An automatic lathe produces rollers for roller bearings, and statistical process control charts are used to monitor the process. The central line of the chart for the sample means is set at 8.50 and for the range at 0.31 mm. The process is in control, as established by samples of size 5. The upper and lower specifications for the diameter of the rollers are (8.50 + 0.25) and (8.50 - 0.25) mm, respectively. a. Calculate the control limits for the mean and range charts. b. If the standard deviation of the process distribution is estimated to be 0.13 mm, is the process capable of meeting specifications? Assume four-sigma perfor- mance is desired. c. If the process is not capable, what percent of the output will fall outside the specification limits?

a. To calculate the control limits for the mean chart:

Upper Control Limit (UCL) = Central Line + 2.66 x Range = 8.50 + 2.66 x 0.31 = 8.50 + 0.83 = 9.33

Lower Control Limit (LCL) = Central Line - 2.66 x Range = 8.50 - 2.66 x 0.31 = 8.50 - 0.83 = 7.67To calculate the control limits for the range chart:

Upper Control Limit (UCL) = D4 x Range = 2.114 x 0.31 = 0.656

Lower Control Limit (LCL) = D3 x Range = 0 x 0.31 = 0b. The process capability index is calculated as Cpk = min[(USL - X̄)/(3σ), (X̄ - LSL)/(3σ)], where USL is the upper specification limit, X̄ is the mean of the process, LSL is the lower specification limit, and σ is the standard deviation of the process.USL = 8.50 + 0.25 = 8.75

LSL = 8.50 - 0.25 = 8.25

X̄ = 8.50

σ = 0.13Cpk = min[(8.75 - 8.50)/(3 x 0.13), (8.50 - 8.25)/(3 x 0.13)] = min[1.92, 1.92] = 1.92Since the process capability index is greater than 1, the process is capable of meeting specifications at a four-sigma level.c. Since the process is capable of meeting specifications, the percent of the output that will fall outside the specification limits is zero.