Question 1183442
The pdf of the exponential distribution is {{{f(x) = lambda*e^(-lambda*x)}}} for {{{x >=0}}} and f(x) = 0 whenever x < 0.  
The mean of the distribution is {{{1/lambda}}}.  
The cdf is equal to {{{(1/400)*int( e^(-t/400), dt, 0, x) = 1-e^(-x/400)}}}.

(a) What is the probability that an assembly on test fails in less than 100 hours?

Ans.  {{{(1/400)*int(e^(-x/400), dx,0, 100) = 1 - e^(-1/4) = 0.2212}}}

(b) What is the probability that an assembly operates for more than 500 hours
before failure?

Ans. {{{(1/400)*int(e^(-x/400), dx,500, infinity) = 1-(1 - e^(-500/400)) = e^(-5/4) = 0.2865}}}

(c) If an assembly has been on test for 400 hours without a failure, what is the probability of a failure in the next 100 hours?

Ans. {{{(1/400)*int(e^(-x/400), dx,400, 500) = e^(-1) - e^(-5/4) = 0.0814}}}


(d) If 10 assemblies are tested, what is the probability that at least one fails in less than 100 hours? Assume that the assemblies fail independently.

{{{1-((1/400)*int(e^(-x/400), dx,100, infinity))^10 = 1-(e^(-1/4))^10 = 0.9179}}}

(e) If 10 assemblies are tested, what is the probability that all have failed by 800
hours? Assume that the assemblies fail independently.

Ans.{{{((1/400)*int(e^(-x/400), dx,0, 800))^10 = (1-e^(-2))^10 = 0.2336}}}