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Home Q & A Board Science-Related Homework Help Mathematics Statistics and Probability
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Catracho Catracho
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Posts: 529
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3 years ago
Auto pistons at Wemming​ Chung's plant in Shanghai are produced in a forging​ process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each​ day, the mean and the range of this diameter have been as​ follows

a) What is the value of x?
- mm (round your response to two decimal places)

b) What IS the value of R?
mm (round your response to two decimal places).

c) What are the UCL- and LCL- using 3.sigma?
Upper Control Limit (UCL-) = C] mm (round your response to two decimal places).
Lower Control Limit (LCL-) = mm (round your response to two decimal places).

d) What are the UCLR and LCLR using 3-sigma?
Upper Control Limit (UCLR) = mm (round your response to two decimal places)
Lower Control Limit (LCL R) = mm (round your response to two decimal p/aces)_

e) If the true diameter mean should be 155 mm and you want this as your center (nominal) line, what are the new
UCL- and LCL-?
Upper Control Limit (UCL-) = mm (round your response to avo decimal places).
Lower Control Limit (LCL- ) = mm (round your response to two decimal places).
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#1
Educator
3 years ago Edited: 3 years ago, bio_man
(a to c)

(d) attached
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#2
Staff Member
3 years ago
All answers here, the numbers might be different.

Bio_man you can use this document to compare your answers
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- Master of Science in Biology
- Bachelor of Science (Biology)
- Bachelor of Education
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#3
Educator
3 years ago
Quote from: Catracho (3 years ago)
e) If the true diameter mean should be 155 mm and you want this as your center (nominal) line, what are the new UCL- and LCL-? Upper Control Limit (UCL-) = mm (round your response to avo decimal places). Lower Control Limit (LCL- ) = mm (round your response to two decimal places).

155 + (0.308)*4.56 = 156.40 UPPER
155 - (0.308)*4.56 = 153.60 LOWER
wrote...
#4
2 months ago
Thanks!
wrote...
#5
1 months ago
Thanks
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