Error in solving for raw score; incorrect formula used?

Question

According to a study of how long a person is willing to wait for their flight, it is found that the mean time a person is willing to wait is 5.2 hours with a standard deviation of 1.1 hours.

Consider the 20% of people who are least willing to spend time traveling. How many hours traveling is the cut-off at which these people would not consider waiting?

This practice question asks me of the formula used, as well as the answer. The formula I am using (which is confirmed via answer key) is:

$$x = z\sigma + \mu$$ $$x = (0.2*1.1)+5.2$$ $$x = 5.42$$

However, the answer provided does not match, with a hard $x = 4.276$. I'm not entirely sure why this is the answer despite using the same formula it indicates to use.

Answer 1

You want the $z$ such that $Pr(Z<z)=0.2$. This can be found using a standard normal table, or from here. The value is about -0.84, so with this you get

$$ x=(-0.84)*1.1+5.2=-0.924+5.2=4.276. $$

Answer 2

The formula $x=z\sigma+\mu$ is correct, but $z$ is not $0.2$.
Usually, in these problems, you assume a normal distribution.
$z$ is the number for which 20 percent of standard normal random numbers are below $z$.
A graphical calculator or ExCel can tell you what that number is. It will be around -1. I think the command is something like norminv(0.2) Once you have $z$, you can calculate $x$.