The diameter of a wire as measured by a screw gauge was found to be 0.026 cm, 0.028 cm, 0.029 cm, 0.027 cm, 0.024 cm, and 0.027cm . Calculate mean value of the diameter, mean absolute error, relative error, percentage error. Also express the result in term of absolute error and percentage error
Answers
Answer:
Explanation:Here is the answer---
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.
Thus, Relative Error is 0.04296 cm.
Hope it helps.
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Answer:
Explanation:
Given observations are---
0.026,0.028, 0.027, 0.029, 0.024, 0.027
No. of Observations = 6
1]∵ 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.
2] 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| (MODULUS OF ANYTHNG IS ALWAYS POSITIVE)
= 0.003
x₆ = |0.027 - 0.027|
= 0.000
3]∵ 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.
4]∴ Relative Error = Absolute Error/Mean Value
= 0.00116/0.027
= 0.04296 cm.
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