Linear combination of normal distributions 
The distribution of independent random variables X and Y are $X\sim \mathcal{N}(24,2^2)$ and $Y \sim \mathcal{N}(25,3^2)$. Find the distribution of $2X-Y$ and $\mathbb{P}[X-Y<40]$.

I am doing self studying and I just reached the topic Working with normal distribution on the chapter Linear combinations of random variables and I came to this topic and am stuck I don't really know what to do. Any help would be much appreciated. Thank You!
 A: There is an important fact that a linear span of INDEPENDENT normally distributed random variables is still normally distributed.
For your case, let $Z=2X-Y$, then $Z$ is normally distributed with mean $E[Z]=2E[X]-E[Y]$ and variance $\mbox{Var}(Z) = \mbox{Var}(2X)+\mbox{Var}(-Y)$.
However, it is important to point out that if the given normally distributed random variables are NOT independent, their linear combination needs not be normally distributed.
A: HINT
The critical results you need are that if $X,Y$ are independent normal random variables and $a,b \in \mathbb{R}$, then $aX+bY$ is also normal (note that the parameters will depend on the constants $a,b$). You can prove this using characteristic functions, for example, and then you will know how to change the parameters.
Your problem becomes an exercise applying the above statements. Can you finish it?
UPDATE
To estimate the first parameter, let $Z = aX+bY$ then
$$\mathbb{E}[aX+bY] = x\mathbb{E}[X] + b\mathbb{E}[Y]$$
Can you now do this for the variance, and plug the numbers from your problem?
UPDATE 2
You are right. We define $Z = 2X-Y$ then $Z \sim \mathcal{N}(m,s^2)$ with
$$
m = \mathbb{E}[Z] = \mathbb{E}[2X-Y] = 2\mathbb{E}[X] - \mathbb{E}[Y]
  = 2\cdot 24 - 25 = 23.
$$
For the variance, you get
$$
s^2 = \mathbb{Var}[Z] = \mathbb{Var}[2X-Y] = 2^2 \mathbb{Var}[X] + (-1)^2\mathbb{Var}[Y],
$$
so now you can complete the arithmetic and you know the distribution of $Z=2X-Y$.
To compute $\mathbb{P}[X-Y<40]$, you can transform it similarly using $V = X-Y \sim (\mu, \sigma^2)$ into the standard normal since $(V-\mu)/\sigma$ is standard normal.
Can you complete this now?
