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Question
The density function of a continuous random variable x is f(x) = -(3/4)(x^3 - 5x^2 + 7x - 3) for 1<x<3.

a. What is the equation for the cumulative distribution function F(x)?

b. Use it to find P(2<x<2.5)

c. Determine E(X)

d. Determine V(X)

Answer
Hi Ricardo,

Normally I don't like to answer homework problems but let me help you out a little here (in the future, please at least show some of your work in attempting to solve the problem or at least your thinking; cutting and pasting a HW problem is bad form).

To begin, to make our nomenclature clear,  let's call the probability density function (pdf) f(x) = -(3/4)(x^3 - 5x^2 + 7x - 3), with domain 1<x<3.

a. The cumulative distribution function is just the integral (area under the density function) between the smallest value in the random variable's domain (x axis) and an upper value in that domain. This area is in fact the probability of the random variable's value falling between the 2 limits of the integration. For your example, it would be

cumulative distribution, cdf: F(x) = integral{ pdf } = integral{-(3/4)(x'^3 - 5x'^2 + 7x' - 3)dx'} between x0 and x, where x0 = smallest (leftmost) value of the domain x. In your case, x0 = 1. Note that:

1) I left the actual integration up to you (its easy, f(x) being a polynomial)
2) I used x' as the variable of integration since I want x to be the variable at which the cdf is evaluated (this is really just a notational thing so don't get too mixed up about it).

What you'll end up with is a number, F(x), that represents the [area under the pdf] = [probability of the random variable being between x0 and x).

b. To find P(2<x<2.5), you just need to integrate f(x) between 2 and 2.5 (i.e., using 2 as the lower limit of integration and 2.5 as the upper limit). Note that this is the same thing as using the cdfs to calculate

P(2<2.5) = F(2.5) - F(2).   <--- using Fundamental Theorem of Calculus.

c. E(X) = expected value of x, which by definition is obtained by multiplying the pdf, f(x), by x and integrating over the whole domain, 1<x<3; i.e.,

E(X) = integral{ xf(x) } over 1<x<3.

Again, I'll leave the calculus up to you. Note that we are using a big X in the notation as a way of reminding ourselves the we are calculating the expected value of a random variable x over it's domain X, i.e., x ∈ X, where X is a set (a segment of the real line in this case).

d. Also by definition, the variance V(X) is obtained by multiplying the pdf by [x - E(X)]^2 and integrating over the whole domain. Note that E(X) is a constant in this expression (the one you calculated in c. above).

Good luck!  

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