Question

The time to failure (in hours) for a laser in a cytometry machine is modeled by an exponential distribution with . Round the answers to 3 decimal places.

(a) What is the probability that the laser will last at least 20891 hours?
(b) What is the probability that the laser will last at most 30598 hours?
(c) What is the probability that the laser will last between 20891 and 30598 hours?

Answer 1

a)e^-(λ20891)

b)1-e^-(λ30598)

c)1-e^-(λ30598)-e^-(λ20891)

Explanation:

The probability density function (pdf) of an exponential distribution is

f(x,λ)=λe^-(λx) for x>0, and 0 for x<0

The cumulative distribution function is given by

F(x,λ)=1-e^-(λx) for x>0, and 0 for x<0

P(X≥20891)=1-P(X≤20891)=1-F(20891,λ)=e^-(λ20891)

P(X≤30598)=F(30598,λ)=1-e^-(λ30598)

P(20891≤X≤30598)=F(30598,λ)-(20891,λ)=1-e^-(λ30598)-e^-(λ20891)