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 .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?1.No2.Yes

Answer 1

Answer:

a) 0.667

b) Yes

Explanation:

Data provided in the question:

Mean = 0.04 chip

Standard deviation, s = 0.003 chip

Lower Specification Limit, LSL  = 0.034

Upper Specification Limit, USL = 0.046

Now,

a) Capability Index = ( USL - LSL ) ÷ ( 6 × s )

or

Capability Index = ( 0.046 - 0.034 ) ÷ ( 6 × 0.003 )

= 0.667

b) Cpk = min( [( USL - Mean) ÷ ( 3s ) , ( Mean - LSL) ÷ ( 3s ))

or

Cpk = min( ( 0.046 - 0.04) ÷ (3 × 0.003 ),  ( 0.04 - 0.034 ) ÷ ( 3 × 0.003 ))

or

Cpk = min( ( 0.046 - 0.04) ÷ (3 × 0.003 ),  ( 0.04 - 0.034 ) ÷ ( 3 × 0.003 ))

or

Cpk = min( 0.667, 0.667 )

Therefore,

Cpk = 0.667

as Cp and Cpk are equal

Hence, it is ideal condition and process is capable