Continuous Random Variables/distribution functions

[Peachyyy]

I am having MORE problems in my statistics class. Thanks in advance for any help.

Find the distribution function of the random variable X whose probability density is given by:


f(x) = x/2 for 0 < x <= 1
1/2 for 1 < x <= 2
(3 - x)/2 for 2 < x < 3
0 elsewhere.

This is what I know:


F(x) = 0 for x <= 0
(x^2)/4 for 0 < x <= 1




1 for x >= 3

I don't know what to make my upper and lower bound for the integral to figure out:

f(x) = 1/2 for 1 < x <= 2
(3 - x)/2 for 2 < x < 3

Any help you could offer would be greatly appreciated.

 

[pka]

F(x) = 0 for x <= 0
(x<SUP>2</SUP>)/4 for 0 < x <= 1

∫{1,x}[(1/2)]dt+F(1) =(1/2)x−(1/4) for 1<x≤2

∫{2,x}[(3−t)/2]dt+F(2) =(3/2)x−( x<SUP>2</SUP>/4)−(5/4) for 2<x≤3

1 for x >= 3

 

[Peachyyy]

Thank you, that helps. One more:

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

find F(x) for 2 ≤ x < 4 and F(x) for x ≥ 1.

I guess i'm thrown off because of the gap between 1<=x<=2

thanks in advance,

 

[tkhunny]

Thank you, that helps. One more:

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

find F(x) for 2 ≤ x < 4 and F(x) for x ≥ 1.

I guess i'm thrown off because of the gap between 1<=x<=2

thanks in advance,

Just build it. One piece at a time. It's just constant wherever the probability density is zero.

F(x) = x/3 for 0 < x < 1

F(0) = 0/3 = 0
F(1) = 1/3

F(x) = 1/3 for 1 < x < 2

F(x) = 1/3 + (x-2)/3 for 2 < x < 4

F(2) = 1/2 + 0/3 = 1/3
F(4) = 1/3 + 2/3 = 1