I'm sorry but I disagree: The question is regarding the traditional limits and not regarding the "very tight process control". Therefore the basis for the calculation is 3 sigma.
--> The distance from the mean to th UCL is 3 sigma.
--> The distance from the mean to the LCL is -3 sigma.
The question now asks for the distance between LCL and UCL: This is 3 sigma + 3 sigma = 6 sigma.
cnppmp
Tue, 06/11/2013 - 07:06
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C
C...
Normmaly or traditionally the Control chart would have UCL and LCL at 3 Sigma level. Universally followed.
Regards
CN Patil
san
Tue, 06/11/2013 - 07:12
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Yes, I too answered as C. But
Yes, I too answered as C. But as per key it is B(Farndales)
maui01
Tue, 06/11/2013 - 07:24
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B
Yes, it is B: 3 Sigma in both directions.
--> 2 * 3 Sigma = 6 Sigma
cnppmp
Tue, 06/11/2013 - 13:29
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I think I have read the
I think I have read the question incorrectly..it is talking about very tight controls...in that case it should be 6 sigma UCL & LCL.
Regards
CN Patil
maui01
Wed, 06/12/2013 - 08:15
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Solution B
I'm sorry but I disagree: The question is regarding the traditional limits and not regarding the "very tight process control". Therefore the basis for the calculation is 3 sigma.
--> The distance from the mean to th UCL is 3 sigma.
--> The distance from the mean to the LCL is -3 sigma.
The question now asks for the distance between LCL and UCL: This is 3 sigma + 3 sigma = 6 sigma.
Regards
Martin
jvleminc
Thu, 07/04/2013 - 22:15
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B
Tricky question, but I think B too (2*3 = 6) .