Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
5  
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

1 Answer 1

up vote 1 down vote accepted

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.