Finding the 99% of a normally distributed graph The heights of adults are normally distributed with a mean of 187.5 cm and a standard deviation of 9.5 cm. A standard doorway is designed so that 99% of adults have a space of at least 17 cm over their heads when going through a doorway. Find the height of a standard normal doorway. I tried doing invnorm(.99, 204.5, 9.5) but I got the incorrect answer. The correct one is 210 cm. I was wondering if someone could help me fix my method?

Let $x$ be the maximum height of an adult that admits a clearance of at least $17$ cm for the doorway. Let $h$ be the height of such a doorway, so $h = x + 17$. Now we wish to find the value of $x$ such that $99\%$ of adults have height less than or equal to $x$; i.e., this is $\Pr[X \le x] = 0.99$ = `invnorm(0.99,187.5,9.5) = 209.600`. Therefore, $x + 17$ is $226.6$ cm.

Your method gives the same result: $226.6$ cm. I believe the claimed answer of $210$ cm fails to take into account the requirement that the door's height must include a clearance of $17$ cm.