Use Table 9.1 to solve the problem. Consider again our company making tires for

Question

Use Table 9.1 to solve the problem. Consider again our company making tires for bikes is concerned about the exact width of their cyclocross tires. The company has a lower specification limit of 22.8 mm and an upper specification limit of 23.2 mm. The standard deviation is 0.15 mm and the mean is 23 mm. The company now wants to reduce its defect probability to 0.01. To what level would they have to reduce the standard deviation in the process to meet this target? O 0.076923 O 0.083333 between 0.076923 and 0.08333 between 0.80000 and 0.86667

Explanation / Answer

Defect probbility of 0.01 is equally spread on both sides. So probability of too narrow or too wide should be limited to 0.01/2=0.005

Z value corresponding to 0.005 = -2.6

LSL = 23-2.6*s = 22.8. Where s is std dev

Therefore, s = (22.8-23)/-2.6 = 0.076923

standard deviation has to be reduced to 0.076923

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