# Find Distribution Function of X: f(x) = x/2, 1/2, (3-x)/2

• kuahji
In summary, the cumulative distribution function of the random variable X is given by F(x)=\int_{-\infty}^x f(t)dt where the constants are coming out of summing all the parts of the density less than the current piece you're integrating over.
kuahji
Find the distribution function of the random variable X whose probability density is given by

f(x)= x/2 for x < x $$\leq1$$
1/2 for 1< x $$\leq2$$
(3-x)/2 for 2 < x $$\leq3$$
0 elsewhere

Ok, so I know have to take the integral of each piece, but I'm having a hard time figuring out the limits of the integrals.

F(x)= 0 for x $$\leq0$$
x^2/4 for 0 < x $$\leq1$$
1/4(2x-1) 1 < x $$\leq2$$
1/4(6x-x^2-5) for 2 < x < 3
1 for x$$\geq3$$

Which is about what I got when I took the "indefinite" integrals of each piece (again, not sure how to go about setting the limits), but for example in this section 1/4(6x-x^2-5) for 2 < x < 3 where is the -5 coming from? Any help would be appreciated.

I suppose you mean cumulative distribution function? The cumulative distribution function is given as $$F(x)=\int_{-\infty}^x f(t)dt$$. Because it's a piecewise function, the constants are coming out of summing all the parts of the density less than the current piece you're integrating over. For the part you appear to have questions with, if you were to find F(x) where 2<x<3, you would find F(3) (definite integral) and then add it to $$\int_{2}^x f(t)dt$$

(Post isn't updating quickly for me so I don't know if this turned out right with my rampant typos)

Last edited:
Right, that much was in the text, but I'm still a bit confused.

x^2/4 for 0 < x $$\leq1$$ I believe from the original step I just take the integral from 0 to x.

But for the next step, I'm not sure what to do. I mean I know its suppose to add up to one, but figuring out the second and consecutive steps is where I'm having the problem.

The cumulative distribution function keeps track of the total probability found between $$-\infty$$ and x. In a sense it is a "running total". If you want to find F(x) for 1<x<2, you have F(1)=1/4, then $$F(x)= \int_{-\infty}^x f(t)dt = \int_{-infty}^{1}f(t)dt+\int_{1}^{x}f(t)dt = \frac{1}{4}+\int_{1}^{x}\frac{1}{2}dt = \frac{1}{4}+\frac{x}{2}-\frac{1}{2} = \frac{1}{4}(2x-1)$$

Edit: fixing typos

Last edited:
Ok, I see how it works now. Thanks for the help, it was really appreciated. ^_^

## 1. What is the distribution function of X given f(x) = x/2, 1/2, (3-x)/2?

The distribution function of X is the cumulative probability function that specifies the probability of X being less than or equal to a certain value. In this case, the distribution function of X would be:

F(x) = 0 for x < 0

F(x) = x^2/4 for 0 ≤ x ≤ 1

F(x) = 1 - (3-x)^2/4 for 1 < x ≤ 3

F(x) = 1 for x > 3

## 2. How do you calculate the probability of a specific value for X using the given distribution function?

To calculate the probability of a specific value for X, simply plug in the value into the distribution function. For example, if you want to find the probability of X being equal to 2, you would plug in x = 2 into the distribution function and solve for F(2).

## 3. Can you graph the distribution function of X for the given f(x)?

Yes, the distribution function can be graphed by plotting the points (x, F(x)) for all values of x. The resulting graph would be a piecewise function with a horizontal line at y = 0 for x < 0, a parabola for 0 ≤ x ≤ 1, another parabola for 1 < x ≤ 3, and a horizontal line at y = 1 for x > 3.

## 4. How do you interpret the distribution function of X?

The distribution function of X can be interpreted as the probability that X takes on a value less than or equal to a specific value. For example, F(2) would represent the probability that X is less than or equal to 2.

## 5. Can the distribution function of X be used to find the expected value and variance of X?

Yes, the expected value of X can be calculated by integrating x*f(x) from -∞ to ∞, where f(x) is the given probability density function. The variance of X can be calculated by integrating (x - μ)^2 * f(x) from -∞ to ∞, where μ is the expected value of X.

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