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(D) Step $1$: Calculate the mean value of the observations.Mean value $\bar{x} = \frac{2.62 + 2.68 + 2.58 + 2.57 + 2.54 + 2.55}{6} = \frac{15.54}{6} =

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$0.04  mPa \cdot s$

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(D) Step $1$: Calculate the mean value of the observations.Mean value $\bar{x} = \frac{2.62 + 2.68 + 2.58 + 2.57 + 2.54 + 2.55}{6} = \frac{15.54}{6} = 2.59 \ mPa \cdot s$.Step $2$: Calculate the absolute errors for each observation $|\Delta x_i| = |x_i - \bar{x}|$.$|\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$$|\Delta x_4| = |2.57 - 2.59| = 0.02$$|\Delta x_5| = |2.54 - 2.59| = 0.05$$|\Delta x_6| = |2.55 - 2.59| = 0.04$Step $3$: Calculate the mean absolute error.Mean absolute error $\Delta \bar{x} = \frac{0.03 + 0.09 + 0.01 + 0.02 + 0.05 + 0.04}{6} = \frac{0.24}{6} = 0.04 \ mPa \cdot s$.

In an experiment, the coefficient of viscosity (in $mPa \cdot s$) 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

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