Question

Repeated measurements of a certain quantity in an experiment gave the following values:1.29, 1.33,1.34,1.35,1.32,1.36,1.30,1.33
Calculate the mean value,Absolute error, the relative error and the percentage error.
With Explanation​

Answer 1

Answer:

Mean value of quantity measured, V=  

8

1.29+1.33+1.34+1.35+1.32+1.36+1.30+1.33

​  

x=1.3275=1.33(round off to two places of decimal).Absolute errors in measurement are:Δx  

1

​  

=1.33−1.29=0.04;   Δx  

2

​  

=1.33−1.33=0.00Δx  

3

​  

=1.33−1.34=−0.01;   Δx  

4

​  

=1.33−1.35=−0.02Δx  

5

​  

=1.33−1.32=+0.01;   Δx  

6

​  

=1.33−1.36=−0.03Δx  

7

​  

=1.33−1.30=+0.03;   Δx  

8

​  

=1.33−1.33=0.00

Mean absolute error,    

Δx

=  

n

∑  

i=l

i=n

​  

∣(Δx)  

i

​  

∣

​  

=  

8

0.04+0.00+0.01+0.02+0.01+0.03+0.03+0.00

​  

 

8

0.14

​  

=0.0175=0.02

Relative error=±  

x

Δx

 

​  

=±  

1.33

0.02

​  

=±0.015=±0.02

Percentage error=±0.015×100=1.5%