Please show the process and equation for \"C\" esign speci cations require that

Question

Please show the process and equation for "C"

esign speci cations require that a key dimension on a product measure 104 ± 14 units. A rocess bein considere tor producir this product as a standard deviation o se en units a. What can you say (quantitatlvely) regarding the process capablity? Assume that the process is centered wth respect to spectfications. (Round your answer to 4 declmal places.) Process capability index 0.6657 b. Suppose the process average shifts to 95. Calculate the new process capability. (Round your answer to 4 decimal places.) New process capability index 2380 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 Probability of detective output

Explanation / Answer

When the process average shifts to 95 :

Upper specification Limit of the product = USL = 104 + 14 = 118

Lower specification Limit of the product = LSL = 104 – 14 = 90

Process average = m = 95

Process standard deviation = Sd = 7

An output will be considered defective if :

Let Z value corresponding to probability that output will be maximum 90 units = Z1

Therefore ,

M + Z1 xSd = 90

Or, 95 + 7,Z1 = 90

Or, Z1 = ( 90 – 95)/ 7

Or, Z1 = - 0.71 ( rounded to 2 decimal places )

Corresponding probability value for Z1 of – 0.71 as derived from normal distribution table 0.23885

Let Z value corresponding to probability that output will be maximum 118 units = Z2

Therefore ,

M + Z2 x Sd = 118

Or, 95 + 7x Z2 = 118

Or, 7.Z2 = 118 – 95

Or Z2 = 23/7

Or, Z2 = 3.29 ( rounded to 2 decimal places )

Taking help of standard normal table as well as NORMSINV ( ) function , probability corresponding to Z value of 3.29 = 0.99950

Therefore probability that output is more than 118

= 1 – Probability that output will be maximum 118

= 1 – 0.99950

= 0.0005

Hence probability of defective output

= Probability of process output < 90 + Probability of process output > 118

= 0.23885 + 0.0005

= 0.23935

PROBABILITY OF DEFECTIVE OUTPUT = 0.23935

PROBABILITY OF DEFECTIVE OUTPUT = 0.23935

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