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?
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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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