Standard Normal Distribution $\Phi^{−1}(z)$ and $\Phi(z)$

Question

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.

Answer 1

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*}

Answer 2

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 $0<p<0.5$