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

Answer:

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  $P(X\geq 13 | X>9)\approx 0.0273237376$

Step-by-step 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 $P(X\geq 13 | X>9)$.

$P(X\geq 13 | X>9)=\frac{P(X\geq 13)}{P(X>9)} \\\\P(X\geq 13 | X>9)=\frac{\int\limits^{\infty}_{13} {0.9e^{-0.9x}} \, dx}{\int\limits^{\infty}_{9} {0.9e^{-0.9x}} \, dx}\\ \\P(X\geq 13 | X>9)=\frac{8.29382\times10^{-6}}{0.000303539} \\\\P(X\geq 13 | X>9)\approx 0.0273237376$