My question reads: Given that the probability function of X is as follows:
$$ {\rm f}\left(x\right) \equiv {\sqrt{\,x + 9\,} \over 10}\,,\qquad x = -8, -5, 0, 7. $$
(a) Find the mean of $X$.
(b) Find the expected value of $\sqrt{\,Y\,}$ where $Y = {\rm g}\left(x\right)$ and ${\rm g}\left(x\right)= x + 9$.
So, I know because I was told that the mean of $X$ is equal to $1$, but how am I supposed to arrive at that conclusion?.
What specific steps do I take to find the mean? and I'm totally clueless about part (b.).
In your explanation please use baby steps, because I have never felt so crushingly unintelligent in my mathematical life. -Tim
(a)
A probability function gives the probability that a discrete random variable is equal to a value.
Here: $\operatorname{P}(X=x)=f(x)=\dfrac{\sqrt{x+9}}{10}, \quad \forall x\in \Omega: \Omega=\{-8,-5,0,7\}$
The (arithmetric) mean of a random variable is the sum of the product of all possible values and their probabilities. It's the expected value of the variable.
$\overline{X} $ $= \operatorname{E}[X] = \sum\limits_{x\in\Omega} x\cdot\operatorname{P}(X=x) \\ = \sum\limits_{x\in\Omega} x\cdot f(x) \\ = \dfrac{\mathbf{-8}\sqrt{\mathbf{-8}+9}}{10}+\dfrac{\mathbf{-5}\sqrt{\mathbf{-5}+9}}{10}+\dfrac{\mathbf{0}\sqrt{\mathbf{0}+9}}{10}+\dfrac{\mathbf{7}\sqrt{\mathbf{7}+9}}{10} \\ = \ldots$
(b)
The expected value of a function of a variable is the sum of the product of the image of all possible the values and their probabilities.
$\operatorname{E}[\operatorname{fun}(X)] = \sum\limits_{x\in\Omega} \operatorname{fun}(x)\cdot \operatorname{P}(X=x)$
Now since $Y=g(X)$ and $g(x)=x+9$ then:
$\therefore \operatorname{E}[\sqrt Y]$ $= \operatorname{E}[\sqrt{X+9}] \\ = \sum\limits_{x\in\Omega} \sqrt{x+9}\cdot f(x) \\ = \sum\limits_{x\in\Omega} \dfrac{|x+9|}{10} \\ = \ldots$