# The difference of two normal random variables

Below is a problem that I did. I would like somebody to check it for me.
Thanks,
Bob

Problem:
Suppose that $$z_1$$ and $$z_2$$ are two independent random variables that are normally distributed with mean $$2$$ and standard deviation of $$4$$. Now if $$z = |z_1 - z_2|$$, what is the probability that $$z$$ will be greater than 10?
Let $$u_z$$ by the mean of the variable $$z$$. Let $$p$$ be the probability we seek. \begin{align*} u_z &= 0 \\ p &= 2P(z_1 - z_2 \geq 10) \\ \text{Let } z_3 &= z_1 - z_2 \\ p &= 2P(z_3 \geq 10) \\ \end{align*} The variance of $$z_3$$ is $$4^2 + 4^2 = 32$$ and the standard deviation fo $$z_3$$ is $$\sqrt{32} = 4\sqrt{2}$$. Now we need to ask, how many standard deviations does $$10$$ represent. The number $$10$$ represents $$\frac{10}{4\sqrt{2}} = 1.7678$$ standard deviation. The Z-score of $$1.7678$$ is $$0.9614528$$. \begin{align*} p &= 2( 1 - 0.9614528) \\ p &= 0.0770944 \\ \end{align*}
• OK, except $u_z$ (the mean of $|z_1 - z_2|$) is not $0$; rather, the mean of $z_1-z_2$ is $0$. – r.e.s. Jun 6 at 4:06
Your approach is correct. The variance of the $$z_3$$ is indeed equal to the sum of the variances of $$z_1$$ and $$z_2$$, and you correctly computed both tails of the distribution corresponding to $$z_1-z_2<-10$$ and $$z_1-z_2>10$$.