Calculating the mean and variance of a distribution 
*

*Suppose $$P(x) = \frac{1}{\sqrt{2\pi\cdot 36}}e^{-\frac{1}{2}\cdot (\frac{x-2}{6})^2}$$


What is the mean of $X$? What is the standard deviation of $X$?


*Suppose $X$ has mean $4$ and variance $4$. Let $Y = 2X+7$.


What is the mean of $Y$? What is the standard deviation of $Y$?
Are there resources that can help me answers these question? I have the answers, but I honestly have no idea how to solve them. Typically when I've seen mean and standard deviation, it's been in the context of finding them for a specific data set.

1) Since: $X \sim N(\mu, \sigma^2)$ you have that:
$$\mu=2 \qquad \sigma^2=6^2$$
2) $$E[Y]=E[2X+7]=E[2X]+E[7]=2E[X]+E[7]$$
Substituting $E[X]$ with $\mu=4$, we get: $$E[Y]=2 \cdot 4+7=15$$
For what it concerns the variance: $$Var[Y]=\cdots=Var[2X]+Var[7]=2^2\cdot Var[X]+Var[7]$$
Substituting $Var[X]=\sigma^2=4$, and since the variance of a number is always zero we get that:
$$4\cdot4=16$$
