Design specifications require that a key dimension on a product measure 99 + 11

Question

Design specifications require that a key dimension on a product measure 99 + 11 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 .7333 O .7333 b. Suppose the process average shifts to 96. Calculate the new process capability. (Round your answer to 4 decimal places.) New process capability index .5333 o .5333 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 .94268 .9426

Explanation / Answer

Given are following data :

Upper Specification Limit = USL = 99 + 11 = 110 units

Lower Specification Limit = LSL = 99 – 11 = 88 units

Process mean = m = 99 units

Standard deviation = S = 5 units

Therefore ,

Process Capability Index, Cpk

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

= Minimum ( ( 110 – 99) / 3 x 5 , ( 99 – 88)/3 x 5)

= Minimum ( 11/15, 11/15)

= 11/15

= 0.7333

PROCESS CAPABILITY INDEX = 0.7333

The new process average = m1 = 96

Revised process Capability Index, Cpk

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

= Minimum ( ( 110 – 96) / 3 x 5 , ( 96 – 88)/3 x 5)

= Minimum ( 14/15 , 8/ 15)

= 8/15

= 0.5333

NEW PROCESS CAPABILITY INDEX = 0.5333

With m1 = 96

Let Z value corresponding to probability that output parameter will be less than USL of 110 units = Z1

Similarly,

Let Z value corresponding to probability that output parameter will be less than LSL of 88 units =Z2

Thus,

‘m1 + Z1 x S = USL

OR, 96 + 5.Z1 = 110

Or, 5.Z1 = 14

Or, Z1 = 2.8

Corresponding probability for Z1 = 2.8 as derived from standard normal distribution table = 0.99744

I.e., Probability that dimension of output will be less than 110 = 0.99744

Therefore , probability that dimension of output will be more than ( DEFECTIVE) 110 units

= 1 – 0.99744

= 0.00256

Similarly ,

‘m1 + Z2 x S = LSL

Or, 96 + 5.Z2 = 88

Or, 5.Z2 = - 8

Or, Z2 = - 8/5 = - 1.6

  Corresponding probability for Z2 = -1.6 as derived from standard normal distribution table 0.05480

Thus ,
Probability of defective output

= Probability that dimension of output will be more than 110 + Probability that dimension of output will be less than 88

= 0.00256 + 0.05480

= 0.05736

PROBABILITY OF DEFECTIVE OUTPUT = 0.05736

PROCESS CAPABILITY INDEX = 0.7333

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