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

Question

Problem 9-22 (modified). A quality control inspector accepts shipments of 500 precision .5" steel rods if the mean diameter of a sample of 64 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=  

What is the probability the inspector will reject a near-perfect shipment having mu=.4999?
Probability=  

Explanation / Answer

A) P(0.4995 < X < 0.5003)

= P((0.4995 - 0.5003)/(0.003 /sqrt(64)) < (x - mean)/(SD/sqrt(n)) < (0.5005 - 0.5003)/(0.003 /sqrt(64))

= P(-2.13 < Z < 0.53)

= P(Z < 0.53) - P(Z < -2.13)

= 0.7019 - 0.0166 = 0.6924

B) P(Z < 0.4995) =P((X - mean)/(SD/sqrt(n)) < (0.4995 - 0.4999)/(0.003/sqrt(64))

= P(Z < -1.07)

=0.1423

P(X > 0.5005) = P((X - mean)/(SD/sqrt (n) > (0.5005 - 0.4999)/(SD/Sqrt (64))

= P(Z > 1.6)

= 1 - P(Z < 1.6)

= 1 - 0.9452 = 0.0548

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