Probability that $f(x,y,z)>0$ given the variables follow normal distribution

Question

Assuming that variables $x,y$ and $z$ follow the Gaussian distribution with $\mu_x=\mu_y=\mu_z=1000000$ and $\sigma_x=\sigma_y=\sigma_z=200000$, what is the probability that $$f(x,y,z) = \frac{x}{1.1}+\frac{y}{1.1^2}+\frac{z}{1.1^3}-2000000>0$$

My try: $P(x)=\frac{e^{\frac{-(x-\mu)^2}{2\sigma^2}}}{\sigma \sqrt{2\pi}}$ The required probability is $$\int\int\int P(x)P(y)P(z) dx dy dz $$ with $(x,y,z)$ such that $f(x,y,z)>0$. After this I don't know how to integrate that too over such a difficult domain.

Answer 1

We have to assume independence. Then your random variable has normal distribution.

The mean of $aX+bY+cZ-d$ is $aE(X)+bE(Y)+cE(Z)-d$. In our case we have $a=\frac{1}{1.1}$, $b=\frac{1}{(1.1)^2}$, and $c=\frac{1}{(1.1)^3}$.

The variance of $aX+bY+cZ-d$ is $ a^2\text{Var}(X)+b^2\text{Var}(Y)+c^2\text{Var}(Z)$.

Now you have everything needed for the computation of your probability. You know how to find the probability that a normal random variable with known mean and variance is $\gt 0$.