Time intervals measured by a clock give the f | vedclass

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(A) The arithmetic mean of the given values is taken as the true value.$t_{	ext{mean}} = \frac{1.25 + 1.24 + 1.27 + 1.21 + 1.28}{5} = \frac{6.25}{5}

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$1.6$

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(A) The arithmetic mean of the given values is taken as the true value.$t_{	ext{mean}} = \frac{1.25 + 1.24 + 1.27 + 1.21 + 1.28}{5} = \frac{6.25}{5} = 1.25 \; s$.The absolute errors in each measurement are:$|\Delta t_1| = |1.25 - 1.25| = 0 \; s$$|\Delta t_2| = |1.24 - 1.25| = 0.01 \; s$$|\Delta t_3| = |1.27 - 1.25| = 0.02 \; s$$|\Delta t_4| = |1.21 - 1.25| = 0.04 \; s$$|\Delta t_5| = |1.28 - 1.25| = 0.03 \; s$The mean absolute error is:$\Delta t_{	ext{mean}} = \frac{0 + 0.01 + 0.02 + 0.04 + 0.03}{5} = \frac{0.10}{5} = 0.02 \; s$.The percentage relative error is:$	ext{Percentage error} = \frac{\Delta t_{	ext{mean}}}{t_{	ext{mean}}} 	imes 100 = \frac{0.02}{1.25} 	imes 100 = 1.6 \%$.

Time intervals measured by a clock give the following readings: $1.25 \; s, 1.24 \; s, 1.27 \; s, 1.21 \; s$,and $1.28 \; s$. What is the percentage relative error of the observations?

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