Question 354335: A supplier contract calls for a key dimension of a part to be between 1.95 and 2.05 cm. The supplier has determined that the standard deviation of its process which is normally distributed, is 0.10 cm.
a. If the mean is 1.98, what fraction of parts will meet specifications?
b. If the mean is adjusted to 2.00, what fraction of the parts will meet specifications?
c. How small must the standard deviation be to ensure that no more than 2% of parts are nonconforming, assuming the mean is 2.oo?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A supplier contract calls for a key dimension of a part to be between 1.95 and 2.05 cm. The supplier has determined that the standard deviation of its process which is normally distributed, is 0.10 cm.
a. If the mean is 1.98, what fraction of parts will meet specifications?
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z(1.95) = (1.95-1.98)/0.1 = -.3000
z(2.05) = (2.05-1.98)/0.1 = 0.7
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P(1.95< x <1.98) = P(-0.3 < z < 0.7) = normalcdf(-0.3,0.7) = 0.3759
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b. If the mean is adjusted to 2.00, what fraction of the parts will meet specifications?
Same procedure as used in part a.
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c. How small must the standard deviation be to ensure that no more than 2% of parts are nonconforming, assuming the mean is 2.oo?
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z-value with a left-tail of 1% and a right tail of 1%
invNorm(0.01) = +/-2.3264
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Solve ::
(1.95-2)/s = -2.3264
s = 0.0215
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(2.05-2)/s = 2.3264
s = 0.0215
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Cheers,
Stan H.
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