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?
Accepted Solution
A:
Answer:a)e^-(λ20891)b)1-e^-(λ30598)c)1-e^-(λ30598)-e^-(λ20891)Step-by-step explanation:The probability density function (pdf) of an exponential distribution isf(x,λ)=λe^-(λx) for x>0, and 0 for x<0The cumulative distribution function is given byF(x,λ)=1-e^-(λx) for x>0, and 0 for x<0P(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)