Question 1113053: Your lumber company has bought a machine that automatically cuts lumber. The seller of the machine claims that the machine cuts lumber to a mean length of 7 feet (84 inches) with a standard deviation of 0.7 inch. Assume the lengths are normally distributed. You randomly select 45 boards and find that the mean length is 84.15 inches. Complete parts (a) through (c).
(a) Assuming the seller's claim is correct, what is the probability that the mean of the sample is 84.15 inches or more?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Your lumber company has bought a machine that automatically cuts lumber. The seller of the machine claims that the machine cuts lumber to a mean length of 7 feet (84 inches) with a standard deviation of 0.7 inch. Assume the lengths are normally distributed. You randomly select 45 boards and find that the mean length is 84.15 inches.
Complete parts (a) through (c).
(a) Assuming the seller's claim is correct, what is the probability that the mean of the sample is 84.15 inches or more?
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z(84.15) = (84.15-84)/(0.7/sqrt(45) = 1.437
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P(u >=84.15) = P(z >= 1.437) = normalcdf(1.437,100) = 0.0753
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Cheers,
Stan H.
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