Help for Chapter 5, SHW Problem 7

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Presentation transcript:

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

z = ? 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.) LSL .0301 X Mean z = ? USL

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 From Appendix B, the z value is z = 1.88. See next slide.

z Table (Text, p. 652) z .00 .01 .02 . .08 .09 0.0 0.1 0.2 1.8 .4699

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.

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.