Answer:
See steps below
Step-by-step explanation:
a)
[tex]\bf f(x)=3(1-x)^2\;(0<x<1);\;f(x)=0 \;elsewhere[/tex]
[tex]\bf F(x)=\int_{0}^{x}f(t)dt=\int_{0}^{x}3(1-t)^2dt=3\int_{0}^{x}(1-t)^2=1-(1-x)^3[/tex]
The cdf associated with f is
[tex]\bf \boxed{F(x)=1-(1-x)^3}[/tex] for 0<x<1
See picture 1
The median is a point x such that
F(x) = ½
so, the median is
[tex]\bf 1-(1-x)^3=1/2\rightarrow (1-x)^3=1/2\rightarrow \boxed{x=1-\sqrt[3]{2}}[/tex]
The 25th percentile equals the 1st quartile and is a point x such
F(x) = ¼
and the 25th percentile is
[tex]\bf 1-(1-x)^3=1/4\rightarrow (1-x)^3=3/4\rightarrow \boxed{x=1-\sqrt[3]{3/4}}[/tex]
b)
[tex]\bf f(x)=\frac{1}{x^2}\;(1<x<\infty)\;f(x)=0\;elsewhere[/tex]
[tex]\bf F(x)=\int_{1}^{x}f(t)dt=\int_{1}^{x}\frac{dt}{t^2}=1-\frac{1}{x}[/tex]
The cdf associated with f is
[tex]\bf \boxed{F(x)=1-\frac{1}{x}}[/tex] for x>1
See picture 2
The median is
[tex]\bf 1-\frac{1}{x}=\frac{1}{2}\rightarrow \boxed{x=2}[/tex]
The 25th percentile is
[tex]\bf 1-\frac{1}{x}=\frac{1}{4}\rightarrow \boxed{x=4/3}[/tex]
c)
f(x) = 1/3 for 0<x<1 or 2<x<4
[tex]\bf F(x)=\int_{0}^{x}\frac{dt}{3}=\frac{x}{3}\;(0<x<1)[/tex]
[tex]\bf F(x)=\frac{1}{3}+\int_{2}^{x}\frac{dt}{3}=\frac{1}{3}+\frac{x-2}{3}=\frac{x-1}{3}\;(2<x<4)[/tex]
The cdf associated with f is
[tex]\bf F(x)=\frac{x}{3}[/tex] for 0<x<1
[tex]\bf F(x)=\frac{x-1}{3}[/tex] for 2<x<4
See picture 3
The median is
[tex]\bf \frac{x-1}{3}=1/2\rightarrow x=1+3/2\rightarrow \boxed{x=5/2}[/tex]
The 25th percentile is
[tex]\bf \frac{x}{3}=1/4\rightarrow \boxed{x=3/4}[/tex]
