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

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

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