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 + 2.71 + 2.80)m)/(5) = (13.12)/(5) m= 2.624m= 2.62 m`
As the lengths are measured to a resolution of 0.01m, 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.
In the first measurement
Error `= Delta x_(1) = 2.63 m- 2.63 m - 2.62 m = + 0.01 m`
Absolute error `= 0.01 m`
Relative error `= 0.01//2.62 = 0.0038`
Percentage error = relative error `xx 100 = 0.38%`
In the second measurement
Error `= Deltax_(2) = 2.56 m-2.62 m = - 0.06m`
Absolute error `= 0.06m`
Relative error `= 0.06//2.62 = 0.023`
Percentage error = relative error `xx 100= 2.3%`
In the third measurement
Error `= Delta x_(3) = 2.42 m- 2.62 m = 0.2 m`
Absolute error `=0.2`m
Relative error `=0.2//2.62 = 0.076`
Percentage error = relative error `xx 100 = 7.6%`
In the fourth measurement
Error `= Delta x_(4) = 2.71 m - 2.62 m = + 0.09 m`
Absolute error `= 0.09m`
Relative error `=0.09 //2.62 = 0.034`
Percentage error = relative error `xx 100=3.4%`
In the fifth measurement
Error `= Delta x_(5) = 2.80 m-2.62m =+0.18m`
Absolute error `=0.18m`
Relative error `=0.18//2.62 = 0.068`
Percentage error = relative error `xx 100 = 6.8%`
Mean or final absolute error `=(sum_(1)^(5) |Delta x_(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 `(2.62 pm 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`.