Question

Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with mean 1/0.41/0.4. What is (a) the probability that a repair time exceeds 33 hours? (b) the conditional probability that a repair takes at least 1010 hours, given that it takes more than 99 hours?

Answer 1

Answer:

a) P(t>3)=0.30

b) P(t>10|t>9)=0.67

Step-by-step explanation:

We have a repair time modeled as an exponentially random variable, with mean 1/0.4=2.5 hours.

The parameter λ of the exponential distribution is the inverse of the mean, so its λ=0.4 h^-1.

The probabity that a repair time exceeds k hours can be written as:

$P(t>k)=e^{-\lambda t }=e^{-0.4t}$

(a) the probability that a repair time exceeds 3 hours?

$P(t>3)=e^{-0.4*3}=e^{-1.2}= 0.30$

(b) the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours?

The exponential distribution has a memoryless property, in which the probabilities of future events are not dependant of past events.

In this case, the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours is equal to the probability that a repair takes at least (10-9)=1 hour.

$P(t>10|t>9)=P(t>1)$

$P(k>1)=e^{-0.4*1}=0.67$