Question

Auto pistons at Wamming 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 diamatar have been as follows Mean x Range R Day (mm) 156.9 153.2 153.6 157.5 158.6 4.2 4.8 4.8 4.3 a) What is the value of x? Xmm (round your response to two decimal places) b) What is the value of R? R-mm (round your response to two dscimal places. c) What are the UCL and LCL, using 3-sigma? Uppor Control Limit (mm (rund your respanse to two decimal placos) mm (round your respanss to two decimal piaces) Lower Control Limit (LCL mm (round your response to two decimel places). d) What are the UCLR and LCLR using 3-sigma? Upper Control Limit (UCLR)-mm(round your rosponse to two docimal places) Lower Control Limit (LCLRmm mm (round your response to two decimal places) e) If the true; diameter mean should be 155 mm and you want this as your center (nominal) line, what are the new UCLİ and LCL? Upper Control Limit (UCL.)- 「 mm (round your response to two decimal pleces). Lower Control Limit (LCLİ>. mm (round your rosponse to tvo decima,places)

Answer 1

a)

$\bar{\bar{X}}$ = 155.56

b)

$\bar{R}$ = 4.4

c)

For n = 10, A₂ = 0.31

$UCL_{X}=\bar{\bar{X}}+A_{2}\bar{R}$

$LCL_{X}=\bar{\bar{X}}-A_{2}\bar{R}$

d)

For n = 10

D₃ = 0.22

D₄ = 1.78

$LCL_{R}=D_{3}\bar{R}$

$UCL_{R}=D_{4}\bar{R}$

e)

$\bar{\bar{X}}$ = 155

For n = 10, A₂ = 0.31

$UCL_{X}=\bar{\bar{X}}+A_{2}\bar{R}$

$LCL_{X}=\bar{\bar{X}}-A_{2}\bar{R}$