Question #152734

In an experiment with a common balance the mass of a ring found to be 2.52g, 2.5g, 2,51g, 2.49g and 2.54g in successive measurements. Calculate a) The mean value of the mass of the ring b) The absolute error in each measurement c) Mean absolute error d) Relative error e) Percentage error 


1
Expert's answer
2020-12-25T14:06:59-0500

Answer

a) mean mass is given

ma=2.52+2.5+2.51+2.49+2.545m_a=\frac{2.52+2.5+2.51+2.49+2.54}{5}

=2.512g=2.512g

b) absolutele error is as below

m1=2.5122.52=0.008gm_1=2.512-2.52=-0.008g

m2=2.5122.5=0.012gm_2=2.512-2.5=0.012g

m3=2.5122.51=0.002gm_3=2.512-2.51=0.002g

m4=2.5122.49=0.022gm_4=2.512-2.49=0.022g

m5=2.5122.54=0.028gm_5=2.512-2.54=-0.028g

c)mean absolute error

<ma>=m1+m2+m3+m4+m55<m_a>=\frac{m_1+m_2+m_3+m_4+m_5}{5}

=0

d) percentage error 0%



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