If $X\sim N(y,\sigma^2)$ and $\mathbb P(X<x)=k$, find $\mathbb P(X<2y-x)$ in terms of $k$.
Note that $y$ is the mean and $\sigma$ is the standard deviation.
I have tried expressing each probability using the standard normal distribution, but I just end up with two terms that don’t help me,$Z<\frac{(x-y)}\sigma$ and $Z<\frac{(y-x)}\sigma$. I’m not sure where to go from here. If you can offer any advice, thanks.
Let $X = y + \sigma Y$, where $Y \sim N(0,1)$.
We know that $k = \mathbb P(X<x) = \mathbb P(Y < \frac{x-y}\sigma)$, therefore $$\mathbb P(X < 2y-x) = \mathbb P(Y < \frac{y-x}\sigma) = 1- \mathbb P(Y \geq \frac{y-x}\sigma) = 1 - \mathbb P(Y \leq \frac{x-y}\sigma) = 1 - k$$