It is estimated that 80% of all eighteen-year-old women... Homework question here, I'm not sure how to finish it out.
It is estimated that $80\%$ of all 18-year-old women have weights ranging from 103.5 to 144.5 lb. Assuming the weight distribution can be adequately modeled by a normal distribution and assuming that 103.5 and 144.5 are equidistant from the average weight $\mu$, calculate $\sigma$.
So I have:
$$\mu = \frac{103.5+144.5}{2} = 124 = E(x)$$
$$P\left(\frac{103.5 - E(x)}{\sigma} \leq \frac{x - E(x)}{\sigma} \leq \frac{144.5 - E(x)}{\sigma}\right) = 0.8$$
$$P\left(-\frac{20.5}{\sigma} \leq z \leq \frac{20.5}{\sigma}\right) = 0.8$$
This is where I get stuck. I found this solution, but I have no clue where the 1.28 comes from. 
Thanks for the help.
 A: From the rule of complement which says that $P(A^c) = 1 - P(A)$ it follows that: $P(z<-\frac{20.5}{\sigma}~~or~~z>\frac{20.5}{\sigma}) = 1 - 0.8 = 0.2$
From the symmetry of the normal distribution then, $P(z<-\frac{20.5}{\sigma}) = 0.1$
From here, either we look at a table of probabilities associated with different values of $z$ or we use calculus to solve it directly.  Given that the wording of the question seems to be the first time you are taught this topic I'll assume you don't know the integral yet.  So, looking at a table for what value of $z$ gives us that score, it corresponds to a value of $-1.282$.

To read the table to find which value of $z$ gives 10%, find the entry which looks as close to 0.100 as possible.  In this table, the number on the left with the extra hundredths place digit from the top, so in this case somewhere between -1.28 and -1.29 is the number we want.  With a larger table you will find more accurate results.
From here you can finish calculating for $\sigma$ by the equation $-\frac{20.5}{\sigma} = -1.282$
