Question

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.

Answer 1

(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