Problem 9-22 (modified). A quality control inspector accepts shipments or 500 pr

Question

Problem 9-22 (modified). A quality control inspector accepts shipments or 500 precision.5" steel rods if the mean diameter of a sample of 81 falls between.4995" and .5005". Previous evaluations have established that the standard deviation for individual rod diameters is .003 What is the probability the inspector will accept an out-of-tolerance shipment having mu.5003? (Note: we aren't told the tolerance, but for simplicity assume that it is.0002 so that mu.5003 is out-of-tolerance. The acceptance standard of between.4995 and.5005" relates to the sample mean, not to the population mean mu.) Probability-0.7448X what is the probability the inspector will reject a near-perfect shipment having mu=.49997 Probability

Explanation / Answer

solution=

So far, I have used the formulas .0005/((.003/9)*sqrt(419/499)) and -.0005/((.003/9)*sqrt(419/499)) to get the z-values. By doing so, I get z-values of +/- 1.63694663, signifying that the probability that the sample mean is accepted is .9495 - .0505 = 0.899. I'm not sure how to proceed with that.

Basically i'm not sure how to incorporate accepting an out-of-tolerance shipment with mu=.5003 into this.

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