Problem 10-21 Design specifications require hat a key dimension on a product mea

Question

Problem 10-21 Design specifications require hat a key dimension on a product measure 106 ± 12 unit A process being considered or producang h s product has a standar de iation of e units. a. What can you say (quantitatively) regarding the process capabty?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. decimal places (0.####).) Probability of defective output

Explanation / Answer

Answer to question a :

Given are following values :

Upper Specification Limit = USL = 106 + 12 = 118 units

Lower Specification Limit = LSL = 106 – 12 = 94 Units

Process mean = m = 106 ( since process is centered with respect to specification )

Process Standard deviation = S = 5 units

Thus,

Process Capability Index , Cpk

= Minimize ( ( USL – m) / 3.S , ( m – LSL) /3.S)

= Minimize ( ( 118 – 106) / 3 x 5 , ( 106 – 94)/3x5))

= Minimize ( 12/15, 12/15)

= Minimize ( 0.8, 0.8)

= 0.8

PROCESS CAPABILITY INDEX = 0.8

Answer to question b :

The revised process average = m1 = 98

Therefore , Revised

Process Capability Index, Cpk

= = Minimize ( ( USL – m) / 3.S , ( m – LSL) /3.S)

= Minimize ( ( 118 – 98) / 3 x 5 , ( 98 – 94)/3x5))

= Minimize ( 20/15 , 4 /15)

= Minimize ( 1.333, 0.2666)

= 0.2666

Answer to question c :

An item will be defective if its dimension will be :

Let corresponding Z value for probability that dimension of output less than 94 = Z1

And

Z value for probability that dimension of output will be more than 118 = Z2

Therefore,

M1 + Z1xS = 94

Or, 98 + 5.Z1 = 94

Or, Z1 = - 4/5 = - 0.8

Corresponding value of probability for Z1 = - 0.8 as derived from standard normal distribution table will be 0.21186

Similarly,

M1 + Z2x S = 118

Or, 98 + 5.Z2 = 118

Or, Z2 = 4

Corresponding probability value for Z2 = 4 will be 0.999969

Thus probability of dimension less than 118 units = 0.999969

Hence probability of dimension more than 118 ( defective item ) = 1 – 0.999969 =0.000031

Hence probability of defective output after process shift

= Probability that dimension is less than 98 units + Probability that dimension is more than 118 units = 0.21186 + 0.000031 = 0.211891

PROBABILITY OF DEFECTIVE OUTPUT = 0.211891

PROCESS CAPABILITY INDEX = 0.8

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