The acceptable tolerance on a shaft product is 9 ± 0.005 in. From past experience, it is known that σ= 0.003 in. What is the percentage of scrap parts?
Solution:-
P(scarp park) = P(X < 8.995 or X > 9.005)
= P(X < 8.995) + P(X > 9.005)
Z for 8.995 = (8.995-9)/0.003 = -1.667
P value = 0.047757
=> P(X < 8.995) = 0.047757 = 0.0478
Z for 9.005 = (9.005-9)/0.003 = 1.667
P value = 0.9522
P(X < 9.005) = 0.9522
P(X > 9.005) = 1-0.922 = 0.0478
=> P(Score park) = P(X < 8.995 or X > 9.005)
= P(X < 8.995) + P(X > 9.005)
= 0.0478 + 0.0478
= 0.0956
The acceptable tolerance on a shaft product is 9 ± 0.005 in. From past experience, it...
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