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A certain process is under statistical control and has a mean value of 130 and a...

A certain process is under statistical control and has a mean value of 130 and a standard deviation of 8. The specifications for the process are: a. USL (upper specification limit) = 150 b. LSL(lower specification limit) =100 a. Calculate the cp and cpk b. Which of these indices is a better measure of process capability and why? c. Assuminng a normal distribution what percentage of output is expected to fall ourside the specification. Why is it important to know this? d. What would you recommend?

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Answer #1

Given: Upper Specification limit, USL= 150

Lower Specification limit, LSL =100

Actual Mean, M = 130, Standard deviation: sd= 8

Answer a: Process capability ratio: Cp= (upper specification-lower specification)/6sd= 1.042

Answer b: Process capability index: Cpk= Min [(M-LSL)/3sd; (USL-M)/3sd]= Min [1.25; 0.833]= 0.833

Cpk is better because it considers the mean value as compared to Cp which does not.

Answer c:

Upper Specification limit USL 150
Lower Specification limit LSL 100
Mean, m 130
Standard deviation, sd 8
z ROUND((USL-m)/sd, 2) 2.5
Probability NORMSDIST(z) 0.9938
Probability of rejection 2*(1-probability) 1.24%

it is important to know this because it gives us an estimation of number of defected products which will be a loss for an organization.

Answer d: Because rejection is 1.24%, it is pretty high, therefore the company should think about reusing the rejected part. The company should also increase the precision of their machines.

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