A fiber spinning process currently produces a fiber whose st

A fiber spinning process currently produces a fiber whose strength is normally distributed with a mean of 75 N/m2. The minimum acceptable strength is 65N/m2. Ten percent of the fiber produced by the current method fails to meet the minimum specification. What is the standard deviation of the fiber strengths in the current process? If the mean remains at 75N/m2 and the minimum acceptable strength remains at 65N/m2, what must the standard deviation be so that only 5% of the fiber will fail to meet the specification? If the mean remains at 75N/m2 and the minimum acceptable strength remains at 65N/m2, what must the standard deviation be so that only 1% of the fiber will fail to meet the specification? If the standard deviation is 5N/m2, to what value must the mean be set so that only 1% of the fiber will fail to meet the minimum specification of 65N/m2? If the standard deviation is 3N/m2, to what value must the mean be set so that only 1% of the fiber will fail to meet the specification of 65N/m2?

Solution

11.

For a left tailed area of 0.10, the z corresponding score is, by table/technology,

z = -1.281551566

Hence,

sigma = (x-u)/z = (65-75)/(-1.281551566) = 7.803041458 [ANSWER]

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12.

For a left tailed area of 0.05, the z corresponding score is, by table/technology,

z = -1.644853627

Hence,

sigma = (x-u)/z = (65-75)/(-1.644853627) = 6.079568319 [ANSWER]

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13.

For a left tailed area of 0.01, the z corresponding score is, by table/technology,

z = -2.326347874

Hence,

sigma = (x-u)/z = (65-75)/(-2.326347874) = 4.298583248 [ANSWER]

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14.

For a left tailed area of 0.01, the z corresponding score is, by table/technology,

z = -2.326347874

Hence,

u = x - z*sigma = 65 - (-2.326347874)*3 = 71.97904362 [ANSWER]