A cylindrical hole is drilled in a block, and a cylindrical piston is placed in the hole. The clearance is equal to one-half the difference between the diameters of the hole and the piston. The diameter of the hole is normally distributed with mean 15 cm and standard deviation 0.025 cm, and the diameter of the piston is normally distributed with mean 14.88 cm and standard deviation 0.015 cm. The diameters of hole and piston are independent.
a. Find the mean clearance.
b. Find the standard deviation of the clearance.
c. What is the probability that the clearance is less than 0.05 cm?
d. Find the 25th percentile of the clearance.
e. Specifications call for the clearance to be between 0.05 and 0.09 cm. What is the probability that the clearance meets the specification?
f. It is possible to adjust the mean hole diameter. To what value should it be adjusted so as to maximize the probability that the clearance will be between 0.05 and 0.09 cm?
Let X be the diameter of a hole, and let Y be the diameter of a piston. Then
The clearance is given by C= 0.5X? 0.5Y.
(a) 
(b) 
(c) Since Cis a linear combination of normal random variables, it is normally distributed, with 
as given in parts (a) and (b).
The z-score of 0.05 is (0.05 ? 0.06)/0.01458 = ?0.69. The area to the left of z= ?0.69 is 0.2451.
Therefore P(C<0.05) = 0.2451.
(d) The z-score of the 25th percentile is ? ?0.67.
The 25th percentile is therefore ? 0.06 ? 0.67(0.01458) = 0.0502 cm.
(e) The z-score of 0.05 is z= (0.05?0.06)/0.01458 = ?0.69. The z -score of 0.09 is z= (0.09?0.06)/0.01458 = 2.06.
The area between z= (0.05-0.06)/0.01458= ?0.69. The z-score of 0.09 is z = (0.09-0.06)/0.01458= 2.06.
The area between z=0.69 and z=2.06 is 0.9803 ? 0.2451 = 0.7352.
Therefore P(0.05
(f) The probability is maximized when 
= 0.07cm , the midpoint between 0.05 and 0.09 cm. Now
To find the probability that the clearance will be between 0.05 and 0.09:
The z-score of 0.05 is z= (0.05 ? 0.07)/0.01458 = ?1.37.
The z -score of 0.09 is z= (0.09 ? 0.07)/0.01458 = 1.37.
The area between z= ?1.37 and z= 1.37 is 0.9147 ? 0.0853 = 0.8294.
Therefore P(0.05
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