A quality-control inspector accepts shipments of 500 precision 1/2 cm (i.e. 0.50

Question

A quality-control inspector accepts shipments of 500 precision 1/2 cm (i.e. 0.5000 cm) steel rods if the mean diameter of a sample of n = 100 falls between 0.4995 cm and 0.5005 cm. The population standard deviation for individual rod diameters is known to be 0.00300 cm.

a. Determine the probability that the inspector will accept an almost out-of-tolerance shipment that actually has = 0.5003 cm.

b. Determine the probability that the inspector will (mistakenly) reject a near-perfect shipment that actually has = 0.4999 cm.

Explanation / Answer

A) P(0.4995 < x < 0.5005) = P((0.4995 - 0.5003)/(0.003/sqrt(100) < (x - mean)/(SD/sqrt (n)) < (0.5005 - 0.5003)/(0.003/sqrt(100))

= P(-2.67 < Z < 0.67)

= P(Z < 0.67) - P(Z < -2.67)

= 0.7486 - 0.0038 = 0.7448

B)  P(0.4995 < x < 0.5005) = P((0.4995 - 0.4999)/(0.003/sqrt(100) < (x - mean)/(SD/sqrt (n)) < (0.5005 - 0.4999)/(0.003/sqrt(100))

= P(-1.33 < Z < 2)

= P(Z < 2) - P(Z < -1.33)

= 0.9772 - 0.0918

= 0.8854

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