Trying to understand when to use z-scores and how to identify them in a question, for example Here was an example question my teacher gave. Didn't know it required a mapping to the z-scores graphing arena.
question:
The scores from the first statistics exam of the year followed a normal distribution with a mean of 69 and a standard deviation of 6. If a 60 is the lowest passing grade, what is the probability that a student failed the exam? (Use a table or technology. Round your answer to four decimal places.)
My reasoning:
I thought this would be a 68-97-99.5 Empirical rule problem because that was being covered recently. But this is actually a z-scores transformation mapping problem. I am having trouble understanding just what exactly it means to go from a raw data to mapping it to the z-score arena, what that actually means once it's turned into a z-score. 
z = (score - x-bar)/s.d
So,
z = (60 - 68)/6 = - 1.5
At this point I don't really know what -1.5 represents. This is the amount of deviation from the mean, right? So essentially now that we have it mapped in the z-score arena we can from their look to the z-chart to see all the data that falls on either side? I don't understand the logic here behind it. This is where I get lost. Once it's transformed it's like, left side? right side? I am not sure what to do after this.
Please help
Thank you 
 A: I don't quite follow the logic of your transformation (maybe you have a typo in your Question), but you can solve the problem stated using z scores.
You have $X \sim \mathsf{Norm}(\mu = 69, \sigma=6)$ and you seek
$P(X \le 60) = 0.0668.$ 
Software. Using technology such as R statistical software (where pnorm is a normal
cdf), you can solve this
directly as follows:
pnorm(60, 69, 6)
## 0.0668072

Also, some statistical calculators are programmed to do this kind of computation.
Printed standard normal tables. If you use standard normal CDF tables, you can get very close to this answer
by standardizing and looking up a probability in the tables:
$$P(X \le 60) = P\left(\frac{X - \mu}{\sigma} \le \frac{60 - 69}{6} \right)
= P(Z \le -1.5) = 0.0668.$$
The z-score $-1.5$ indicates that the borderline failing score 60 is 1.5 standard deviations below the mean 69. [Standardization allows a vast
variety of problems involving different values of $\mu$ and $\sigma$ to
be solved using the same printed table.]
I don't know exactly what style of printed normal tables you are using, so
I can't tell you exactly how to get the answer 0.0668 from the table.
If your standard normal CDF table has negative z-values, look for -1.5 in
the margin of the table and then look in the body of the table for the corresponding probability. If your table has only positive z-values, you
can use the symmetry of the standard normal distribution about 0. 
Graphs. The left-hand panel below shows the PDF of the standard normal distribution. The area under the density curve to the left of the vertical red dashed line
is your answer. Using tables you may need to get the area under the curve
between 0 and +1.5 or the area below +1.5 (depending on the style of the table),
and then use symmetry and rules of probability to deduce the required probability.
The right-hand panel shows the PDF of $\mathsf{Norm}(\mu = 69, \sigma=6).$
The area under the density curve to the left of the red line is the what
I found using R software.

