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?

The Cpk for the process is 1.61

Step-by-step explanation:

The formula for Cpk is:

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: 23.546

USL: 23.23+0.612=23.842

LSL: 23.23-0.612=22.618

standard dev=0.092

So,

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The Cpk for the process is 1.61, so is a good indicator even with the control of two standard dev.

goog may have it for you

i got 142 as my answer

step-by-step explanation: