1. 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, x<100 c/x2, x>=100, where c is a constant. i. What value must c take in order for this to define a valid PDF? ii. What is the probability that the bulb lasts no longer than 150 hours? iii. Given that a bulb lasts longer than 150 hours, what is the probability that it lasts longer than 200 hours? iv. Evaluate the cumulative distribution function.

(i) Any valid PDF must integrates to 1, so

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

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

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

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

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

So, $P(X<200|X>150)={\frac {{\frac 1 6}} {\frac 2 3}}=0.25$

(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

$F(x)=100\int_{100}^x {\frac 1 {t^2}}dt=-100*{\frac 1 t}|_{t=100}^{t=x}=1-{\frac {100} x}$ when x ≥ 100