Tolerances for a new part require weights between 247.0 and 265.5 pounds. the part is made using a process that has a mean of 256.4 pounds with a standard deviation of 3.5 pounds. the process population is normally distributed. if the process can be adjusted so that it is centered, what will be the sigma capability for the adjusted process?

The sigma capability of the process is 5.2857

Step-by-step explanation:

The sigma capability of a process is defined by the number of times that the distance between the lower specification limit and the upper specification limit contains the value of the standard deviation. In this case:

(USL - LSL) / sd = (247 - 265.5) / 3.5 = 5.2857

Then, the sigma capability of the process is 5.2857

maybe i can

the answer is 150.6

step-by-step explanation:

you multiply .02 by 30 and get .6.

and then you multiply 5.00 by 30 and get 150.

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