Answer:
Kindly check explanation
Step-by-step explanation:
Xbar = 0.3015 ; Rbar = 0.00154
USL = 0.295 ; LSL = 0.305
For a subgroup size of 3 ; k = 3 ; d2 = 1.693
The mean = xbar = 0.3015 mm
The standard deviation = Rbar / d2 = 0.00154 / 1.693 = 0.0009096
b) Suppose that parts below the LSL can be reworked, but parts above the USL must be scrapped. Estimate the proportion of scrap and rework produced by this process.
P(X < 0.295)
(x - mean) / standard deviation
(0.295 - 0.3015) / 0.0009096 = 0.00003
0.00003 should be reworked.
P(X > 0.305)
1 - P(X < 0.305)
(x - mean) / standard deviation
(0.305 - 0.3015) / 0.0009096 = 3.847
1 - P(Z < 3.847)
1 - 0.99994
= 0.00006
0.00006 should be reworked.
c) Suppose that the mean of this process can be reset by fairly simple adjustments. What value of the process mean would you recommend? Estimate the proportion of scrap and rework produced by the process at this new mean.
Recommended mean = 0.3000 mm
P(X < 0.295)
(x - mean) / standard deviation
(0.295 - 0.3000) / 0.0009096 = 0.00003
0.00003 should be reworked.
P(X > 0.305)
1 - P(X < 0.305)
(x - mean) / standard deviation
(0.305 - 0.3000) / 0.0009096 = 5.497
1 - P(Z < 5.497)
1 - 0.99997
= 0.00003
0.00003 should be reworked.
