Suppose that $X_1$, $X_2$, and $X_3$ are independent, normal distributed with same mean and standard deviation. How do I find $P(X_1+X_2 > 1.9X_3)$?
What I did so far:
$$P(X_1+X_2 > 1.9X_3) = 1- P\left(\dfrac{X_1+X_2}{X_3}\le1.9\right)$$
Adding $X_1$ and $X_2$ together will yield a new random variable with different normal distribution:
$$X_1+X_2\sim N(\mu_{X_1}+\mu_{X_2},\sigma^2_{X_1}+\sigma^2_{X_2})$$ Then how do I calculate a random variable with ratio distribution of $\dfrac{X_1+X_2}{X_3}$? Cauchy distribution would not work here because the means are not zero. Is there other way to do this?