Requirements for the width of a tractor engine component are 23.23 +/- 0.612 millimeters. The current process produces components with an average width of 23.546 and a population standard deviation of 0.092. The process is normally distributed. If the control limits are set at +/- 2 standard deviations instead of +/- 3 standard deviations from the mean, what is the Cpk of this process?

Answer:The Cpk for the process is 1.61Step-by-step explanation:The formula for Cpk is:[tex]Cpk=min(\frac{USL-mean}{2*stdDev},\frac{mean-LSL}{2*stdDev} )[/tex]In this case in specific that we can test the capability of the process for 2 standard dev we modify the formula to adjust it:so for this formula:average: 23.546USL: 23.23+0.612=23.842LSL: 23.23-0.612=22.618standard dev=0.092So,[tex]Cpk=min(\frac{23.842-23.546}{2*0.092} , \frac{23.546-22.618}{2*0.092} )\\[tex]Cpk=min(1.61 ,5.04 )\\\\Cpk=1.61[/tex]\\\\[/tex]The Cpk for the process is 1.61, so is a good indicator even with the control of two standard dev.