The manufacturing tolerance for the length of steel bars produced by a mill is a 3.6 mm. If the mean length is 1200 mm and the standard deviation is 2 mm, what is the expected percentage of acceptable bars?
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Given : The manufacturing tolerance for the length of steel bars produced by a mill is a 3.6 mm.
mean length is 1200 mm and the standard deviation is 2 mm
To Find : expected percentage of acceptable bars?
Solution:
Assuming tolerances both sides
Required length 1200 ± 1.8 m
Z score = ( Value - Mean)/SD
Z score = ± 1.8/2 = ± 0.9
- 0.9 = 18.41 %
0.9 = 81.59 %
81.59 - 18.41 = 63.18 %
expected percentage of acceptable bars = 63.18 %
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