A) 0.1 s
B) 0.11 s
C) 0.01 s
D) 1.0 s
Correct Answer: B
Solution :
Average value \[=\frac{2.63+2.56+2.42+2.71+2.80}{5}\] \[=2.62\,\sec \] Now \[|\Delta {{T}_{1}}|\,=\,2.63-2.62=0.01\] \[|\Delta {{T}_{2}}|\,=\,2.62-2.56=0.06\] \[|\Delta {{T}_{3}}|\,=\,2.62-2.42=0.20\] \[|\Delta {{T}_{4}}|\,=\,2.71-2.62=0.09\] \[|\Delta {{T}_{5}}|\,=\,2.80-2.62=0.18\] Mean absolute error \[\Delta T=\frac{|\Delta {{T}_{1}}|+|\Delta {{T}_{2}}|+|\Delta {{T}_{3}}|+|\Delta {{T}_{4}}|+|\Delta {{T}_{5}}|}{5}\]\[=\frac{0.54}{5}=0.108\,=0.11sec\]
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