Normal distribution percentile calculation I'm working out the following problem and there is a part that I am not understanding clearly. * The weight distribution of parcels sent is normal with mean value $12$ lbs and standard deviation 3.5lbs. * The service wishes to establish a weight value c beyond which there will be a surcharge. * What value $c$ is such that $99\%$ of all parcels that are at least 1 lb under the surcharge weight? I feel like I get the general task presented, however there is one part that I am stuck on. $$ P(x < c-1) = 0.99 $$ $$P\left(z \le \frac {c-1-12}{3.5} \right) = 0.99 $$ $$\frac {c-13}{3.5} = 0.99 $$ I know I have to solve for $c$ but I'm not sure why the answer that I get is wrong from the book's at this point. The book's answer to this question is 21.155 and I'm getting 16.465.

You need $\Phi\left(\dfrac{c-13}{3.5}\right) = 0.99$, not $\dfrac{c-13}{3.5} = 0.99$.

Hence $\dfrac{c-13}{3.5} = \Phi^{-1}(0.99)$, and so $\dfrac{c-13}{3.5} \approx 2.33$.

$\Phi$ is the function tabulated in the back of every book.