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

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

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

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

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
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.