Can I get help with this multiple step problem?

A process that is in control has a mean of # = 18.5 and a standard deviation of o = 0.8. (a) Compute the upper and lower control limits if samples of size 4 are to be used. UCL = LCL = Construct the x chart for this process. 21.0+ 21.0+ 21.0+ 20.5+ 20.5 + 20.5+ UCL 20.0+ UCL 20.0+ 20.0 19.5 19.5 + UCL 19.5 - 19.0- 19.0 19.0+ Sample Mean x 18.5 18.5 Sample Mean x 18.5 18.0 18.0 18.0- 17.5 Sample 17.5 LCL 17.5 17.0 LCL 17.0+ 17.0 16.5 16.5+ 16.5+ LCL 2 4 6 8 10 2 4 6 8 10 2 4 6 8 10 O Sample Number Sample Number Sample Number 21.0 + 20.5 + 20.0+ UCL 19.5- 19.0+ Sample Mean x 18.5 18.0 17.5 17.0 LCL 16.5 2 4 6 8 10 C Sample Number(b) Compute the upper and lower control limits if samples of size 8 are to be used. (Round your answers to three decimal places.) UCL = LCL = Construct the x chart for this process. 20.5 + 20.5 + 20.5 + 20.0+ 20.0+ UCL 20.0+ UCL UCL 19.5- 19.5 19.5- 19.0 19.0 19.0+ Sample Mean X 18.5 Sample Mean x 18.5 Sample Mean X 18.5 - 18.0 18.0 - 18.0 17.5 LCL 17.5 17.5 17.0 17.0 LCL 17.0+ LCL 4 8 10 2 4 6 8 10 2 4 6 8 10 C Sample Number OC Sample Number Sample Number 20.5+ 20.0 UCL 19.5- 19.0- Sample Mean X 18.5 18.0- 17.5 17.0 LCL N 4 6 8 10 O Sample Number Compute the upper and lower control limits if samples of size 16 are to be used. UCL = LCL =Construct the x chart for this process. 20.5+ 20.5+ 20.5+ 20.0+ 20.0+ 20.0+ 19.5 UCL 19.5 UCL 19.5 + 19.0 19.0 UCL 19.0- Sample Mean x 18.5 Sample Mean x 18.5 Sample Mean x 18.5 18.0 18.0 18.0 17.5 LCL 17.5 LCL 17.5 - LCL 17.0 17.0 17.0- 2 4 6 8 10 4 6 8 10 2 4 6 8 10 O Sample Number Sample Number OC Sample Number 20.5 + 20.0 19.5 UCL 19.0 Sample Mean x 18.5 18.0 17.5 17.0 LCL 2 4 6 8 10 C Sample Number (c) What happens to the limits of the control chart as the sample size is increased? Discuss why this is reasonable. The limits become ---Select--- as n increases. If the process is in control, the larger samples should have ( ---Select--- C variance and should fall ( ---Select--- C # = 18.5