6 7 Requirements for the width of a tractor engine component are 23.07+/-0.634 m

Question

6 7 Requirements for the width of a tractor engine component are 23.07+/-0.634 millimeters. The current process produces components with an average width of 23.500 and a population standard deviation of 0.079. The process is normally distributed. What is the Cok of this process? (Use four decimal places) 8) Requirements for the width of a tractor engine component are 23.61 -0.639 milimeters. The current process produces components with an average width of 23.538 and a population standard deviation of 0.053. 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 Cok of this process? (Use four decimal places)

Explanation / Answer

1. Cpk = min. [ USL-mean / 3sigma, mean -LSL/ 3sigma ]

Here USL = 23.07+0.634 = 23.704

LSL = 23.07-0.634 = 22.436

Cpk = min [ 23.704-23.5 / 3x0.079 , 23.5-22.436 / 3x0.079 ]

= min. [ 0.860, 4.489] = 0.860

8. 2 sigma limits of the control limits means that the width of USL -LSL band is equal to 4 sigma. The Cpk is a measure of the process spread across the mean that compares the distance of the nearest limit with the half of the process spread ( 3 sigma on each side of process spread). Thus value of cpk is given by

Cpk = min. [ USL-mean / 3sigma, mean -LSL/ 3sigma ]

USL = 24.249

LSL = 22.971

Mean =23.538

cpk = min[ 24.249-23.538 / 3x0.053 , 23.538 -22.971 / 3x0.053 ]

= min [ 4.471, 3.566 ] = 3.566

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