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.16.516.5-16.516.516.0-16.0-16.0-16.0-15.515.5-UCL15.515.5-UCLUCL15.0UCL15.015.015.014.514.514.514.514.0-14.014.014.0-LCLLCL13.513.5LCL13.5-13.5LCL13.0+13.013.0+13.0-24610246.810246810246810Sample NumberSample NumberSample NumberSample 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-- vvariance and should fall ---Select--- vu = 14.5.Sample Mean XSample Mean XSample Mean XSample Mean X A process that is in control has a mean of u = 14.5 and a standard deviation of o = 0.6.(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.17.0f16.516.0-15.515.014.514.013.517.0-16.516.0-15.515.017.0-16.5-* 16.0-15.515.017.0-16.5-* 16.0UCLUCLUCL15.515.0UCL14.514.013.5-* 13.012.514.514.013.513.014.514.013.5-* 13.0LCLLCLLCL13.0-12.5+LCL12.5-12.56 8481021024681024 6 810eticatheerttheSample NumberSample NumberSample NumberSample 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.16.5+16.0-16.5-16.516.516.0-16.0-16.0-UCLUCLUCL15.515.515.515.5UCL15.015.0-15.015.014.514.514.514.514.0-14.0-14.0-14.013.513.513.513,5-LCLLCLLCLLCL13.0-13.013.013.0-6 86 86 82410210468.102410Sample NumberSample NumberSample NumberSample NumberCompute the upper and lower control limits if samples of size 16 are to be used.UCL =LCL =Sample Mean XSample Mean XSample Mean XSample Mean xSample Mean XSample Mean XSample Mean XSample Mean X

The X -bar chart is a type of SHEWART control chart that is used to monitor the arithmetic means…

process that is in control has a mean of u = 14.5 and a standard deviation of o = 0.6. (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. 17.0 16.5 17.0 16.5 * 16.0 15.5 15.0 14.5 14.0 13.5 13.0 12.5 17.0 16.5 * 16.0- 15.5 15.0 14.5 17.0 16.5 * 16.0 E 15.5- UCL 16.0 15.5 15.0 UCL. UCL UCL 14.5 14.0 13.5 i 13.0 12.5 14.0 13.5 13.0 12.5 15.0 14.5 214.0- 13.5 A 13.0- 12.5 LCL LCL LCL. LCL 4 8. 10 4 10 2 8 10 2 6 10 Sample Number Sample Number Sample Number 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. 16.5 16.0- 16.5 16.0- 16.5 16.0- 15.5 16.5 16.0 UCL UCL UCL 15.5 15.5 15.5 UCL. 15.0 15.0 15.0 15.0 14.5 14.0 13.5 13.0 14.5 14.5 14.5 14.0 14.0 14.0 13.5 13.0 13.5 13.5 LCL LCL LCL LCL. 13.0- 13.0 2 4 6. 10 2 4 6 10 4 8 10 10 Sample Number Sample Number Sample Number Sample Number Compute the upper and lower control limits if samples of size 16 are to be used. UCL - Sample Mean x Sample Mean X Sample Mean X Sample Mean X Sample Mean X Sample Mean X Sample Mean X

process that is in control has a mean of u = 14.5 and a standard deviation of o = 0.6. (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. 17.0 16.5 17.0 16.5 * 16.0 15.5 15.0 14.5 14.0 13.5 13.0 12.5 17.0 16.5 * 16.0- 15.5 15.0 14.5 17.0 16.5 * 16.0 E 15.5- UCL 16.0 15.5 15.0 UCL. UCL UCL 14.5 14.0 13.5 i 13.0 12.5 14.0 13.5 13.0 12.5 15.0 14.5 214.0- 13.5 A 13.0- 12.5 LCL LCL LCL. LCL 4 8. 10 4 10 2 8 10 2 6 10 Sample Number Sample Number Sample Number 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. 16.5 16.0- 16.5 16.0- 16.5 16.0- 15.5 16.5 16.0 UCL UCL UCL 15.5 15.5 15.5 UCL. 15.0 15.0 15.0 15.0 14.5 14.0 13.5 13.0 14.5 14.5 14.5 14.0 14.0 14.0 13.5 13.0 13.5 13.5 LCL LCL LCL LCL. 13.0- 13.0 2 4 6. 10 2 4 6 10 4 8 10 10 Sample Number Sample Number Sample Number Sample Number Compute the upper and lower control limits if samples of size 16 are to be used. UCL - Sample Mean x Sample Mean X Sample Mean X Sample Mean X Sample Mean X Sample Mean X Sample Mean X

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Times for a surgical procedure are normally distributed. There are two methods. Method A has a mean of 31 minutes and a standard deviation of 2 minutes, while method B has a mean of 35 minutes and a standard deviation of 1.0 minutes. (a) Which procedure is preferred if the procedure must be completed within 30 minutes? O Method A O Method B O Either Method (b) Which procedure is preferred if the procedure must be completed within 38.0 minutes? O Either Method O Method A O Method B (c) Which procedure is preferred if the procedure must be completed within 37 minutes? O Method B O Either Method O Method A < Prev 19 of 19 Next to search 22 立

* Question Completion Status: OC. 0.0099 d. 0.0918 QUESTION 3 EPA fuel economy estimates for automobile models tested recently predicts a mean of 24 mpg and a standard deviation of 6 mpg. Assume that a Normal model applies, N(24, 6 ). What mpg for the automobiles would be the cutoff value for the lowest 10% of the mpg's (the 10th percentile of gas mileage)? O a. 12.80 mpg O b. 10.00 mpg O c. 31.38 mpg O d. 16.32 mpg QUESTION 4
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Paramesan cheese is aged for an average of 24 months to allow it to develop the desire richness in flavor the standard deviation for the length of time Parmesan cheese is aged is 3.7 months How long would Parmesan cheese have to be aged in order to be in the 90th percentile? Assume that the data is normally distributed
I attached.
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