A company making tires for bikes is concerned about the exact width of their cyclocross tires. The company has a lower specification limit of 19.930 mm and an upper specification limit of 20.070 mm. The standard deviation is 0.18 mm and the mean is 20.000 mm. b. The company now wants to reduce its defect probability and run a "six-sigma process." To what level would they have to reduce the standard deviation in the process to meet this target? (Round the answer to 5 decimal places.)

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Question:

A company making tires for bikes is concerned about the exact width of their cyclocross tires. The company has a lower specification limit of 19.930 mm and an upper specification limit of 20.070 mm. The standard deviation is 0.18 mm and the mean is 20.000 mm. b. The company now wants to reduce its defect probability and run a "six-sigma process." To what level would they have to reduce the standard deviation in the process to meet this target? (Round the answer to 5 decimal places.)

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Answer:Missing word "a. What is the process capability index for the process?"a. Cpk = Min[USL-μ/3σ, μ-LSL/3σ]Cpk = Min[(20.070-20)/(3*0.18) ,(20-19.930)/(3*0.18]Cpk = Min[0.1296, 0.1296]Hence, process capability index is 0.1296b. 2 = Min[USL-μ/3σ, μ-LSL/3σ]2 = Min[(20.070-20)/(3*σ) ,(20-19.930)/(3*σ]2 = Min[(0.07/3*σ, 0.07/3*σ)]= Min[0.07/6, 0.07/6]σ = 0.07/6σ = 0.0117Hence, reduced standard deviation level is 0.0117

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