Question

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 14. cm and standard deviation 0.02 cm, and the diameter of the piston is normally distributed with mean 13.88 cm and standard deviation 0.012 cm. 1 Specifications call for the clearance to be between 0.07 and 0.09 cm. What is the probability that the clearance meets the specification? 2) It was 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 0.07 and 0.09 cm? Show your explanations. Displaying only the final answer is not enough to get credit. Note: round calculated numerical values to the fourth decimal place where applicable.

Answer 1

Answer:

Given,

Mean = 1/2(u1 - u2)

= 14/2 - 13.88/2

= 0.06

Standard deviation = sqrt((1/4)*s1^2 + (1/4)*s2^2))

= sqr((1/4)*0.02^2 + (1/4)*0.012^2)

= 0.01166

a)

P(0.07 < X < 0.09) = P((0.07 - 0.06)/0.01166 < (x-u)/s < (0.09 - 0.06)/0.01166)

= P(0.86 < z < 2.57)

= P(z < 2.57) - P(z < 0.86)

= 0.994915 - 0.8051054 [since from z table]

= 0.1898

b)

u = 1/2(u1 - u2)

1/2(x - 13.88) = (0.07 + 0.09)/2

x = 14.04

So,

Clearance = (0.07 + 0.09)/2 = 0.08

P(0.07 < X < 0.09) = P((0.07 - 0.08)/0.01166 < z < (0.09 - 0.08)/0.01166)

= P(-0.86 < z < 0.86)

= P(z < 0.86) - P(z < - 0.86)

= 0.8051054 - 0.1948945 [since from z table]

= 0.6102