Question

18-9. In Example 18.3-1, determine the following: (a) The average number of failures in 1 week, assuming the service is offered 24 hrs a day, 7 days a week. (b) The probability of at least one failure in a 2-hr period. (c) The probability that the next failure will not occur within 3 hrs (d) If no failure has occurred 3 hrs after the last failure, what is the probability that interfailure time is at least 4 hrs?

Answer 1

(a). averge number of failures in a week (working in hours =24*7=168 hours a week)

Calculating E(X) where x is no of failures in a week.

r.f (x)dr

168 0.2te-02tdt

solving this by integration by parts (0.2t as first e^(-0.2t) as second), on solving we get mean number of failures is 5.

(b) Atleast one failure (one or more failures) P(X>1) = 1- P(X<1)

= = 0.8187

0.2e-02t dt

(c) Not failing withing next 3 hours = 1- P(failing withing next 3 hours)

= 0.73

0.2e-02t dt