Question

A process that produces computer chips has a mean of .04 defective chip and a standard deviation of .003 chip. The allowable variation is from .03 to .05 defective. a. Compute the capability index for the process. b. Is the process capable?

Answer 1

Answer:

A. 1.111

B. The process is not capable

Explanation:

Part A

Capacity index help to determine the performance of a process and how it could perform in the future. A capacity index of above 1.33 means that the process is capable but a capacity index below 1.33 means that the process is not capable. The capacity index can be calculated using equation 1;

From the mean  which is 0.5, it can be determined that the process is a centered process.

For centered process, the mean = 0.5 x (Upper s. - Lower S.) = 0.5 x 0,02 = 0.04

so the capacity index for centered mean will be used

$C_{p} =\frac{Upper Specification-Lower Specification}{6 * standard deviation}$ ................................................1

Given standard deviation = 0.003

upper specification = 0.05

lower specification = 0.03

$C_{p} =\frac{0.05- 0.03}{6 * 0.003}\\\\C_{p} = \frac{0.02}{0.018} \\\\C_{p} = 1.111$

Therefore the capacity index of the process is 1.111

Part B

The capacity index of the process is 1.111 and it is less than 1.33, this means that the process is not capable.