A student measures the time period …
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Payal Singh 6 years, 9 months ago
Let’s first find the mean value of time:
{tex}t_{mean}={t_1+t_2+t_3+t_4\over 4}{/tex}
= {tex}{90+91+95+92\over 4}= 92 {/tex}
Then, we can find the absolute error for each measurement:
{tex}\Delta t_1=|t_{mean}-t_1|= |92-90|=2{/tex}
{tex}\Delta t_2=|t_{mean}-t_2|= |92-91|=1{/tex}
{tex}\Delta t_3=|t_{mean}-t_3|= |92-92|=0{/tex}
{tex}\Delta t_4=|t_{mean}-t_4|= |92-95|=3{/tex}
Let’s calculate the mean absolute error:
{tex}\Delta t_{mean}={ \Delta t_1+ \Delta t_2+ \Delta t_3+ \Delta t_4\over 4}{/tex}
= {tex}{2+1+0+3\over 4}= 1.5{/tex}{tex}\approx 2{/tex}
Therefore, the reported mean time should be 92 ± 2s
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