Question

The diameter of the thin wire is measured in the screw gauge as 2.04 mm, 2.06 mm 2.08 mm and 2.07 mm find absolute error and mean absolute error

Answer 1

Dear Student,

◆ Answer -

∆1 = -0.0225 mm

∆2 = -0.0025 mm

∆3 = 0.0175 mm

∆4 = 0.0075 mm

∆(mean) = 0.0125 mm

◆ Explanation -

# Given -

d1 = 2.04 mm

d2 = 2.06 mm

d3 = 2.08 mm

d4 = 2.07 mm

# Solution -

Mean of given observations is -

d = (d1 + d2 + d3 + d4) / n

d = (2.04 + 2.06 + 2.08 + 2.07) / 4

d = 2.0625 mm

Absolute errors for individual observations are -

∆1 = d1 - d = 2.04 - 2.0625 = -0.0225 mm

∆2 = d2 - d = 2.06 - 2.0625 = -0.0025 mm

∆3 = d3 - d = 2.08 - 2.0625 = 0.0175 mm

∆4 = d4 - d = 2.07 - 2.0625 = 0.0075 mm

Mean absolute error is calculated by -

∆(mean) = (|∆1| + |∆2| + |∆3| + |∆4|) / n

∆(mean) = (0.0225 + 0.0025 + 0.0175 + 0.0075) / 4

∆(mean) = 0.0125 mm

Hope this helps you...

Answer 2

Answer:

it HEPS you. ..............