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