January 2001

Vol. 3, No. 1

Quality Quiz from Professor Cleary

You're right!


If a process has a Cp equaling 1.0, a Cpk equaling -1.0, is positive and is negative, then the entire distribution will be to the right of the USL.


  1. For to be negative, has to be greater than USL.

  2. For Cp to be equal to 1, the width of the distribution has to be equal to the width between USL and LSL.

  3. Thus, the distribution will start at the USL and be the width that is between USL and LSL. (See above.)

Note: There are other ways to come to the same conclusion.

Copyright 2001 PQ Systems.

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