Six Sigma Quality at Flyrock Tires Executive Summary The process of creating tires at Flyrock Tires involves 20 different steps to take the rubber from bales to final curing. Given this complexity and the high production volume (the factory produces about 10,000 tires per hour), it takes only a small margin of error in each of these steps to begin to compound and result in a high defective rate. For both public safety and their reputation, Flyrock strives to minimize the number of defects. The answers to the questions asked by this case form a good base for evaluating the production and extrusion process at Flyrock. The company begins by setting expectations for what defect rates should be under ideal conditions as well as setting …show more content…
However to find the proportion of defective sheets we must take 1-.9593 which gives us a defective rate of 4.07%. Assuming the three sigma control limits of 403.795 and 396.205, we can calculate the probability that a sample is out of control by first calculating the percentage of in control samples. z = (x - µ)/σ z = (393.205-403)/1.265 = -2.5 → z(-5.37) = 0.0000 z = (403.795-403)/ 1.265= 2.5 → z(.628) = 0.7357 P in control (0<z<.7357) = 0.7357 P Out of control = 1-.7357 = .2643 = 26.43% Using the proportion of defective units we can calculate the average time it will take to discover a worn bearing. 1/.2643 = 3.77 hours. On average it will take an operator about 4 hours to detect that the process is out of control. Question 4 Assuming our mean is still 400, but a new standard deviation of 1.667 based on a six-sigma process, we find the proportion of the rubber extruded that is within our specifications by calculating the probability that rubber sent through the extruder will be outside our limits: z = (x - µ)/σ P(x<410) = P(z<410-400/1.667) = P(z < 6.0) = 1 --- based on our normal tables P(x<390) = P(z<390-400/1.667) = P(z < -6.0) = 0 --- based on our normal tables To then find the probability of our extruded rubber being within specifications, we subtract
You work in marketing for a company that produces work boots. Quality control has sent you a memo detailing the length of time before the boots wear out under heavy use. They find that the boots wear out in an average of 208 days, but the exact amount of time varies, following a normal distribution with a standard
Decision Rule: The calculated test statistic of -3.024 does fall in the rejection region of z<-1.645, therefore I would reject the null and say there is sufficient evidence to indicate mu<50.
First we look for the area of both by doing “2nd ,Vars, NORMALCDF” and inputting “-1000, “Z,” 0, 1 then find the difference between both.
(a) Then mean of the sample and the value of Z with an area of 10% in right tail.
C. An unknown, rectangular substance measures 3.6 cm high, 4.21 cm long, and 1.17 cm wide.
5. Assuming that operators will continue to take samples of 10 sheets each hour to check if the process is in control, what control limits should Douglas set for the case when extrusion is a Six Sigma process?
Quality Associates, Inc., a consulting firm advises its clients about sampling and statistical procedures that can be used to control their manufacturing processes. In one particular application a client gave Quality Associates a sample of 800 observations taken during a time in which the client’s process was operating satisfactorily. The sample standard deviation of this data was 0.21; hence with so much data, the population standard deviation was assumed to be 0.21. Quality Associates then suggested that random samples of size 30 be taken periodically to monitor the process on an ongoing basis. By analyzing the new samples, the client could quickly learn whether the process was operating satisfactorily. When the process was not operating satisfactorily, corrective action could be taken to eliminate the problem. The design specification indicated the mean for the process should be 12. The hypothesis test suggested by Quality Associates follows.
From the calculation (See Appendix I), we get the 3-sigma control limits for the process, i.e. UCL=0.091, LCL=0.014. These control limits indicate that if the error proportion is within the range of [0.014, 0.091], the process is under control; if not, the process is out of control.
Since the expected demand is 2000, thus, the mean µ is 2000. Through Excel, we get the z value given a 95% probability is 1.96. Thus, we have: z= (x-µ)/ σ=(30000-20000)/
3) Compute limits for the sample mean X around μ=12 such that, as long as a new sample mean is within those limits, the process will be considered to be operating satifactorily. If X exceeds the upper limit or if X is below the lower limit, corrective action will be taken. These limits are refferred to as upper and lower control limits fro quality control purposes.
Quality Associates, Inc. is a consulting firm that advises its clients about sampling and statistical procedures that can be used to control manufacturing processes. In one case, a client provided Quality Associates with a sample of 800 observations that were taken during a time when the client's process was operating satisfactorily. The sample standard deviation for these data was .21, hence, the population standard deviation was assumed to be .21. Quality Associates then suggested that random samples of size 30 be taken periodically to monitor the process on an ongoing basis. By analyzing the new samples, the client could quickly learn whether the process was operating satisfactorily. When the process was not operating
Symmetry is the backbone to the bible. It provides a comforting lull to the reader and it proves time and again to be effective at subtlety emphasizing the important motifs of the stories. The balance in the bible is found, also, in the characters. The authors of the New Testament used symmetry to their advantage and placed Mosaic motifs along with the story of Jesus of Nazareth. In doing so, the parallel between Moses and Jesus is a dominant theme that makes the New Testament both slightly predictable while also reinforcing Jesus’s role as the “lawgiver.” This archetype is interesting because religion is dominated by rules, and at the most basic level, the prophets created the rules in which followers of religion live by. The link between Moses and Jesus is inextricable and undeniable, causing the relationship to be so obvious even to the least experienced bible reader. The parallelism between the figures of Moses and Jesus is important because it emphasizes the necessity to have a leader in a religious group and illustrates the quiet difference between being the figurehead who receives the law of G-d versus the one who gives the law. In the symmetry, the figures of Moses and Jesus act as a balance where Jesus’ actions are representative, but not repetitive, of those biblical actions of Moses. Through water, numerical, and Passover motifs the story of Jesus is connected to the passages of Moses.
“Where there is no imagination there is no horror.” This idea is discussed in “What is the Horror Genre” and this essay will analyze “The Tell-Tale Heart” according to the criteria put forth by the author S. Russel. Edgar Allen Poe was one of the most famous horror story writers because he used his life experience and imagination to give us suspenseful characters and events. “The Tell-Tale Heart” provides suspense through the events of murder and disposing the body.
Ash Carter in The Pentagon on Thursday, announced that the transgender ban is ended. Now, transgender people will be able to serve in the military. The transgender people would be allowed to serve in U.S. military because in the United States everyone has equal opportunity, let alone join the army as a volunteer to serve the country. As long as they have the ability to serve in the army, and it will not cause any impact on the country and the people, we should end the transgender ban. After the abolition of the ban, transgender people will enjoy the same treatment as the others. For example, the medical coverage.
The area under the curve to the left of the unknown quantity must be 0.7 (70%). So, we must first find the z value that cuts off an area of 0.7 in the left tail of standard normal distribution. Using the cumulative probability table, we see that z=0.53.