Question

The repair time of a telecommunication antenna has an exponential distribution, with an average of 22 minutes. It is requested:

a) Find the probability that the repair time of the antenna is less than 10 minutes.

b) To perform a scheduling, how much time should be assigned to each repair so that the probability of the repair time exceeding the assigned time is only 0.1?

c) Knowing that the repair time has already exceeded 5 minutes, what is the probability that the repair will take at least 15 minutes?

Answer 1

Here repair time is let say t mins

so here exponential parameter = 1/22 min-1

t ~ EXP(1/22)

f(t) = (1/22) e-t/22 ; t > 0

cumulative distribution functionis

F(t) = 1 - e-t/22 ; t > 0

(a) Here we have to find

P(t < 10 mins) = F(10) = 1 - e-10/22 = 0.3653

(b) Here let say that time is equals to t0

so here

P(t < t0) = 1 - 0.1 = 0.9

so here

F( t0) = 0.9

1 - e-t/22  = 0.9

e-t/22 = 0.1

t = 50.66 mins

(c) As the exponentrial distribution is memoreyless that means

P(t > x + s l t > s) = P(t > x)

so here we have to find

P(t > 15 l t > 5)

so that would be equal tp P(t > 10)

as we calculated in part (a)

P(t < 10) = 0.3653

then,

P(t > 10) = 1 - 0.3653 = 0.6347