Pin Acceptance

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 inches. 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. The 0.02 inch is known as the tolerance.

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.

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 re-engineering the process.

Following are the parameters

SD= 0.018 ,

Mean = 1.012,

Targeted Length = 1 inch,

Permissible error 0.02 inch both side.

The importance of the standard normal distribution derives from the fact that any normal random variable may be transformed to the standard normal random variable.

What percentage of the pins will be acceptable to the consumer

In this case the upper acceptable limit is 1.02 inch and lower acceptable limit is 0.98 inch. Customer will accept the pins which are falling in this range. By using excel templates for normal distribution we can find out the probability of pins in this range. This would be the percentage acceptable to customer-

P(x1