Measurements and Calculating Measurement Error

Question

If a thin piece of wire is measured with a Vernier calliper and produces the following lengths: 1.23±0.05mm, 1.19±0.04mm and 1.21±0.05mm, how can the length of wire be determined along with its appropriate error?

Answer 1

l = (l₁ + l₂ + l₃)/3 = 1.21 mm.

Δl = l.√((Δl₁/l1)²+(Δl₂/l₂)²+(Δl₃/l₃)²) = 0.08 mm

So : l = (1.21±0.08) mm.