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.
Best Answer
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*}