Question

For the given set of measurements, find the relative error. \[ 20.00,\ 19.75,\ 18.25,\ 17.01 \]

• A. \(0.12\)
• B. \(0.06\)
• C. \(0.09\)
• D. \(0.17\)

Hint:

Relative error is dimensionless and gives a better sense of accuracy than absolute error alone.

Answer 1

The Correct Option is B

Alternative Method (Assumed Mean Method):

To simplify calculations, we take a value close to all observations as an assumed mean. This reduces arithmetic complexity while giving the same result.

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Step 1: Choose an assumed mean

Let the assumed mean be:

a = 18.75

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Step 2: Find deviations from the assumed mean

d_i = x_i − a

20.00 − 18.75 = 1.25
19.75 − 18.75 = 1.00
18.25 − 18.75 = −0.50
17.01 − 18.75 = −1.74

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Step 3: Calculate the corrected mean

Mean deviation correction:

\(\bar{x} = a + \dfrac{\sum d_i}{n}\)

\[ \bar{x} = 18.75 + \frac{(1.25 + 1.00 − 0.50 − 1.74)}{4} \]

\[ \bar{x} = 18.75 + \frac{0.01}{4} = 18.7525 \]

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Step 4: Mean absolute error

\[ \Delta x = \frac{\sum |x_i − \bar{x}|}{n} \]

\[ = \frac{1.2475 + 0.9975 + 0.5025 + 1.7425}{4} = 1.1225 \]

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Step 5: Relative error

\[ \text{Relative error} = \frac{\Delta x}{\bar{x}} = \frac{1.1225}{18.7525} \approx 0.06 \]

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Final Answer:

Relative error ≈ 0.06