Given the data below, a lower specification of 62.6, and an upper specification of 101.8, what...

Given the data below, a lower specification of 62.6, and an upper specification of 101.8, what is the long term process performance (ppk)?

Data
66.06284
82.57716
78.64111
92.72893
76.18137
71.46201
76.24239
74.83622
69.87486
77.90479
82.39439
79.18856
84.34492
77.32829
80.50536
83.36017
97.34745
84.56226
87.95131
65.64412
70.73183
74.28879
89.07007
78.50745
77.51397
89.04946
73.75787
91.30598
87.12589
89.29855
81.398
86.52962
84.33249
80.48321
81.87089
83.54964
71.19464
80.02001
90.00112
82.29257
77.55125
88.07639
88.95467
83.92542
88.33509
84.36723
77.89679
82.38985
67.81415
80.68263
87.25767
81.1521
82.15546
72.52171
67.58353
86.11663
75.5958
69.29909
77.69888
88.10717
84.43768
76.63519
76.67074
73.78486
79.98661
72.25349
88.68449
87.50085
75.20974
83.26245
86.24998
82.80463
81.16292
81.38507
83.01762
80.03256
88.0504
79.60369
72.79961
76.64304
78.34641
76.24377
80.96636
82.47478
77.07063
84.55949
78.45641
86.03345
80.5294
81.23737
86.94495
70.80997
76.14143
90.86433
71.27545
63.78769
69.48347

Expert Solution

We calculate the following values using the data given in the above problem with the help of EXCEL SHEET.

Data sheet is longer ,so i unable to upload data file.

	Mean =	80.25095928
	Standard deviation =	6.703880411
	Lower Specification Limit or LSL =		62.6	(given)
	Upper Specification Limit or USL =		101.8	(given)

	*cpu=(USL-Mean)/(3standard deviation)**
	*cpl=(Mean-LSL)/(3standard deviation)**
	*process performance(ppk)=min(cpu,cpl)/(6standard deviation)**


	cpu =	1.071471019
	cpl =	0.877648874
	min(cpu,cpl) =	0.877648874
	ppk =	0.021819424

RESULT:
	long term process performance (ppk) for the given data with lower specification 62.6
	and upper specification 101.8 is equal to 0.0218194.

milcah answered 1 year ago

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