1. Problem 21 Since the capability index is greater than 1, the process is capable.

2. Problem 21 Process fraction defective =.00003, or 3 out of every 100,000 packages will be outside of the specification limits.

3. NORMSDIST(-4.17) = Area under curve to left of z Problem 21.000015 0-4.17 z.000015 4.17 USLLSL Fraction defective = 2 NORMSDIST(-4.17) = 2(.000015) =.00003

4. Problem 21 The C p index assumes that the process is on target. However, the process is not on target. The mean is 9.8 and the target is 10. The fraction defection will therefore be greater than 3 out of 100,000. Therefore, we should compute C T and the actual fraction defective.

5. Problem 21

6. Since the capability index is less than 1, the process is not capable. The actual process fraction defective is.006, or 6 out of 1,000 packages (see next slide). If process is center on target, the capability index would increase to 1.39, and the process fraction defective would decrease to 3 out of 100,000 packages.

7. Problem 21 F1 9.89.5 F2 10.5 LSL USL Fraction defective = F1 + F2 = 1 – [Area(z L ) + Area(z U )] Area(z L ) Area(z U )

8. Problem 21