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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.


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You're using the pdf instead of the cdf. – anon Apr 23 '12 at 10:58
P(X<=12)=P(Z<=-2/3). So look for the probability in the standard normal table for the cumulative probability correcponding to x=-2/3. – Michael Chernick May 21 '12 at 18:03
up vote 1 down vote accepted

look up the answer in a z-table, after finding the z-score.

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