A company has just negotiated a contract to produce a part for another firm. In the process of manufacturing the part, the inside diameter of successive parts becomes smaller and smaller as the cutting tool wears. However, the specs are so wide relative to machine capabilities that it is possible to set the diameter initially at a large value and let the process run for a while before replacing the cutting tool. The inside diameter decreases at an average rate of .001 cm per part, and the process has a standard deviation of .05 cm. The variability is approximately normal. Assuming a three-sigma buffer at each end, how frequently must the tool be replaced if the process specs are 3 cm and 3.5 cm?

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Solution to Three Sigma:

The sigma in this question is 0.5cm

To get the mean of the start process you have tosubtract 3 sigma from the upper and lower specs limits.

In our case 3 sigma is 3 by 0.05 cm (sigma) = 0.15 cm

So

The mean for the lower spec would be 3 cm + 0.15 cm= 3.15 cm (lower spec mean)

The mean for the upper spec would be 3.15cm – 0.15cm= 3 cm (Upper spec mean)

To get the parts before replacement

= (Upper spec mean – Lower spec mean)/ Rate of inside diameter decrease

= (3.15 – 3)/ 0.001cm

= 150Related: Step by step on how to do three sigma calculation

Answer:The tool should be replaced after producing 150 parts.Was this answer helpful? Ask your question