Applying a Normal Distribution to Another Function to Find Probability

Question

Suppose that the number of hours students spend studying for an exam is approximately normally distributed with $\mu=10$ and $\sigma=\sqrt{2}$. If a student spends $t$ hours studying, he/she receives a grade of M(t): $$ M(t) = \frac{1}{1 + e^{7-t}} $$

1. What is the probability that a student receives at least 90%?

2. What is the probability that a student receives at least 90%, given that he/she studied for at least 12 hours?

1. I think I want to find $\mathbb{P}(X \geq 0.9)$, where $X = M(T)$, and $T \sim \mathcal{N}(10, \sqrt{2})$. Then, $$\begin{align*} \mathbb{P}(X \geq 0.9) &= 1 - \mathbb{P}(X \leq 0.9)\\ &= 1 - \mathbb{P}(M(T) \leq 0.9)\\ &= 1 - \mathbb{P}(T \leq M^{-1}(0.9)) \end{align*}$$ I can then use the CDF of the normal distribution to calculate $\mathbb{P}(T \leq 9.197)$ with $\mu=10$ and $\sigma=\sqrt{2}$, giving me 0.715. Is this the right idea?

2. Here, I am trying to find $\mathbb{P}(X \geq 0.9 \mid T \geq 12)$. I'm not sure how to proceed from here, though. Could someone provide me perhaps with a pointer to what to do next? Thanks!

Answer 1

Your handling of 1) is correct. It works fine because function $M$ is strictly increasing.

$X=M(T)$ is completely determined by $T$ and $M(t)\geq 0.9$ is true for every value $t\geq12$. This allready justifies the conclusion: $$P(X=M(T)\geq0.9\mid T\geq12)=1$$

Working alternatively with conditional probability would lead to $$P(X\geq0.9\mid T\geq12)=P(X\geq0.9\wedge T\geq12)P(T\geq12)^{-1}$$

Here : $$X\geq0.9\wedge T\geq12\iff T\geq12$$ and consequently: $$P(X\geq0.9\mid T\geq12)=P(T\geq12)P(T\geq12)^{-1}=1$$