Flyrock Tires

Six Sigma Quality at Flyrock Tires Executive Summary The physical process of creating tires at Flyrock Tires involves 20 different steps to take the surface from bales to final exam curing. Given this complexity and the elevated production volume (the factory produces nearly 10,000 tires per hour), it takes only a small margin of error in distributively of these steps to begin to compound and result in a high uncollectible rate. For both public safety and their reputation, Flyrock strives to minimize the number of defects. The answers to the top dogs asked by this case form a good base for evaluating the production and jut process at Flyrock.The company begins by set expectations for what defect range should be under ideal conditions as well as setting expectations for defect rates when machinery has become languid out. This allows them to detect things like irresolute bearings in machinery. The case also begins the framework for evaluating the process from a six-sigma s tatus and how this might help reign defects. Under ideal conditions, the extruder should produce tires that amass specifications 98. 67% of the time, meaning only 1. 33% of tires are defective. When the bearing is worn out, the defect rate increases collar-fold to 4. 7%. This means that more than one in twenty-five tire sheets is defective. If testing samples of ten sheets per hour, the machine operators could expect to fetch a defective every two and a half hours. In testing whether the extrusion process is in control or not, the three sigma control limits recommended by Susan Douglas will narrow the bandwidth of accept able-bodied tires from cd 10 thou to 400 3. 795 thou. By implementing a stricter six-sigma system and decreasing the hackneyed deviation, the likelihood of producing tires within accep flurry levels of oppressiveness increases to 100%.This becomes a near-perfect process. With this information, Susan Douglas should straight off take appropriate steps to cou nt the actual number of defects that materialise from the extrusion process. Having that will allow her to analyze if the process is actually in control, using various control charts. Using that, she will be able to adjust the process as needed to reduce the number of defective tires, improve quality and efficiency, and ultimately reduce costs for Flyrock. Question 1 The mean, confidence separations, and standard deviation are provided to us. Mean () = 400 thouConfidence interval = 10 thou Standard deviation (? ) = 4 thou This psyche is a simple normal probability distribution problem. It can be restated as What is the probability that rubber sent finished the extruder will be between 390 thou and 410 thou in thickness? P(390 ? x ? 410) We first gear need to find the z evaluate for each boundary wherefore find the corresponding probability in the normal distribution table z = (x )/? z = (390-400)/4 = -2. 5 z(-2. 5) = . 0062 z = (410-400)/4 = 2. 5 z(2. 5) = . 9938 P(-2. 5 ? z ? 2. 5) = . 9938 . 0062 = . 9867 Therefore, there is a 98. 7% probability that the rubber will be extruded with the specifications. Question 2 To find the upper and decline control limits, the following formulas apply Upper Control Limit (UCL) = + z? Lower Control Limit (LCL) = z? where ? = ? /n = 400 ? = 4 n = 10 z = 3 ? = 4/10 = 1. 265 UCL = 400 + 3(1. 265) = 403. 795 LCL = 400 3(1. 265) = 396. 205 Question 3 If a bearing is worn out, the extruder produces a mean thickness of 403 thou even though the setting is at 400. This means that has shifted to 403 and the confidence level will be lower than the 98. 7% we calculated when the bearings are not worn out. We can restate the question for number 1 to ask What is the probability that rubber sent through the extruder will be outside 390 thou and 410 thou in thickness when the mean thickness has shifted to 403 thou? Again, we need to find the z value for each boundary then find the corresponding probability in the normal di stribution table z = (x )/? z = (390-403)/4 = -3. 25 z(1. 75) = 0. 9599 z = (410-403)/4 = 1. 75 z(-3. 25) = 0. 0006 P(-3. 25 z(. 628) = 0. 7357 P in control (0