Question

In an experiment, the coefficient of viscosity (in mPa s) of a liquid was determined as 2.62, 2.68, 2.58, 2.57, 2.54 and 2.55. The mean absolute error in the determination of the coefficient of viscosity of the liquid is

• A. 0.08 mPa s
• B. 0.12 mPa s
• C. 0.06 mPa s
• D. 0.04 mPa s

Hint:

The process for finding mean absolute error is: 1. Calculate the mean (average) of all measurements. 2. Find the absolute difference (error) between each measurement and the mean. 3. Calculate the mean (average) of these absolute errors.

Answer 1

The Correct Option is D

Step 1: Calculate the mean value of the measurements.
Let the measurements be \(x_1, x_2, ..., x_6\).
Mean \(\bar{x} = \frac{2.62 + 2.68 + 2.58 + 2.57 + 2.54 + 2.55}{6}\).
\(\bar{x} = \frac{15.54}{6} = 2.59\) mPa s.
Step 2: Calculate the absolute error for each measurement.
The absolute error is \( |\Delta x_i| = |x_i - \bar{x}| \).
\(|\Delta x_1| = |2.62 - 2.59| = 0.03\).
\(|\Delta x_2| = |2.68 - 2.59| = 0.09\).
\(|\Delta x_3| = |2.58 - 2.59| = |-0.01| = 0.01\).
\(|\Delta x_4| = |2.57 - 2.59| = |-0.02| = 0.02\).
\(|\Delta x_5| = |2.54 - 2.59| = |-0.05| = 0.05\).
\(|\Delta x_6| = |2.55 - 2.59| = |-0.04| = 0.04\).
Step 3: Calculate the mean absolute error.
The mean absolute error is the average of the individual absolute errors.
\(\overline{\Delta x} = \frac{0.03 + 0.09 + 0.01 + 0.02 + 0.05 + 0.04}{6}\).
\(\overline{\Delta x} = \frac{0.24}{6} = 0.04\) mPa s.