You're looking for the 15th percentile of a normal distribution with <em>µ</em> = 242 and <em>σ</em> = 13, which is to say you want to find <em>x</em> * such that
P(<em>X</em> ≤ <em>x</em> *) = 0.15
Transform the wait-time random variable <em>X</em> to <em>Z</em>, which follows the standard normal distribution with mean 0 and s.d. 1 :
<em>Z</em> = (<em>X</em> - <em>µ</em>) / <em>σ</em> → <em>X</em> = <em>µ</em> + <em>σ</em> <em>Z</em> = 242 + 13 <em>Z</em>
Now,
P(<em>X</em> ≤ <em>x</em> *) = P(242 + 13<em> Z</em> ≤ <em>x</em> *) = P(<em>Z</em> ≤ (<em>x</em> * - 242)/13) = 0.15
Use a calculator (something with an inverse CDF function) or <em>Z</em>-table to look up the <em>z</em>-score, <em>z</em> = (<em>x</em> * - 242)/13, associated with a probability of 0.15. You would find that
<em>z</em> = (<em>x</em> * - 242)/13 ≈ -1.03643
Solve for <em>x</em> * :
<em>x</em> * ≈ 242 + 13 (-1.03643) ≈ 228.526 ≈ 229