INDU6331 ASIGNMENTS All the assignments from one to seven are found in this package Unit being Advanced Quality Control All the questions have their workings(Answers) in each assignment.
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4.1 The inside diameters of bearings used in an aircraft landing gear assembly are known to have a standard deviation of σ= 0.002cm. A random sample of 15 bearings has an average inside diameter of 8.2535cm a. Test the hypothesis that the mean inside bearing diameter is 8.25cm Use a two-sided a...
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Add to cart4.1 The inside diameters of bearings used in an aircraft landing gear assembly are known to have a standard deviation of σ= 0.002cm. A random sample of 15 bearings has an average inside diameter of 8.2535cm a. Test the hypothesis that the mean inside bearing diameter is 8.25cm Use a two-sided a...
6.25 Samples of n= 5 units are taken from a process every hour. The x and R values for a particular quality characteristic are determined. After 25 samples have been collected, we calculate x avg.= 20 and R= 4.56 a. What are the three sigma control limits for x and R? n= 5 Three sigma control l...
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Add to cart6.25 Samples of n= 5 units are taken from a process every hour. The x and R values for a particular quality characteristic are determined. After 25 samples have been collected, we calculate x avg.= 20 and R= 4.56 a. What are the three sigma control limits for x and R? n= 5 Three sigma control l...
What are the three primary technical tools used for quality control and improvement? The three primary technical tools used for quality control and improvement are: 1. Statistical process control (SPC): On-line tool useful for process monitoring and reduction of variability. 2. Acceptance sampli...
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Add to cartWhat are the three primary technical tools used for quality control and improvement? The three primary technical tools used for quality control and improvement are: 1. Statistical process control (SPC): On-line tool useful for process monitoring and reduction of variability. 2. Acceptance sampli...
A mechatronic assembly is subjected to a final functional test. Suppose that defects occur at random in these assemblies, and that defects occur according to a Poisson distribution with parameter λ= 0.02 a. What is the probability that an assembly will have exactly one defect? λ=0.02 p ( x )=...
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Add to cartA mechatronic assembly is subjected to a final functional test. Suppose that defects occur at random in these assemblies, and that defects occur according to a Poisson distribution with parameter λ= 0.02 a. What is the probability that an assembly will have exactly one defect? λ=0.02 p ( x )=...
Consider the hole diameter data. Estimate process capability using ´x and R charts. If specifications are at 0±0.01, calculate Cp, Cpk, and Cpkm. Interpret these ratios. n= 5 ´x= ∑ ´x m = (8+0+…+20+18) 20 =10,90 R´ = ∑Ri m = (80+110+…+60+50) 20 =63,50 σ^= R´ d2 ...
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Add to cartConsider the hole diameter data. Estimate process capability using ´x and R charts. If specifications are at 0±0.01, calculate Cp, Cpk, and Cpkm. Interpret these ratios. n= 5 ´x= ∑ ´x m = (8+0+…+20+18) 20 =10,90 R´ = ∑Ri m = (80+110+…+60+50) 20 =63,50 σ^= R´ d2 ...
The data in table represent individual observations on molecular weight taken hourly from a chemical process. The target value of molecular weight is 1050 and the process standard deviation is thought to be about σ= 25. a) Set up a tabular cusum for the mean of this process. Design the cusum to...
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Add to cartThe data in table represent individual observations on molecular weight taken hourly from a chemical process. The target value of molecular weight is 1050 and the process standard deviation is thought to be about σ= 25. a) Set up a tabular cusum for the mean of this process. Design the cusum to...
INDU 6331 Assignment 7 Advanced Quality Control As an example considering p= 0.02, the value of Pa is Pa=P{d ≤ c }=∑ d=0 c n! d !(n−d)! p d (1−p) n−d=∑ d=0 c nCd∗p d (1−p) n−d Pa=P{d ≤2}=100C 0∗0.020 (1−0.02) 100−0 +100C1∗0.021 (1−0.02) 100−1 +...
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Add to cartINDU 6331 Assignment 7 Advanced Quality Control As an example considering p= 0.02, the value of Pa is Pa=P{d ≤ c }=∑ d=0 c n! d !(n−d)! p d (1−p) n−d=∑ d=0 c nCd∗p d (1−p) n−d Pa=P{d ≤2}=100C 0∗0.020 (1−0.02) 100−0 +100C1∗0.021 (1−0.02) 100−1 +...
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