Finding expected value and variance from CDF $1 - \frac{81}{x^4}$ if $x \geq 3$ The CDF is as follows:
$$F(x) =\begin{cases}
0, & \text{if } x \leq  3;\\
 1 - \frac{81}{x^4}, & \text{if } x \geq 3.\end{cases}$$
From this I derived for the PDF, which would be $324/x^5$, but I'm having a hard time visualizing this and calculating the expected value from it. For example, I attempted to calculate the answers for several values of x, hoping to simply insert them into the $E(X)$ formula,
$$E(X) = \sum_x  p(x),$$
But when I enter values like $1, 2$, or $3$ in the PDF I obviously get crazy high numbers ($324, 20$, and $4$) which doesn't make sense for a probability problem. Shouldn't they be between $0$ and $1$ like they are for the CDF?
Furthermore, how can I get the variance? Can I get it through the CDF instead, via subbing in values for $E(X)$ and $E(X^2)$, or can I only get there from the PDF?
 A: There seem to be a bunch of problems with your understanding of probability and expectation. First, this problem is continuous, rather than discrete. What this means is that your random variable $X$ can take values on the range from $[3,\infty)$ rather than on some discrete set of values, e.g., $\{3,4,\ldots,\infty\}.$
Now, the first question you have in there is why are the values you're getting above one in PDF? The answer is because, for a continuous problem, the PDF doesn't give you probabilities. Understanding this can be a question of its own, but the short answer is, integrating under the PDF gives you probabilities on ranges of values. For example, if we call $f(x)$ the PDF of $X$ and want to know the probability that $X$ is between $5$ and $10$ we can integrate, $$\mathbb{P}(5\leq X \leq 10) = \int_5^{10}f(x)dx.$$
This means that the probability of getting any individual value is zero, that is, $$\mathbb{P}(X = a)=\mathbb{P}(a\leq X \leq a) = \int_a^{a}f(x)dx = 0.$$
So in this way, the PDF doesn't give probabilities and there is no reason to think that it will have a range of $[0,1]$. Rather, you can think about areas where the PDF is high being ranges of values which are likely.
Now, as for the expectation calculation, again you need to compute it differently because $X$ is continuous. In particular, $$\mathbb{E}(X) = \int_{-\infty}^{\infty}xf(x)dx = \int_3^{\infty}x\cdot \frac{324}{x^5}dx = \left. -\frac{108}{x^3}\right|_3^{\infty} = 4.$$ Where the first equality is simply by definition, and the following equalities are for your particular problem.
Similarly, you can compute $$\mathbb{E}(X^2) = \int_{-\infty}^{\infty}x^2f(x)dx = \int_3^{\infty}x^2\cdot \frac{324}{x^5}dx = \left. -\frac{162}{x^2}\right|_3^{\infty} = 18.$$
To be complete, we can then take these numbers to compute the variance, $$\text{Var}(X) = \mathbb{E}\left[\left(X - \mathbb{E}(X)\right)^2\right] = \mathbb{E}(X^2) - \left(\mathbb{E}(X)\right)^2 = 18 - 4^2 = 2.$$
A: You say the density function is $f(x) = 324/x^4,$ for $x \ge 3.$
You should really start by verifying that $\int_3^\infty f(x)\,dx = 1,$
a required property for a density function.
Then (as commented by @Max) $E(X)$ is not a sum, but the integral $\mu_X = E(X) = \int_3^\infty xf(x)\,dx.$
Also. $E(X^2) = \int_3^\infty x^2 f(x)\,dx.$
Finally, you can get the variance as $\sigma_X^2 = V(X) = E(X^2) - \mu_X^2
= \int_3^\infty (x - \mu_X)^2 f(x)\,dx.$
Here are sketches of the CDF and PDF.

