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Understanding Z-Values in Quality Control: Defect Analysis and Specification Limits

This guide helps in finding the z-value corresponding to the upper specification limit given a defect rate of 6.02% for a product. Using the provided probabilities, such as 0.0301 for the upper tail area, we determine that the z-value is 1.88, indicating how many standard deviations the upper limit is from the mean. The concept of z-values is essential in quality control as it helps establish relationships between the mean, standard deviation, and specification limits. This understanding is crucial for effective product quality management.

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Understanding Z-Values in Quality Control: Defect Analysis and Specification Limits

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  1. Finding the z value Help for Chapter 5, SHW Problem 7

  2. .0301 .0301 LSL Mean USL X z = ? 0 Assume that 6.02% of a company product is defective. Find the z value corresponding to the upper spec. limit. (Use .0602 and split it between the two tail areas beyond spec limits.)

  3. Use Appendix B, p. 652 to find z. Since the tail area above USL is .0301 and Appendix B gives the area between 0 and z, we Look up the area between 0 and z, which is .5000 -.0301. =.4699 .4699 .0301 .0301 LSL USL z = 1.88 0 From Appendix B, the z value is z = 1.88. See next slide.

  4. z Table (Text, p. 652)

  5. Meaning of z • The z value tells us that the upper spec. limit is 1.88 standard deviations above the mean. • Because the normal distribution is symmetrical, the z value corresponding to the lower spec. limit is -1.88. • This indicates that the lower spec. limit is 1.88 standard deviations below the mean.

  6. The expression for z is: If we know any 3 of the terms in z, we can solve for the 4th. For example, if know z, Mean, and the estimated standard deviation, we can solve for USL.

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