Originally Posted by

**halfdome**
Given the reliability at 2100 hours =0.992, then lambda(2100 hours) = -Ln(0.992)=0.008. I need to convert the hazard rate (lambda) for 2100 hours to the normalized value in terms of failure rate per million hours (fpmh, 10^6 hours).

I am foggy on the process from this point; should I use the pdf function to convert such as:

F(t)=1-(1+0.008*10^6)^-1

Then take the derivative f(t)=0.008 (1+0.008*10^6)^-2

Then h(t): f(t)/1-F(t)

Can someone point me in the right direction?