Reference no: EM131694008

Problem: A company supplies pins in bulk to a customer. The company uses an automatic lathe to produce the pins. Factors such as vibration, temperature, and wear and tear affect the pins, so that the lengths of the pins made by the machine are normally distributed with a mean of 1.008 inches and a standard deviation of 0.045 inch. The company supplies the pins in large batches to a customer. The customer will take a random sample of 50 pins from the batch and compute the sample mean. If the sample mean is within the interval 1.000 inch ± 0.010 inch, then the customer will buy the whole batch.

Question: 1. What is the probability that a batch will be acceptable to the consumer? Is the probability large enough to be an acceptable level of performance?

To improve the probability of acceptance, the production manager and the engineers discuss adjusting the population mean and standard deviation of the lengths of the pins.

2. If the lathe can be adjusted to have the mean of the lengths at 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 continues to be 1.008 inches.)

4. Repeat part 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 (x1,000) inch. Find the cost of reducing the standard deviation to the values found in parts 3 and 4.

7. Now assume that the mean has been adjusted to the best value found in part 2 at a cost of $80. Calculate the reduction in standard deviation necessary to have 90%, 95%, and 99% of the parts acceptable. Calculate the respective costs, as in part 6.

8. Based on your answers to parts 6 and 7, what are your recommended mean and standard deviation to which the machine should be adjusted?