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LeoInc2475
09.11.2019 •
Mathematics
Suppose that engineering specifications on the shelf depth of a certain slug to be turned on a cnc lathe are from .0275 in. to .0278 in. and that values of this dimension produced on the lathe can be described using a normal distribution with mean μ and standard deviation σ . (a) if μ = .0276 and σ = .0001, about what frac- tion of shelf depths are in specifications? (b) what machine precision (as measured by σ ) would be required in order to produce about 98% of shelf depths within engineering spec- ifications (assuming that μ is at the midpoint of the specifications)?
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Ответ:
0.8188 or 82%
0.0000643
Step-by-step explanation:
Given that engineering specifications on the shelf depth of a certain slug to be turned on a CNC lathe are from .0275 in. to .0278 in.
If x is the dimension then X is N(0.0276, 0.0001)
Or
is Normal (0,1)
a) Fraction of shelf depths are in specifications
=![P(0.0275](/tpl/images/0366/7286/5a1e8.png)
b) For 98% to be within the specifications,
Margin of error =![0.0278-0.02775\\=0.00015](/tpl/images/0366/7286/57bed.png)
For 98% critical value Z two tailed is 2.33
Hence std error =![\frac{0.00015}{2.33} \\=0.0000643](/tpl/images/0366/7286/49895.png)
So sigma should be = 0.0000643
Ответ: