Shafts manufactured for use in optical storage devices have diameters that are normally distributed with mean µ = 0.652 cm and standard deviation ? = 0.003 cm. The specification for the shaft diameter is 0.650 ± 0.005 cm.

Question

a. What proportion of the shafts manufactured by this process meet the specifications?
b. The process mean can be adjusted through calibration. If the mean is set to 0.650 cm, what proportion of the shafts will meet specifications?
c. If the mean is set to 0.650 cm, what must the standard deviation be so that 99% of the shafts will meet specifications?

Answer 1

(a) The specification interval is 0.645 to 0.655 cm.
The z-score of 0.645 is (0.645 ? 0.652)/0.003 = ?2.33.
The z-score of 0.655 is (0.655 ? 0.652)/0.003 = 1.00.
The area between z= ?2.33 and z= 1.00 is 0.8413 ? 0.0099 = 0.8314.
Therefore 83.14% of shafts will meet the specification.

(b) The z-score of 0.645 is (0.645 ? 0.650)/0.003 = ?1.67.
The z-score of 0.655 is (0.655 ? 0.650)/0.003 = 1.67.
The area between z= ?1.67 and z= 1.67 is 0.9525 ? 0.0475 = 0.9050.
Therefore 90.50% of shafts will meet the specification.

(c) Let?be the required standard deviation. Since the mean of 0.650 is in the center of the specification interval,the proportion of diameters that are too thin will be equal to the proportion of diameters that are too thick.Therefore ?must be chosen so that the proportion of diameters that are greater than 0.655 is 0.005.
The z-score for which 0.005 of the area is to the right is approximately z= 2.58. Therefore we find ?so that the z-score of 0.655 is 2.58 by solving 2.58 = (0.655 ? 0.650)/?, so ?= 0.00194 cm.Note that if z= 2.57 is used instead of z= 2.58, then ?= 0.00195 cm.