Cumulative probabilities - when to take inverse probability? Tasked with:

Given EPS is distributed with $\mu = 6$ and $\sigma = 2$, what is the
probability that EPS will be $9.70 or more?

Answer, convert to z-score and input into $F()$:
$$ \frac{9.70-6}{2} = 1.85$$
A z-table gives us $F(1.85) = .9678$ but this is $P(EPS\le9.70)$, we want $P(EPS\gt9.70)$, which is $1-P(EPS\gt9.70)$, or .0322.
However, the following nearly identical question in terms of wording seems to have a totally different interpretation of the probability calculation:

A survey on annual income finds that $\mu = 175,000$ and $\sigma =
> 25,000$. What is the percentage that respondents of the study have
incomes greater than $150,000?

I thought, surely, we'd do the same as above. After all, calculating probability for "9.70 or more" sounds like exactly the same task as finding "greater than 150,000". However, according to my textbook, we do not. We simply use $1-F(-1)=F(1)=.8413.$ Formally:
$$ \frac{150,000-175,00}{25,000}=-1$$
Question
Wouldn't the answer be $1-.8413$ like the initial example? And if I'm wrong on that, then why is the wording is the same but the technique not?

I think you've missed a minus sign when working something out. Using the same method as your first example,
$$
\frac{150000-175000}{25000} = -1\\
F(-1) = P(inc \leq 150000) = 0.1587\\
P(inc > 150000) = 1 - P(inc \leq 150000) = 1 - 0.1587 = 0.8413
$$
Which is exactly the same as $F(1) = 0.8413$.
It looks like the answer key uses a trick that $F(a) = 1-F(-a)$ to simplify the calculation. But both methods give the same answer.
(Some Z-tables only give values for positive $a$ or for negative $a$, so that's a very useful trick to remember.)
