P212 A driveshaft will suffer fatigue failure with a mean ti

P.212

A driveshaft will suffer fatigue failure with a mean time-to-failure of 40,000 hours of use. If it is known that the probability of failure before 39,000 hours is 0.04 and that the distribution governing time-to-failure is a normal distribution, what is the standard deviation of the time-to-failure distribution?

? = hours

Solution

Mean time of failure = mu = 40000

P(X<39000) = 0.04

Let us find corresponding z value for x from std normal table

z = -1.75

Find x using the formula

x = mu+z(sigma)

= 40000-1.75(sigma) = 39000

From this we can solve for sigma

1000 = 1.75 sigma

Or std dev = 1000/1.75 = 571.429

Hence answer is 571.429