Question 543895
A component of a precision instrument is subject to random failure. The time to failure of the component, or its productive life, is normally distributed with a mean of 440 hours and a standard deviation of 25 hours. The production engineer would like to replace the component prior to failure after a certain number of productive hours. He is willing to take a 1% chance that the component fails prior to this scheduled replacement time.
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(a) Should the replacement time be less than or greater than the mean time to failure?
Less than.
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(b) What replacement time should he select in this case?
Find the z-value with a left-tail of 1%
invNorm(0.01) = -2.3263
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Find the corresponding x-value:
x = zs+u
x = -2.3263**25+440 = 381.84 hours
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Cheers,
Stan H.
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