆)/6
= (0.001 + 0.001 + 0.000 + 0.002 + 0.003 + 0.000)/6
= 0.007/6 = 0.00116 cm

.∴ Relative Error = Absolute Error/Mean Value = 0.00116/0.027 = 0.04296 cm.

Percentage error = Relative error x 100 = 0.04296 = 4.296 %

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