Question

sorry it is blurry

The time between failures of a laser in a machine, X, is exponentially distributed with a mean of 25,000 hours. In other words, X= (failures/hour). 25,000 Exponential Distribution (pdf): f(x) = 1.e-r, for 2 > 0. (a) What is the probability that the next failure occurs in 27,000 hours? (b) What is the expected time until the third failure? (c) What is the probability that the time until the third failure exceeds 25,000 hours?

Answer 1

a)

P(X<27000)=$\int_{0}^{27000}$ f(x) dx =$\int_{0}^{27000}$(1/25000)*e^-x/25000 dx =(-e^-x/25000)|²⁷⁰⁰⁰₀=1-e^-27000/25000 =0.6604

b)

expected time E(x) =$\int_{0}^{\infty }$xf(x) dx=$\int_{0}^{\infty }$(x/25000)*e^-x/25000 dx

=(1/25000)*(-25000xe^-x/25000-25000^2*e^-x/25000) |^$\infty$₀ =25000 hours

C)

P(X>25000)= $\int_{25000}^{\infty }$ f(x) dx= $\int_{25000}^{\infty }$ (1/25000)*e^-x/25000 dx = (-e^-x/25000)|^$\infty$₂₅₀₀₀=e^-25000/25000

=e^-1 =0.3679