Topics: Arithmetic mean, Standard deviation, Normal distribution, Statistics, Root mean square, Inch / Pages: 2 (386 words) / Published: Sep 28th, 2014

Case 5: Acceptable Pins
A company supplies pins in bulk to a customer. The company uses an automatic lathe to produce the pins. Due to many causes-vibrations, temperature, wear and tear, and the like-the lengths of the pins made by 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 ± 0.02 inch. In other words, the customer wants the length to be 1.00 inch but will accept up to 0.02 inch deviation on either side. This 0.02 inch is known as the tolerance.
1. What percentage of the pins will be acceptable to the consumer?
In order to improve percentage accepted, the production manager and the engineers discuss adjusting the population mean and standard deviation of the length of the pins.
2. If the lathe can be adjusted to have the mean of the lengths to any desired value, what should it be adjusted to? Why?
3. Suppose the mean cannot be adjusted, but the standard deviation can be reduced. What maximum value of the standard deviation would make 90% of the parts acceptable to the consumer? (Assume the mean to be 1.012.)
4. Repeat question 3, with 95% and 99% of the pins acceptable.
5. In practice, which one do you think is easier to adjust, the mean or the standard deviation? Why?
The production manager then considers the costs involved. The cost of resetting the machine to adjust the population mean involves the engineers' time and the cost of production time lost. The cost of reducing the population standard deviation involves, in addition to these costs, the cost of overhauling the machine and reengineering the process.
6. Assume it costs $150x2 to decrease the standard deviation by (x /1000) inch. Find the cost of reducing the standard deviation to the values found in question 3 and 4.
7. Now assume that the mean has been adjusted to the best value found in question 2 at a cost of $80. Calculate the reduction in

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