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

Concept:

Mean value of measurements: \[ \bar{x} = \frac{\sum x_i}{n} \]
Mean absolute error: \[ \Delta x = \frac{\sum |x_i - \bar{x}|}{n} \]
Relative error: \[ \text{Relative error} = \frac{\Delta x}{\bar{x}} \]
Step 1: Calculate the mean value. \[ \bar{x} = \frac{20.00 + 19.75 + 18.25 + 17.01}{4} = \frac{75.01}{4} = 18.7525 \]
Step 2: Calculate absolute deviations. \[ |20.00 - 18.7525| = 1.2475 \] \[ |19.75 - 18.7525| = 0.9975 \] \[ |18.25 - 18.7525| = 0.5025 \] \[ |17.01 - 18.7525| = 1.7425 \]
Step 3: Mean absolute error. \[ \Delta x = \frac{1.2475 + 0.9975 + 0.5025 + 1.7425}{4} = \frac{4.49}{4} = 1.1225 \]
Step 4: Relative error. \[ \text{Relative error} = \frac{1.1225}{18.7525} \approx 0.06 \]