Question

Design specifications require that a key dimension on a product measure 101 ± 8 units. A process being considered for producing this product has a standard deviation of five 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 98. 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            

Hints

Answer 1

Given are following data :

Upper Specification limit ( USL ) = 101 + 8 = 109 units

Lower Specification Limit ( LSL) = 101 – 8 = 93 units

Process mean = m = 101

Process standard deviation = Sd = 5 units

1. Process capability index

= Minimum ( ( USL – m) / 3 x sd , ( m – LSL) / 3 x sd)

= Minimum ( ( 109 – 101)/ 3 x 5 , ( 101 – 93) / 3 x 5)

= Minimum ( 8/15 , 8/15)

= Minimum ( 0.5333, 0.5333)

= 0.5333

1. With process average shifting to 98 , the revised details as follows :

USL = 109 units

LSL = 93 units

Process mean = m1 = 98

Process standard deviation = Sd = 5 units

Process capability index

= Minimum ( ( ( USL – m1) / 3 x Sd , ( m1 – LSL) / 3 x Sd)

= Minimum ( ( 109 – 98)/ 3 x 5 , ( 98 – 93) / 3 x 5 )

= Minimum( 11/15 , 1/3)

= Minimum ( 0.733, 0.333)

= 0.3333

1. Probability of defective output after the process shift

= 1 – Probability of non defective output after process shift

= 1 – ( Probability that output is less than 109 – Probability that output is less than 93 )

= 1 – Probability that output is less than 109 + Probability that output is less than 93

Let Z value corresponding to the probability that output is less than 109 = Z1

Z value corresponding to probability that output is less than 93 = Z2

M1 + Z1 x Sd= 109

Or, Z1 = 2.2

M1 + Z2 x Sd = 93

Or, 98 + 5 x Z2 = 93

Or, 5x Z2 = - 5

Or, Z2 = - 1

Probability for Z1 = 2.2 as derived from standard normal distribution table = 0.98610

Probability for Z2 = -1 as derived from standard normal distribution table = 0.15866

Probability of defective output after the process shift

= 1 – 0.98610 + 0.15866

= 0.17256 ( 0.1726 rounded to 4 decimal places )

PROBABILITY OF DEFECTIVE OUTPUT AFTER THE PROCESS SHIFT = 0.1726