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Question

The arithmetic mean of given values is taken as true value.

$t _{	ext {mean }}=\frac{t_{1}+ t _{2}+ t _{3}+ t _{4}+ t _{5}}{5}$

$t_{	ext {mean

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

The arithmetic mean of given values is taken as true value.

$t _{	ext {mean }}=\frac{t_{1}+ t _{2}+ t _{3}+ t _{4}+ t _{5}}{5}$

$t_{	ext {mean }}=\frac{1.25+1.24+1.27+1.21+1.28}{5}$

$t_{	ext {mean }}=1.25 s$

$\Delta t_{	ext {mean }}=\frac{\left|\Delta t_{1}ight|+\left|\Delta t_{2}ight|+\left|\Delta t_{3}ight|+\left|\Delta t_{4}ight|+\left|\Delta t_{5}ight|}{5}$

$\Delta t_{	ext {mean }}=\frac{0+0.01+0.02+0.04+0.03}{5}=0.02 s$

Percentage error $=\frac{\Delta t_{\operatorname{mean}}}{t_{	ext {mean }}} 	imes 100=\frac{0.02}{1.25} 	imes 100$

$=1.6 \%$

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