Question

The manufacturer of a certain kind of light bulb claims that the lifetime of the bulb in hours, X can be modeled as a random quantity with PDF.

fX(x)=0,c/x^2

x< 100

x> 100, where e is a constant. i. What value must e take in order for this to define a valid PDF? ii. What is the probability that the bulb lasts no longer than 150 hours? ii. Given that a bulb lasts longer than 150 hours, what is the probability that it lasts longer than 200 hours? Evaluate the cumulative distribution function.

Answer 1

$i)$

$\int_{-\infty}^{\infty} f(x)dx=c\int_{100}^{\infty}{\frac 1 {x^2}}dx=-c\times{\frac 1 x}|_{100}^{\infty}={\frac c {100}}=1\implies c=100$

$ii)$

(ii) $P(X<150)=100\int_{100}^{150}{\frac 1 {x^2}}dx=-100\times{\frac 1 x}|_{100}^{150}=-100({\frac 1 {150}}-{\frac 1 {100}})={\frac 1 3}$

$iii)$

 $P(X<200|X>150)={\frac {P(150<X<200)} {P(X>150)}}$

$P(X>150)=1-P(X<150)=1-{1\over3}={\frac 2 3}$

$P(150<X<200)=100\int_{150}^{200}{\frac 1 {x^2}}dx=-100\times{\frac 1 x}|_{150}^{200}=-100\times({\frac 1 {200}}-{\frac 1 {150}})={\frac 1 6}$

So,

 $P(X<200|X>150)={\frac {{\frac 1 6}} {\frac 2 3}}={1\over4}$

$iv)$

Let F(x) be a cumulative distribution function, then $F(x)=\int_{-\infty}^x f(t)dt$.

So, $F(x) = 0$ when $x < 100$ and,

$F(x)=100∫ ^{ x}_ {100} ​{1\over t^ 2} ​ dt=−100\times{1\over t } ​ ∣ ^{ x}_ {100} ​ =1− {100\over x} ​$ when $x \ge100$