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

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*}

  • $\begingroup$ The approach looks correct to me. Didn't go into the calculations tho. $\endgroup$ – Vizag Jun 6 '19 at 1:42
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    $\begingroup$ OK, except $u_z$ (the mean of $|z_1 - z_2|$) is not $0$; rather, the mean of $z_1-z_2$ is $0$. $\endgroup$ – r.e.s. Jun 6 '19 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 $z_1-z_2>10$.


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