How Do You Calculate Probability Using a Density Function?

[mrvirgo]

A continuous random variable X has the density function
f(x)=x for 0<x<1
2-x for 1 _<x<2
0 elsewhere.
a. Show that P(0<X<2)=1
B. Find P(X<1.2).


Please see the attached file.
Thank

 

[HallsofIvy]

1. A lot of people will not open files for fear of viruses.

2. Here, there was no point in attaching the file since it doesn't say anything you haven't written here! In particular, there is not attempt at the problem youself.
How is P(a< X< b) defined for any probability density function f(x)? Where exactly is your difficulty?

 

[tacman]

you for sharing this density function.
a. To show that P(0<X<2)=1, we can integrate the density function over the interval of 0 to 2. This will give us the probability of the random variable X being between 0 and 2, which should equal to 1.
So, we have:
P(0<X<2) = ∫0^1 xf(x) dx + ∫1^2 (2-x)f(x) dx
= ∫0^1 x(x) dx + ∫1^2 (2-x)(2-x) dx
= ∫0^1 x^2 dx + ∫1^2 (4-4x+x^2) dx
= 1/3 + (4x-2x^2+(x^3)/3)|1^2
= 1/3 + (8-8+8/3) - (4-2+1/3)
= 1
Therefore, P(0<X<2) = 1, which means that there is a 100% chance that the random variable X will fall between 0 and 2.
b. To find P(X<1.2), we can use the same approach as above and integrate the density function over the interval of 0 to 1.2.
So, we have:
P(X<1.2) = ∫0^1.2 xf(x) dx + ∫1.2^2 (2-x)f(x) dx
= ∫0^1.2 x(x) dx + ∫1.2^2 (2-x)(2-x) dx
= ∫0^1.2 x^2 dx + ∫1.2^2 (4-4x+x^2) dx
= 1/3 + (4x-2x^2+(x^3)/3)|1^1.2
= 1/3 + (4.8-2.88+1.728) - (4-2+1/3)
= 0.928
Therefore, P(X<1.2) = 0.928, which means that there is a 92.8% chance that the random variable X will be less than 1.2.