Standard Normal Distribution $\Phi^{−1}(z)$ and $\Phi(z)$ 
Scores on an examination are assumed to be normally distributed with
  mean 78 and variance 36.
a) What is the probability that a person taking the examination scores
  higher than 72?
b) Suppose that students scoring in the top 10% of this distribution
  are to receive an A grade. What is the minimum score a student must
  achieve to earn an A grade?
Express your answer in terms of $\Phi^{-1}(z)$ for $0.5 \lt p \lt 1$

What does the above text mean? I know how to express my answer in terms of $\Phi(z)$, but how do I express my answer in terms of $\Phi^{-1}(z)$? 
EDIT: I think $\Phi^{-1}(z)$ is the quantile function but I'm not sure.
 A: a) is easy, we have, denoting the random variable "score" in question by $S$, that: 
\begin{align*}
  \def\P{\mathbb P}\P(S > 72) &= \P(S - 78 > -6)\\
       &= \P\left(\frac{S-78}6 > -1\right)\\
       &= 1 - \Phi(-1).
\end{align*}
b) Asks us to solve the equation 
$$ \P(S > s) = 0.1 $$
for $s$. We first do as in a), writing 
$$ \P(S > s) = \P \left(\frac{S-78}6 > \frac{s-78}6\right)  = 1 - \Phi\left(\frac{s-78}6\right) $$
So, 
\begin{align*} 1 - \Phi\left(\frac{s-78}6\right) = 0.1 &\iff 
    &\Phi\left(\frac{s-78}6\right) &= 0.9\\ &\iff& 
   \frac{s-78}6 &= \Phi^{-1}(0.9)\\& \iff& s &= 78 + 6\Phi^{-1}(0.9) 
\end{align*}
A: The required form   is already given in the answer  of martini $78+6\Phi^{-1}(0.9)$. But there is an alternative form.
Because of the symmetry of the standard normal distribution around $0$ it is  
$$\Phi^{-1}(p)=-\Phi^{-1}(1-p)$$. 
As I have mentioned in the comment the arguement should be $p$ not $z$. Finally we have the equality
$$78+6\Phi^{-1}(0.9)=78-6\Phi^{-1}(0.1)$$ where $0<p<0.5$
