Question

The lifetime of a machine is continuous on the interval (0, 40) with probability density function f, where f(t) is proportional to (t + 10)−2, and t is the lifetime in years. Calculate the probability that the lifetime of the machine part is less than 10 years. Hint: Show that f(t) is legitimate and find the proportionality constant.

Answer 1

$f(t)=\alpha \frac{1}{(t+10)^2}$, $t \in [0,40]$ for some real $\alpha.$

$f(t)$ is density, so $\int_{-\infty}^{+\infty} f(t)dt =1$.

Find $\alpha$:

$\int_{-\infty}^{+\infty} f(t) dt= \alpha\int_{0}^{40} (t+10)^{-2}dt=\{ -\alpha (t+10)^{-1}\} \bigg|^{40}_{0}$

$=(-\alpha(40+10)^{-1}) - (-\alpha(0+10)^{-1})=-\alpha/50 + \alpha/10=2\alpha/25$

$2\alpha/25=1 => \alpha=25/2.$

Probability the lifetime between 0 and 10 years is $\int_{0}^{10} f(t)dt.\\$$\\ \text{Then} \int_{0}^{10} f(t)dt = \alpha\int_{0}^{10} (t+10)^{-2} dt$

$=25/2\cdot\int_0^{10}(t+10)^{-2}dt=25/2\cdot \{-(t+10)^{-1}\}\bigg|_{0}^{10}=25/2\cdot(-\frac{1}{20}-(-\frac{1}{10}))=\frac{5}{8}.$

Answer: $\frac{5}{8}.$