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# Relaibility analysis Normal Distribution &Lognormal distribution

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1. Five percent of a certain grade of tires wear our before 25,000 miles, and another 5 percent of tires exceed 35,000 miles. Determine the tire reliability at 24,000 miles if wearout is normally distributed.

2. The wearout of a machine part has a lognormal distribution with s=0.5 and t_med=5,630 hours. What is MTTF?

https://brainmass.com/engineering/power-engineering/relaibility-analysis-normal-distribution-lognormal-distribution-625125

#### Solution Preview

1. Five percent of a certain grade of tires wear our before 25,000 miles, and another 5 percent of tires exceed 35,000 miles. Determine the tire reliability at 24,000 miles if wearout is normally distributed.
The normal distribution is defined by a probability of failure pdf f(t) given by
f(t)=1/(Ïƒâˆš2Ï€) e^(-1/2 ((t-Î¼)/Ïƒ)^2 )

Where Î¼ represents the mean of the distribution plot (ie the horizontal co-ordinate of the peak of the plot because the normal distribution is symmetrical) and Ïƒ represents the standard distribution of the plot. See the plot below

The cumulative distribution (up to and including time t) then represents the probability of failure up to this time and is given by
F(t)=âˆ«_0^tâ–’ã€–(t)dt=âˆ«_0^tâ–’ã€–1/(Ïƒâˆš2Ï€) e^(-1/2 ((t-Î¼)/Ïƒ)^2 ) dtã€—ã€—

The above integral may be simplified by making the substitution
z=(t-Î¼)/Ïƒ
So that
t=Ïƒz+Î¼
Then
dz/dt=1/Ïƒ
dt=Ïƒdz
And we can re-write the integral as
F(z)=âˆ«_0^(Ïƒz+Î¼)â–’ã€–1/(Ïƒâˆš2Ï€) e^(-1/2 z^2 ) ...

#### Solution Summary

Problems and solutions in reliability & failure analysis using Normal Distribution and Lognormal distribution, Q parameters, mean values, standard deviations and other statistics. Mean time to failure and reliability of product is determined from given statistical parameters

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