For an outofround condition smaller the better of a steel sh

For an out-of-round condition (smaller the better) of a steel shaft, the true indicator readings (TIR) for eight shafts are 0.05, 0.04, 0.04, 0.03, 0.04, 0.02, 0.04, and 0.03 mm.

a) If the average loss at 0.03 is $15.00, what is the loss function?

b) What is the loss at 0.05?

c) What is the average loss?

To calculate the k value, please use the k = A/y^2 equation for part a. A = 15.00, and y = 0.03

Solution

a) A= 15 , Y = 0.03..
k = A / y^2 = 16771.16  ..
loss function =  16771.16 * (yi^2) .....

b) loss at 0.05 = 16771.16 * ( 0.05)^2 = 41.9279..


c) average loss= ( 6.708464+2*15 + 4* 26.83386 + 41.9279 ) / 8 = 23.24648....