Product specifications call for a part at Vaidy Jayaraman's Metalworks to have a length of 1.100" ±.070". Currently, the process is performing at a grand average of 1.100" with a standard deviation of 0.015". Calculate the capability index of this process. The Cpk of this process is Is the process "capable"? O No O Yes (round your response to two decimal places).
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- Linda Boardman, I nc., an equipment manufacturerin Boston, has submitted a sample cutoff valve to improve yourmanufacturing process. Your process engineering department basconducted experiments and found that the valve has a mean (J.L)of 8.00 and a standard deviation (cr) of .04. Your desired performanceis f..L = 8.0 ±3cr, where cr = .045. What is the Cpk of theBoardman valve?< Refer to Table S6.1 - Factors for Computing Control Chart Limits (3 sigma) for this problem. Sampling 4 pieces of precision-cut wire (to be used in computer assembly) every hour for the past 24 hours has produced the following results: Hour X 1 2 3 4 5 6 R 3.25" 0.65" 3.00 1.23 3.32 1.43 3.39 1.26 3.07 1.17 2.96 0.37 Hour X R Hour R Hour 7 2.95" 0.48" 13 X 3.11" 0.90" 2.73 1.36 19 20 8 2.55 1.18 14 9 0.71 3.12 1.11 21 3.12 2.85 15 16 10 1.38 2.74 0.50 22 11 2.83 1.22 17 2.86 1.38 23 12 3.07 0.45 18 2.64 1.29 24 X 4.51" 2.89 2.65 3.18 3.04 2.54 R 1.56" 1.14 1.13 0.51 1.58 1.02 Based on the sampling done, the control limits for 3-sigma x chart are (round all intermediate calculations to three decimal places before proceeding with further calculations): Upper Control Limit (UCL) = inches (round your response to three decimal places).24. Each of the processes listed is non-centered with respect to the specifications for that process. Compute the appropri ate capability index for each, and decide if the process is capable. Process Mean Standard Deviation Lower Spec Upper Spec H 15.0 0.32 14.1 16.0 K 33.0 1.00 30.0 36.5 18.5 0.40 16.5 20.1 25. An appliance manufacturer wants to contract with a repair shop to handle authorized repairs in Indianapolis. The company has set an acceptable range of repair time of 50 minutes to 90 minutes. Two firms have submitted bids for the work. In test trials, one firm had a mean repair time of 74 minutes with a standard deviation of 4.0 minutes and the other firm had a mean repair time of 72 minutes with a standard deviation of 5.1 minutes. Which firm would you choose? Why?
- 1. The data shown in Table 1 are x and R values for 20 samples of size n= 5 taken from a process producing bearings. The measurements are made on the inside diameter of the bearing, with only the last three decimals recorded (i.e., 31.6 should be 0.50316). Please show all your work for full credit. (a) Set up x and R charts on this process. Does the process seem to be in statistical control? If necessary, revise the trial control limits. (b) Assume that diameter is normally distributed. Estimate the process standard deviation. Sample R Sample R 1 31.6 4 11 29.8 4 33.0 3 12 34.0 4 35.0 4 13 33.0 10 4 32.2 4 14 34.8 4 5 33.8 38.4 31.6 15 35.6 7 3 16 30.8 7 4 17 33.0 5 8 36.8 10 18 31.6 3 9. 35.0 15 19 28.2 9 10 34.0 6 20 33.8 Table 1: Bearing Diameter Data6. The defect rate for your product has historically been about 4.50%. For a sample size of 500, the upper and lower 3-sigma control chart limits are: Part 2 UCLp = _______ (enter your response as a number between 0 and 1, rounded to four decimal places). The Standard deviation of defect rate of product is given __________. The Upper and Lower 3-sigma control charts limits are: UCLp ________ LCLp ________1) Titan Industries manufactures parts for an aircraft company. It uses a computerized numerical controlled (CNC) machining center to produce a specific part that has a design (nominal) target of 1.275 inches with tolerances of +/- 0.024 inch. What are the Lower Specification Limit and the Upper Specification Limit? Show your work below.
- At Gleditsia Triacanthos Company, a certain manufactured part is deemed acceptable if its lengthis between 12.45 to 12.55 inches. The process is normally distributed with an average of 12.49inches and a standard deviation of 0.014 inches. A) Is the process capable of meeting specifications? B) Does the process meet specifications?3. Refer to Table S6.1 - Factors for Computing Control Chart Limits (3 sigma) LOADING... for this problem. Thirty-five samples of size 7 each were taken from a fertilizer-bag-filling machine at Panos Kouvelis Lifelong Lawn Ltd. The results were: Overall mean = 57.75 lb.; Average range R = 1.54 lb. a) For the given sample size, the control limits for 3-sigma x chart are: Part 2 Upper Control Limit (UCLx) = _______ lb. (round your response to three decimal places). Lower Control Limit (LCLx) = _______ lb. (round your response to three decimal places). B. the control limits for 3-sigma r chart are: Upper Control Limit (UCLr)= _______ lb. (round your response to three decimal places). Lower Control Limit(LCLr)= _______ lb. (round your response to three decimal places).Refer to Table S6.1 - Factors for Computing Control Chart Limits (3 sigma) LOADING... for this problem. Thirty-five samples of size 7 each were taken from a fertilizer-bag-filling machine at Panos Kouvelis Lifelong Lawn Ltd. The results were: Overall mean = 60.75 lb.; Average range R = 1.64 lb. a) For the given sample size, the control limits for 3-sigma x chart are: Upper Control Limit (UCLx) = nothing lb. (round your response to three decimal places).
- Auto pistons at Wemming Chung's plant in Shanghai are produced in a forging process, and the diameter is a critical factor that must be controlled. From sample sizes of 10 pistons produced each day, the mean and the range of this diameter have been as follows: a) What is the value of x? x= 156.76 mm (round your response to two decimal places). b) What is the value of R? Day 1 2 3 4 5 Mean x (mm) 158.9 155.2 155.6 157.5 156.6 R = 4.40 mm (round your response to two decimal places). c) What are the UCL and LCL using 3-sigma? Upper Control Limit (UCL) = mm (round your response to two decimal places). Range R (mm) 4.2 4.4 4.3 4.8 4.3 ÇTwenty samples (K=20) of 200 observations (n = 200) were taken by an operator at a workstation in a production process. The number of defective items in each sample was recorded as follows. Sample Number of Defects 1. 12 18 10 4. 15 16 6. 19 7. 17 8 12 11 10 14 11 16 12 15 13 13 14 16 15 18 16 17 17 18 18 20 19 21 20 22 Management wants to develop ap-chart using 3-sigma limits. What is the Lower Control Limit (LCL)? (calculation 2 sig fig)Specifications for a metal shaft are much wider than the machine used to make shafts is capable of. Consequently, the decision has been made to allow the cutting tool to wear a certain amount before replacement. The tool wears at the rate of 0.004 centimeter per piece. The process has a natural variation, o, of 0.010 centimeter and is normally distributed. Specifications are 15.0 to 15.2 centimeters. A three-sigma cushion is set at each end to minimize the risk of output outside of the specifications. Wear rate 3σ 1 30 Starting Ending Lower specification process Upper specification process mean mean Number of shafts How many shafts can the process turn out before tool replacement becomes necessary? (See diagram.) (Do not round intermediate calculations. Round your final answer to the nearest whole number.) Number of shafts pieces