Question

The refractive index of water as measured by the
Real depth
relation
was found to have the
Apparent depth
values 1.29, 1.33, 1.34, 1.35, 1.32, 1.36, 1.30, 1.33.
Calculate (i) mean value of (ii) mean value of absolute
error (iii) relative error (to) percentage 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%