Answer:
1. Mean = r bar = 0.0154mm
Standard deviation = 0.0009096
2. Proportion of rework produced = 30ppm or 0.0003%
Proportion of scrap = 60ppm or 0.0006%
3 recommended value = 0.3000
Explanation:
The mean = x bar = 0.3015mm
Standard deviation = r bar/d2
R bar = 0.0154 mm
d2 = 1.693mm
= 0.0154mm/1.693mm
= 0.0009096mm
B. Lower specification 0.295
P(x<0.295)
= ϕ(0.295-0.3015/0.0009096)
This gives us 0.00003
So we conclude that 30 ppm or 0.0003% of the parts produced by the process would be reworked
For scrap,
X>0.305 upper specifications
1-ϕ(0.305-0.3015/0.0009096)
= 1-0.99994
= 0.00006
So we conclude that 60 ppm or 0.0006% of the parts produced by the process would be scrapped.
c. to get the recommended value of the process mean
= (0.305+0.295)/2
= 0.6/2
= 0.3000
the recommended value would be 0.3000