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 .034 to .046 defective. a.Compute the capability index (Cp) for the process. (Round your intermediate calculations to 3 decimal places and final answer to 2 decimal places.) Capability index b. Is the process capable? Yes or No

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valetta valetta · Answer 1 · 2019-11-13T08:25:53+01:00

Answer:

a) 0.667

b) Yes

Explanation:

Data provided in the question:

Mean = 0.04

Standard Deviation = 0.003

Upper Specification Limit, USL = 0.046

Lower Specification Limit, LSL = 0.034

Now,

a) Capability Index is given as:

Cp = [tex]\frac{(USL-LSL)}{(6\sigma)}[/tex]

or

Cp = [tex]\frac{(0.046-0.034)}{(6\times0.003)}[/tex]

or

Cp = 0.667

Also,

Cpk = min([tex]\frac{(USL-Mean)}{(3\sigma)},\frac{(Mean-LSL)}{(3\sigma)}[/tex]

or

Cpk = min([tex]\frac{(0.046-0.04)}{(3\times0.003)},\frac{(0.04-0.034)}{(3\times0.003)}[/tex]

or

Cpk = min( 0.667 , 0.667 )= 0.667

Since,

Cp and Cpk are same in this case

therefore, it is ideal condition and process is capable

b) yes