Answer to Question #234810 in Statistics and Probability for talie

Question

The life of an electronic device is known to have the exponential distribution with parameter λ = 1 1000 . (i) What is the probability that the device lasts more than 1000 hours? [2 marks] (ii) What is the probability it will last less than 1200 hours? [2 marks] (iii) Find the mean and variance of the life of the electronic device. 

Answer 1

$\lambda$ is given as 1 100, which is not clear whether it is 11,000 or 1,000. I will use 1,000.

$f(x) = \lambda e^{-\lambda x}, x>0$

$\lambda=1000$

i. $P(x)>1000$

$P(x)<1000=\int_{1000}^{\infin}1000 e^{-1000 x}dx$

$=-e^{-1000x}|_{1000}^{\infin}$

$=0-0=0$

ii $P(x)<1200$

$P(x)<1200=\int_{0}^{1200}1000 e^{-1000 x}dx$

$=-e^{-1000x}|_{0}^{1200}$

=$0--1=1$

iii Mean and variance

The mean and variance of an exponential distribution is $\lambda$ .

Thus the mean of the distribution is 1000 and variance is 1000.