Is there a way to to convert from a normal distribution to a standard normal distribution. For example: $ Y = N(-8,4) $
An answer to the question says that this is the same as writing: $$\frac{Y+8}{2} = N(0,1)$$
Can anyone explain to me how they made this transformation? Thank you for any guidance.
$X$ is random variable is normal distribution with mean $\mu$ and variance $\sigma^2$, $X\sim N(\mu,\sigma^2)$ have probability density function
$$f(x)=\dfrac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}, -\infty<x<\infty$$
$Y$ is standard normal distribution, have mean $0$ and variance $1$, $Y\sim N(0,1)$ have probability density function $$f(y)=\dfrac{1}{\sqrt{2\pi(1)}}e^{-\frac{1}{2}\left(\frac{y-0}{1}\right)^2}=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}y^2}, -\infty<y<\infty.$$
According to the two equations above, the transformation is $Y=\dfrac{X-\mu}{\sigma}$.