Question

17#12

Consider a system with one component that is subject to failure, and suppose that we have 100 copies of the component. Suppose further that the lifespan of each copy is an independent exponential random variable with mean 15 days, and that we replace the component with a new copy immediately when it fails.

(a) Approximate the probability that the system is still working after 1725 days.
Probability ≈

(b) Now, suppose that the time to replace the component is a random variable that is uniformly distributed over (0,0.5). Approximate the probability that the system is still working after 1950 days.
Probability ≈≈

Answer 1

a) normal distribution

mean=15*100 = 1500

std dev = s*√n= 15*√100 = 150

P ( X ≥   1725   ) = P( (X-µ)/σ ≥ (1725-1500) / 150)      
= P(Z ≥   1.50   ) = P(Z<-1.5 ) =    0.0668   (answer)

b)

uniform distribution

mean =    (a+b)/2 =    (0+0.5)/2=   0.25
          
variance =    (b-a)²/12 =    (0.5-0)²/12=   0.0208
          
std dev =   √ variance =        0.144337567

normal distribution,

µ=100*15 +99*0.25

σ= √(100*15²+99*0.1443²)=150.007

P ( X ≥   1950   ) = P( (X-µ)/σ ≥ (1950-1524.75) / 150.007)      
= P(Z ≥   2.83   ) = P(Z<-2.83 ) =    0.0023   (answer)