Suppose I use the following rules to generate random numbers. I roll a fair $D_6$ twice. The first roll I call $\mu$ and the second $\sigma^2$. I then Generate a normal random variable with mean $\mu$ and variance $\sigma^2$. I call this $X$

In more concise notation: $X \sim N(U_1,U_2)$ where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$

Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$

I think $x=3.5$.

Reasoning: Symmetry. More precisely: $\mathbb{P}[X \leq x] = \sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\mathbb{P}[U_1=i]=\sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\frac{1}{6}$

Now examine each term in the sum, $\mathbb{P}[X\leq x|U_1=i]\cdot\mathbb{P}[U_1=i]$. Let us take $i=1 $ and $i=6$ as $3.5$ is in the middle of these two values we have that $\mathbb{P}[N(1,U_2)\leq3.5]=\mathbb{P}[N(6,U_2)\geq3.5]$ from symmetry of normal dist. And then finally $\mathbb{P}[N(6,U_2)\leq3.5] = 1- \mathbb{P}[N(6,U_2)\geq3.5] = 1 - \mathbb{P}[N(1,U_2)\leq3.5]$

Hence: $\mathbb{P}[N(1,U_2)\leq3.5]+\mathbb{P}[N(6,U_2)\leq3.5] = \mathbb{P}[N(1,U_2)\leq3.5] + 1- \mathbb{P}[N(1,U_2)\leq3.5]=1 $

So if we split the sum into the $3$ pairs of integers that have $3.5$ in the middle, i.e sum to $7$ i.e $\{(1,6) , (2,5) , (3,4) \}$ we end up with:

$\mathbb{P}[X \leq x] = \sum\limits_{i=1}^6\mathbb{P}[X\leq x|U_1=i]\cdot\frac{1}{6} = \frac{1}{6}(1+1+1)=\frac{1}{2}$

Is this correct? :)

The second part of the question is to now have $X \sim N(U_1^2,U_2)$ where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$ Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$

And then finally to have $X \sim N(U_1^n,U_2)$ for some $n \in \mathbb{N}$ and where $U_1,U_2$ are identical independent discrete uniform R.V's on $\{1,2,3,4,5,6\}$ Find an $x \in \mathbb{R} $ such that $\mathbb{P}[X\leq x] = \frac{1}{2}$