# Expected number of days?

You’re drawing from a random variable that is normally distributed $$X \sim \text{N}(0,1)$$, once per day. What is the expected number of days that it takes to draw a value that’s higher that two?

Started with calculating the probability $$\mathbb{P}(X \geqslant 2) \approx. 0.0227$$. We have $$E(x) \geqslant n \cdot \mathbb{P}(X \geqslant 2)$$, i.e $$n \leqslant 100$$. Is this the correct answer?

Let $$X_1,X_2,X_3,... \sim \text{IID N}(0,1)$$ be the draws on each day. The probability of interest for a single day is:
$$\theta \equiv \mathbb{P}(X_i>2) = 1- \Phi(2) \approx 0.02275013.$$
Letting $$Y \equiv \min \{ n \in \mathbb{N} | X_n > 2 \}$$ be the first day where we draw a value greater than two, this random variable has a geometric distribution $$Y \sim \text{Geom}(\theta)$$. The expected number of days until we draw a value greater than two is:
$$\mathbb{E}(Y) = \frac{1}{\theta} \approx 43.95579.$$