Hello Mathematics dep!
Variate $X$, which follows the normal distribution, has median $\mu = 14$ and variance $\sigma ^2 = 9$. What are the odds that $X > 12$.
Attempt at a solution: $P(X>12) = 1 - P(X \leq 12)$ \begin{equation} z = \frac{x - \mu}{\sigma} = \frac{12 - 14}{3} = -2/3 \end{equation} so \begin{equation} \frac{1}{\sigma \sqrt{2 \pi}} e^{-z ^2 /2} = \frac{1}{3 \cdot \sqrt{2 \pi}}e^{-(-2/3)^2 /2} = 0.106482669 \end{equation} $1 - 0.106482669 = 0.893517331$ which is the wrong answer. Any pointers would be appreciated; I've grown tired of having this problem defeating me.
Thanks.