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.

tube or goog would be a great resource to with that question

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."