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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?

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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:

$$\mathbb{E}(Y) = \frac{1}{\theta} \approx 43.95579.$$

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