Question

Flextronics Incorporated manufactures circuit boards. A random sample of 5,000 boards is inspected every week for 5 characteristics. During a recent week, 2 defects were found for 1 characteristic, and 1 defect, each, was found for the other 4 characteristics. If these inspections produced defect counts that were representative of the population, what is the overall sigma level for the process? What is the sigma level for the characteristic that showed 2 defects?

Answer 1

Solution: The overall sigma level for the process is given by:
Number of defects = 2+1+1+1+1 = 6 defects
Number of opportunities = 5,000
Sigma level = NORM.S.INV (1- no of defects/no of opportunities) + shift   
= NORM.S.INV (1-6/5000) +1.5 = 3.036 + 1.5 = 4.536
= 4.54 sigma level

Using Excel  NORM.S.INV (1-6/5000)=3.036

NORM.S.INV (1-2/5000)=3.353

Sigma level for the characteristic that showed 2 defects is computed as follows:

Number of defects = 2
Number of opportunities = 5,000
Sigma level = NORM.S.INV (1- no of defects/no of opportunities) + shift   
= NORM.S.INV (1-2/5000) +1.5 = 3.353 + 1.5 = 4.853 = 4.85 sigma level