Finding the Mean of a Probability Distribution

Question

Let $X$ be a random variable with the following probability distribution: $$x: \{−3;6;9\}$$ $$f(x): \{\frac16;\frac12;\frac13\}$$ I'm asked to find $\mu g(X)$, where $g(X) = (2X+1)^2$.

The mean i'm looking is expressed as: Mean $g(x) = E[g(x)] = E[(2x+1)]^2$. How could I use the values given to find what it is?

Answer 1

Guide:

Notice that $\mathbb{E}[(2X+1)]^2$ and $\mathbb{E}[(2X+1)^2]$ are different.

Fill in this table:

$$\begin{array}{|c|c|c|c|} \hline \\ x & -3 & 6 & 9 \\\hline (2x+1)^2 & & & \\ \hline P(X=x) & & & \\ \hline \end{array}$$

Now you should be able to compute the quantity of interest.