# Acceptable Pins Essay

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.

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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. 0343. 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 150*(1000x)^2 to reduce the standard deviation by x inch, the cost of setting the standard deviation to 0. 00624 (90% acceptance) is Rs. 20736.

Similarly, the cost of setting the standard deviation to 0. 00486 (95% acceptance) is Rs. 25899 and to 0. 00343 (99% acceptance) is Rs. 31843. However, if the mean is set to 1. 00 inch (incurring cost Rs. 80) and the standard deviation is then reduced, the maximum value of standard deviation that would make 90% of the parts acceptable is 0. 0121 and the total cost incurred would be Rs. 5302. Similarly, for making 95% of the parts acceptable, the standard deviation needs to be reduced to 0. 0102 which would incur total cost of Rs. 206. For making 99% of the parts acceptable, the standard deviation needs to be reduced to 0. 0077 which would incur a total cost of Rs. 15994. Based on the above calculations of costs involved, we can conclude that the most cost effective way of increasing pins acceptability is to adjust the mean to 1. 00 first and then attempt to reduce the standard deviation. The standard deviation should be reduced to at least 0. 0102 so that 95% of the pins are acceptable. The total cost required for these adjustments would be Rs. 9206.

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