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?
Answer:
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*}