Question

Find the cdf F(x) associated with each of the following probability density functions. Sketch the graphs of f(x) and F(x).

(a) f(x) = 3(1 − x)2, 0 (b) f(x)=1/x2, 1 (c) f(x) = 1/3 , 0 Also, find the median and the 25th percentile of each of these distributions

Answer 1

See steps below

Explanation:

a)

`\bf f(x)=3(1-x)^2\;(0<x<1);\;f(x)=0 \;elsewhere`

`\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`

The cdf associated with f is

`\bf \boxed{F(x)=1-(1-x)^3}` for 0<x<1

See picture 1

The median is a point x such that

F(x) = ½

so, the median is

`\bf 1-(1-x)^3=1/2\rightarrow (1-x)^3=1/2\rightarrow \boxed{x=1-\sqrt[3]{2}}`

The 25th percentile equals the 1st quartile and is a point x such

F(x) = ¼

and the 25th percentile is

`\bf 1-(1-x)^3=1/4\rightarrow (1-x)^3=3/4\rightarrow \boxed{x=1-\sqrt[3]{3/4}}`

b)

`\bf f(x)=(1)/(x^2)\;(1<x<\infty)\;f(x)=0\;elsewhere`

`\bf F(x)=\int_(1)^(x)f(t)dt=\int_(1)^(x)(dt)/(t^2)=1-(1)/(x)`

The cdf associated with f is

`\bf \boxed{F(x)=1-(1)/(x)}` for x>1

See picture 2

The median is

`\bf 1-(1)/(x)=(1)/(2)\rightarrow \boxed{x=2}`

The 25th percentile is

`\bf 1-(1)/(x)=(1)/(4)\rightarrow \boxed{x=4/3}`

c)

f(x) = 1/3 for 0<x<1 or 2<x<4

`\bf F(x)=\int_(0)^(x)(dt)/(3)=(x)/(3)\;(0<x<1)`

`\bf F(x)=(1)/(3)+\int_(2)^(x)(dt)/(3)=(1)/(3)+(x-2)/(3)=(x-1)/(3)\;(2<x<4)`

The cdf associated with f is

`\bf F(x)=(x)/(3)` for 0<x<1

`\bf F(x)=(x-1)/(3)` for 2<x<4

See picture 3

The median is

`\bf (x-1)/(3)=1/2\rightarrow x=1+3/2\rightarrow \boxed{x=5/2}`

The 25th percentile is

`\bf (x)/(3)=1/4\rightarrow \boxed{x=3/4}`