I'm struggling to understand Find P(|X - 100| > 5) due to the modulus.
I have no problem with the standard format, like P(X < 5) but I really struggle when there are inequalities, or additions/subtractions on the left-hand-side like this:
P(|X - u| < 5)
P(X - u < 5)
Could someone please help me understand the general technique, so I can evaluate all types?
Let $d \ge 0$. Then $$|X - c| > d$$ is equivalent to saying "the distance of $X$ from $c$ is greater than $d$." This means $X - c > d$, or $c - X > d$. Since these are disjoint in the sense that $X$ cannot simultaneously satisfy both of these inequalities (why?), then
$$\begin{align} \Pr[|X - c| > d] &= \Pr[(X - c > d) \cup (c-X > d)] \\ &= \Pr[X - c > d] + \Pr[c - X > d] \\ &= \Pr[X > c+d] + \Pr[X < c-d]. \end{align}$$
In your case, $c = 100$ and $d = 5$ gives $$\Pr[|X - 100| > 5] = \Pr[X > 105] + \Pr[X < 95].$$ And this makes intuitive sense, since $|X - 100| > 5$ means "the distance of $X$ from $100$ is greater than $5$." That implies $X$ is more than $105$ or less than $95$.
Notice that none of the above has anything to do with $X$ being normally distributed. It's just algebra. The only property related to probability that we used was that the probability of the union of two disjoint events is the sum of the probabilities of each event; i.e. $$\Pr[A \cup B] = \Pr[A] + \Pr[B] \quad \text{if} \quad \Pr[A \cap B] = 0.$$