Question

Design specifications require that a key dimension on a product measure 102 ± 10 units. A process being considered for producing this product has a standard deviation of two units. a.

What can you say (quantitatively) regarding the process capability? Assume that the process is centered with respect to specifications. (Round your answer to 4 decimal places.) Process capability index;

b. Suppose the process average shifts to 95. Calculate the new process capability. (Round your answer to 4 decimal places.) New process capability index:

c. What is the probability of defective output after the process shift? (Use Excel's NORM.S.DIST() function to find the correct probability. Round "z" values to 2 decimal places. Round probabilities to 4 decimal places (0.####).) Probability of defective output:

Answer 1

Upper Specification Limit(USL) = 102 + 10 = 112

Lower Specification Limit(LSL) = 102 -10 = 92

Process mean = 102

Standard Deviation = 2

a.

Process Capability Index = Minimum of (( (USL - Process mean) / ( 3 x Standard Deviation)), ( ( Process mean - LSL) / ( 3 x Standard Deviation))])

= Minimum of ((112 - 102)/ (3 x2) , (102 - 92) / ( 3 x 2))

= Minimum of (10 / 6, 10 / 6)

= 10 / 6 = 1.6667

Therefore the Process Capability Index = 1.6667

b.

Process Mean = 95

Process Capability Index = Minimum of (( (USL - Process mean) / ( 3 x Standard Deviation)), ( ( Process mean - LSL) / ( 3 x Standard Deviation)))

= Minimum of ( ( 112 - 95) / ( 3 x 2) , ( 95 - 92) / (3 x 2))

= Minimum of (17 / 6 , 3 /6)

= 3 / 6 = 0.5

Therefore the new Process Capability Index = 0.5

c.

z = (x - Process Mean) / Standard Deviation

For the Left hand side:

z = ( 92 - 95) / 2 = -3 / 2 = -1.5

Using NORM.S.DIST() function on Excel with z value as -1.5, the probability is 0.0668

This is the cumulative probability of values going below the LSL

For the Right hand side:

z = (112 - 95) / 2 = 17 / 2 = 8.5

Using NORM.S.DIST() function on Excel with z value as 8.5, the probability is 1.

So, the cumulative probability of values going above the USL is 1 - 1 = 0.

Therefore, the probability of defective output = 0.0668