Question

2.0,2.2,2.4 fts
Fine mean value, absolute error, average absolute error, relative error, persentage error

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%