Normal Distribution -Acceptable Pins

#1
A company supplies pins in bulk to a customer. The company uses an automatic lathe to produce the pins. The machine are normally distributed with a mean of 1.012 inches and a standard deviation of 0.018 inch. The customer will buy only those pins with lengths in the interval 1.00 + or minus 0.02 inch. The customer wants the length to be 1.00 inch but will accept up to 0.02 inch deviation on either side.

Question - If the lathe can be adjusted to have the mean of the lengths to any desired value, what should it.
be adjusted to? Why?

Please help I am stuck answering this question.

I have attempted to answer the problem. I have 1.00 as the answer.
This appears to result in the lowest SSD and MSD. I am just not sure that I have done it right.

.99 - 1.00 div. 0.018 and
1.02 - 1.00 div. 0.018

Resulting in .7335
 
Last edited:

JohnM

TS Contributor
#2
The lathe should be adjusted to produce pins with length 1.00 inch --> centering the lathe in the customer's specifications will minimize the number of pins that are outside of the customer's specs.

No math is really necessary here - since a normal distribution is symmetrical, in order to minimize the tail areas outside the specifications, you should center the process right in the middle of the specs.