Question

the length of a metal plate was measured using a vernier calliper of least count 0.01 cm. The measurements made were 3.11 cm, 3.13 cm, 3.14 cm and 3.14 cm. Find the mean length, mean absolute error and percentage error in measurement of length

Answer 1

Formula used:

Mean length = Sum of all measurements / No. of measurements

Absolute Error = Actual value - Measured value

Mean Absolute error = Sum of all absolute errors / No. of measurements

%Error = Error value / Actual value * 100

Solution :

Mean length =$\frac{3.11+3.13+3.14+3.14}{4}= \frac{12.52}{4}= 3.13 cm$

Taking Mean length as actual value, we have values of absolute error as 0.02 cm, 0 cm , 0.01 cm and 0.01 cm.

Mean Absolute value = $\frac{0.02 + 0 + 0.01 + 0.01}{4}= \frac{0.04}{4} = 0.01 cm$

%Error = $\frac{0.01}{3.13} * 100 = 0.32 %$

Answer 2

Formula used:

Mean length = Sum of all measurements / No. of measurements

Absolute Error = Actual value - Measured value

Mean Absolute error = Sum of all absolute errors / No. of measurements

%Error = Error value / Actual value * 100

Solution :

Mean length =\frac{3.11+3.13+3.14+3.14}{4}= \frac{12.52}{4}= 3.13 cm

4

3.11+3.13+3.14+3.14

=

4

12.52

=3.13cm

Taking Mean length as actual value, we have values of absolute error as 0.02 cm, 0 cm , 0.01 cm and 0.01 cm.

Mean Absolute value = \frac{0.02 + 0 + 0.01 + 0.01}{4}= \frac{0.04}{4} = 0.01 cm

4

0.02+0+0.01+0.01

=

4

0.04

=0.01cm

%Error =