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?
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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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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/(σ√2π) e^(-1/2 ((t-μ)/σ)^2 )
Where μ 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 σ 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)=∫_0^t▒〖(t)dt=∫_0^t▒〖1/(σ√2π) e^(-1/2 ((t-μ)/σ)^2 ) dt〗〗
The above integral may be simplified by making the substitution
z=(t-μ)/σ
So that
t=σz+μ
Then
dz/dt=1/σ
dt=σdz
And we can re-write the integral as
F(z)=∫_0^(σz+μ)▒〖1/(σ√2π) e^(-1/2 z^2 ) ...
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