Normal Distribution with Linear Transformation

Question

I have a random variable $Y \sim \mathsf{N}(2,5)$ and we define $Z = 3Y-4$. I want to find the distribution of $Z$.

Intuitively I can see that it is Normal as well due to the nature of the transformation. To show this, my first thought is to scale the variance by 3 and shift the mean by -4, giving $Z \sim \mathsf{N}(-2, 15)$.

However I am uncomfortable with this as it seems too rudimentary. Another thought of mine is to calculate the following. $$ f_Z(x) = 3f_Y(x) - 4$$ where $f_Z$ and $f_Y$ are the pdfs. I tried subbing in $f_Y$ and manipulating the expression into the usual $\frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(x-\mu)^2}{2\sigma^2}}$ but the $-4$ term is unwieldy.

Are either of these approaches correct, or are there better alternative methods?

Answer 1

$$E(3Y-4)=3E(Y)-4=2$$ $$Var(3Y-4)=9Var(Y)=45$$

So $Z$ has the distribution $ N(2,45)$

Answer 2

A random variable $Y\sim N(\mu, \sigma^2)$ if $Y=\mu +\sigma W$ for some $W$ where $W\sim N(0, 1)$. Thus in your case $$ Z=3Y-4=3(2+\sqrt{5}W)-4=2+3\sqrt{5}W $$ where for some $W\sim N(0,1)$. Thus $Z\sim N(2,(3\sqrt{5})^2)$ i.e. $Z\sim N (2,45)$.