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?
(ii) What is the probability it will last less than 1200 hours?
(iii) Find the mean and variance of the life of the electronic device.

Answer 1

Let X be the lifetime of the given electronic device. X has the exponential distribution with parameter $\lambda = 1/1000$ means that $P(X>t)=e^{-\lambda t}$, for all $t>0$.

(i) $P(X>1000)=e^{-1000/1000}=e^{-1}=0.3679$

(ii) $P(X\leq 1200)=1-e^{-1200/1000}=1-e^{-1.2}=0.6988$

(iii) The probability dense function is

$f_X(t)=\frac{d}{dt}P(X\leq t)=\frac{d}{dt}(1-e^{-\lambda t})=\lambda e^{-\lambda t}$

The mean of the life of the electronic device is

$E(X)=\int\limits_{0}^{+\infty}tf_X(t)dt=\int\limits_{0}^{+\infty}t\lambda e^{-\lambda t}dt=$

$=\frac{1}{\lambda}\int\limits_{0}^{+\infty}\lambda te^{-\lambda t}d(\lambda t)=\frac{1}{\lambda}\int\limits_{0}^{+\infty} te^{- t}dt=\frac{1}{\lambda}=1000$

The mean of the squared X is

$E(X^2)=\int\limits_{0}^{+\infty}t^2f_X(t)dt=\int\limits_{0}^{+\infty} t^2 \lambda e^{-\lambda t}dt=$

$=\frac{1}{\lambda^2}\int\limits_{0}^{+\infty}(\lambda t)^2e^{-\lambda t}d(\lambda t)=\frac{1}{\lambda^2}\int\limits_{0}^{+\infty} t^2e^{- t}dt=\frac{2}{\lambda^2}=2\cdot 10^6$

The variance of the life of the electronic device is

$Var(X)=E(X^2)-E(X)^2=2\cdot 10^6-1000^2=10^6$

The standard deviation of the life of the electronic device is

$\sigma_X=\sqrt{Var(X)}=\sqrt{10^6}=1000$