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.
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?
Answer:
Step 1 of 6:
(a)
In this question, we are asked to find the mean clearance.
Clearance is equal to one-half the difference between the diameters of the hole and the piston.
The diameter of the hole has mean 15 and standard deviation
.
The diameter of the piston has mean 14.88 and standard deviation
The diameter of both hole and the piston are normally distributed.
Let be the diameter of the hole and
be the diameter of piston.
Then we can write and
as,
If are independent random variables and
are constants, then
Mean will be linear combinations of these variables, hence we can write,
…….(1)
Likewise for standard deviation, we can write,
………(2)
The clearance is given by,
C =
Hence the mean of clearance,
(using equation 1)
0.06
Hence the mean of clearance is 0.06.
