# Acceptable Pins

Pages: 2 (574 words) Published: August 3, 2010
Case Problem 2:Acceptable pins

Data Analysis

The lengths of the pins made by the automatic lathe 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, that is, between 0.98 inches to 1.02 inches. The probability of a pin having a length between 0.98 and 1.02 is 0.6339. Therefore, we can say that 63.39% of the pins produced will be acceptable to the customers. This is a very low acceptance level.

In order to improve the percentage accepted, either the mean or the standard deviation needs to be adjusted. In order to maximize the acceptable percentage of pins keeping the standard deviation same, the mean needs to be adjusted to a value that is exactly between the tolerance limits. This will ensure that the tolerance limit will cover the maximum area under the standard normal curve which in turn means that the acceptance probability will be maximum. Thus, the mean needs to be set to 1.00 inches. This will make the probability of creating a pin within the tolerance limit as 0.7335. Thus, 73.35% of the pins would be acceptable when the mean is set to 1.00.

If the mean cannot be adjusted, but the standard deviation can be reduced, the maximum value of standard deviation that would make 90% of the parts acceptable (probability = 0.90) is 0.00624 inches. For making 95% of the parts acceptable, the standard deviation needs to be reduced to 0.00486. Similarly, for making 99% of the parts acceptable, the standard deviation needs to be reduced to 0.00343.

In practice it is easier to adjust the mean as compared to standard deviation. While adjusting the mean, the total cost involves only the engineer’s time and the cost of the 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.

Assuming that it costs...