The readings of a length come out to be 2.63 m , 2.56 m , 2.42 m , 2.71 m and 2.80 m . Calculate the absolute errors and relative errors or percentage errors. What do you think of the actual value of the length and its limits ?

Explanation

The mean value of length L`=((2.63+2.56+2.42+2.71+2.80)m)/5=(13.12)/5m=2.624m=2.62m`
As the lengths are measured to a resolution of `0.01m` because all lengths are given to the second place of decimal , it is proper to round off this mean length also to the second place of decimal .
Thus the errors in the measurement are
`Deltaa_(1)=2.63m-2.62m=+0.01m,` `Deltaa_(2)=2.56m-2.62m=-0.06m`
`Deltaa_(3)=2.42m-2.62m=-0.20m,` `Deltaa_(4)=2.71m-2.62m=+0.09m`
`Deltaa_(5)=2.80m-2.62m=+0.18m,` `Deltasuma_(i)=+0.02m`
Students must note that if rounding off was not done at all, this `+0.01m`should , in fact, have been zero . It has not come out to be zero because rounding off was done to the second place of decimal . Whenever rounding is done , it make the result less precise.
`:."Mean of final absolute error"=(sum_(1)^(5)|Deltaa_(i)|)/5=((0.01+0.06+0.20+0.09+0.18)m)/5`
`=0.54m//5=0.108m=0.11m`
This means that the length is `(2.62+-0.11m)`i.e, it lies between `(2.62+0.11)`and `(2.62+0.11m)` and `(2.62-0.11m)`i.e, between `2.73m "and" 2.51m`.
As the arithmetic mean of all the absolute errors in `0.11m`, there is an error in it already at the first place of decimal because at first decimal place we have got 1 and not zero . Hence our precision cannot be upto the second place of decimal . Hence , we should write
`L =2.6+-0.1m "and not" L=2.62+-0.11m`
This result can be written with relative error (in place of absolute error)as
`L=2.6m +-(0.1)/(2.6)xx 100% =2.6m+- 4%`