**Problem 10-21 (Algo)**A process that produces computer chips has a mean of 0.04 defective chips and a standard deviation of 0.003 chips. The allowable variation is from 0.032 to 0.048 defective.a. Compute the capability index (\(C_p\)) for the process. *(Round your intermediate calculations to 3 decimal places and final answer to 2 decimal places.)*- **Capability index:** [Input box]b. Is the process capable? *Hint: Use 1.33 as the standard capability index.*- ○ Yes- ○ No

Process capacity index (Cp) is a statistical tool which helps to identify whether a process capable…

Answered: A process that produces computer chips…

Practical Management Science

6th Edition

ISBN:9781337406659

Author:WINSTON, Wayne L.

Publisher:WINSTON, Wayne L.

Chapter2: Introduction To Spreadsheet Modeling

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Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...

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. For a process quality characteristic, the specification limits are 0.4037 +/- 0.0013 (that is from 0.4024 to 0.4050). The following data gives the data for 14 measurement data sets, each data set is a subgroup of 5 measurement. The value is recorded in the units of 0.0001 (that is, the specification limit will be considered 24 to 50) Based on the following data, compute Cp,Cpk,Pp and Ppk.(Use MINITAB is ok) and determine if the capability of the process. Subgroup# 1 2 3 4 5 Subgroup# 1 2 3 4 5 1 36 35 34 33 32 8 34 38 35 34 38 2 31 31 34 32 30 9 36 35 37 34 33 3 30 30 32 30 32 10 30 37 33 34 35 4 32 33 33 32 35 11 28 31 33 33 33 5 32 34 37 37 35 12 33 30 34 33 35 6 32 32 31 33 33 13 35 36 29 27 32 7 33 33 36 32 31 14 33 35 35 39 3
2) The specification for a Lindemann’s stainless steel chimney liner requires a thickness of 7.1 mm ± 0.1 mm. The firm’s quality analyst determines that the liner process operates at a mean thickness of 7.13 mm with the standard deviation of the process being 0.02 mm. a) Calculate the upper and the lower limits for the stainless steel chimney liner. b) Calculate the Cpand Cpkfor this process. Hint: Cp = (Upper Specification Limit – Lower Specification Limit)/6sigma Cpk = minimum of [(Upper Specification Limit – Xbar)/3sigma], [(Xbar – Lower Specification Limit)/3sigma]
A process sampled 20 times with a sample of size 8 resulted in = 28.5 and R = 1.8. Compute the upper and lower control limits for the chart for this process. (Round your answers to two decimal places.) UCL = LCL = Compute the upper and lower control limits for the R chart for this process. (Round your answers to two decimal places.) UCL = LCL =
A process screens a certain type of potash granule, resulting in a mean diameter of 0.067 cm and a standard deviation of 0.004 cm. The allowable variation in granule diameter is from 0.048cm to 0.087cm. Is the process capable? a. Cannnot be determined because sample size is not available b. Cannot be determined because USL and LSL information is not available c. Process is NOT capable d. Process is capable
Consider a process with lower process control limit as 1.338 and upper process control limit as 1.342 . The lower specification limit is 1.335 and the upper specification limit is 1.345. Utilizing the information provided, is the process is capable of meeting design specifications?
S Book Hint Print eferences Problem 13-2 (Algo) A metal fabricator produces connecting rods with an outer diameter that has a 1 ± 0.02 inch specification. A machine operator takes several sample measurements over time and determines the sample mean outer diameter to be 1.004 inches with a standard deviation of 0.006 inch. Calculate the process capability index for this example. (Round your answer to 3 decimal places.) Process capability index
Figure 12–1 presents a conceptual picture of the trade-off between process cost and the costs of losses due to poor quality. What are the costs of poor quality and what difficulties might arise when attempting to measure these costs?
A process is in statistical control with = 202,5 y s = 2,0. Specifications are atLSL = 196 and USL = 206.a) Estimate the process capability with an appropriate process capability ratio.b) What is the potential capability of this process?c) Items that are produced below the lower specification limit must bescrapped, while items that are above the upper specification limit can bereworked. What proportion of the process output is scrap and whatproportion is rework?d) Because scrap is more expensive than rework, the process has been centered closer to the upper specification limit. If scrap is twice as expensive as rework, is the process mean at the best possible location? What value of the process target would you recommend?
The central limit theorem:a) is the theoretical foundation of the c -chart.b) states that the average of assignable variations is zero.c) allows managers to use the normal distribution as thebasis for building some control charts.d) states that the average range can be used as a proxy forthe standard deviation.e) controls the steepness of an operating characteristiccurve.
Suppose that a quality characteristic is normally distributed with specifications at 100 ± 20. The process standard deviation is 6.(a) Suppose that the process mean is 100. What are the natural tolerance limits? Calculate Cp and Cpk and interpret these ratios. What fraction of parts do you expect to be defective?(b) Suppose that the process mean is 106. What are the natural tolerance limits? What is thefraction defective? Calculate Cp and Cpk and interpret these ratios.
A process generating attribute data has an estimated sigma capability = 3.3. Assuming the process mean is shifted +/- 1.5 sigma from the target value, what is the estimated dpmo? Estimated dpmo = ____.
When observations can be placed into oneof two categories. Examples include items(observations) that can be classified asa. Good or badb. Pass or failc. Operate or don’t operate

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