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

$P(X>x) = e^{-λx}$

a) $P(X>1000) = exp(-\lambda x)$

$= exp(-0.001 \times 1000) \\ = 0.3678$

b) $P(X<1200)= 1 - exp(-0.001 \times 1200)$

$= 1 -0.3012 \\ = 0.6988$

c) Mean

$\mu = \frac{1}{λ} \\ \mu = \frac{1}{0.001} = 1000$

Variance

$\sigma^2 = \frac{1}{λ^2} \\ \sigma^2 = \frac{1}{(0.001)^2} \\ = 10^6$