Question

Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter λ=0.9. What is (a) the probability that a repair time exceeds 8 hours? 1-e^(1/0.9*8) (b) the conditional probability that a repair takes at least 13 hours, given that it takes more than 9 hours? Note: You can earn partial credit on this problem. Your score was recorded. You have attempted this problem 3 times. You received a score of 0% for this attempt. Your overall recorded score is 0%. You have unlimited attempts remaining.

Answer 1

The probability that a repair time exceeds 8 hours is
`P(X>8)\approx 0.000746586`

The conditional probability that a repair takes at least 13 hours, given that it takes more than 9 hours is
`P(X\geq 13 | X>9)\approx 0.0273237376`

Explanation:

A continuous random variable X is said to have an exponential distribution with parameter
`\lambda > 0` shown as
`X \sim Exponential(\lambda)`, if its probability density function is given by

`f_X(x)=\begin{cases}\lambda e^(-\lambda x) & x > 0\\ 0 &\text{otherwise}\end{cases}`

Let X denote the time require to repair a machine.

(a) The probability that a repair time exceeds 8 hours;

`P(X>8)=1-P(X\leq 8)\\P(X>8)=1-\int\limits^8_0 {0.9e^(-0.9x)} \, dx \\P(X>8)=1-0.999253\\P(X>8)\approx 0.000746586`

(b) The conditional probability that a repair takes at least 13 hours, given that it takes more than 9 hours;

We want to find
`P(X\geq 13 | X>9)`.

`P(X\geq 13 | X>9)=(P(X\geq 13))/(P(X>9)) \\\\P(X\geq 13 | X>9)=\frac{\int\limits^(\infty)_(13) {0.9e^(-0.9x)} \, dx}{\int\limits^(\infty)_(9) {0.9e^(-0.9x)} \, dx}\\ \\P(X\geq 13 | X>9)=(8.29382*10^(-6))/(0.000303539) \\\\P(X\geq 13 | X>9)\approx 0.0273237376`