Answer
(a) (i) The probability that a bulb fails within the first 200 hours is 0.18
(ii) The probability that a bulb burns for more than 800 hours is 0.45
(b) The median lifetime is 693 hours.
Work Step by Step
$f(t) = \frac{1}{\mu}e^{-t/\mu},$ where $t \gt 0$
$f(t) = \frac{1}{1000}e^{-t/1000}$
(a) (i) We can find $P(T \leq 200)$:
$P(T \leq 200) = \int_{0}^{200}f(t)~dt$
$P(T \leq 200) = \int_{0}^{200}\frac{1}{1000}e^{-t/1000}~dt$
$P(T \leq 200) = -e^{-t/1000}~\vert_{0}^{200}$
$P(T \leq 200) = 1-e^{-200/1000}$
$P(T \leq 200) = 1-e^{-0.2}$
$P(T \leq 200) = 0.18$
The probability that a bulb fails within the first 200 hours is 0.18
(ii) We can find $P(T \gt 800)$:
$P(T \gt 800) = \int_{800}^{\infty}f(t)~dt$
$P(T \gt 800) = \int_{800}^{\infty}\frac{1}{1000}e^{-t/1000}~dt$
$P(T \gt 800) = -e^{-t/1000}~\vert_{800}^{\infty}$
$P(T \gt 800) = 0+e^{-800/1000}$
$P(T \gt 800) = e^{-0.8}$
$P(T \gt 800) = 0.45$
The probability that a bulb burns for more than 800 hours is 0.45
(b) To find the median, we can find $t$ such that $P(T \lt t) = 0.5$:
$P(T \lt t) = 0.5$
$\int_{0}^{t}f(t)~dt = 0.5$
$\int_{0}^{t}\frac{1}{1000}e^{-t/1000}~dt = 0.5$
$-e^{-t/1000}~\vert_{0}^{t} = 0.5$
$1-e^{-t/1000} = 0.5$
$e^{-t/1000} = 0.5$
$-t/1000 = ln(0.5)$
$t = -1000~ln(0.5)$
$t = 693$
The median lifetime is 693 hours.
