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)/(5) m =2.624 m =2.62 m`
As the length are measured to a resolution of 0.01 m 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 = 2.63 m =- 2.62 m =+0.01 m , " " Deltaa_2 = 2.56 m - 2.62 m =-0.06 m`
`Delta a_3 = 2.42 m - 2.62 m=-0.20 m , " " Deltaa_4 = 2.71 m - 2.62 m =+0.09 m`
`Delta a_5 = 2.80 m -2.62 m =+0.18 m , " " Delta sum a_i =+0.02 m`
Students must note that if rounding off was not done at all, this + 0.01 m 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 makes the result less precise.
`therefore ` Mean of final absolute error =`(sum_1^5 |Deltaa_i|)/(5) =((0.01+0.06+0.20+0.09+0.18 )m)/(5)`
`=0.54 m//5 = 0.108 m = 0.11 m`
This means that the length is `(2.62pm 0.11m)` i.e. , it lies between `(2.62-0.11m) and (2.62-0.11m) ` i.e. between 2.73 m and 2.51 m.
As the arithmetic mean of all the absolute errors is 0.11 m , 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 pm 0.1 m and` not `L = 2.62 pm 0.11 `m
This result can be written with relative error (in place of absolute error) as
`L = 2.6 m pm (0.1)/(2.6)xx100% = 2.6 m pm 4%`