# Understanding Expected Value

If $X$ is a normally distributed random variable with standard deviation $\sigma=10$, and $P(X>16.34) = .1212$, what is the mean (expected value) of $X$?

Attempt at solution: This problem doesn't make sense... standard deviation is given, by the probability $X>16.34$ has no upper bound, so how can this be computed? The expected value is just the summation of all the values which $=.1212$ here, so I'm not exactly sure what is being asked. please help!

-
"by the probability $X>16.34$ has no upper bound" uh... what? –  leonbloy May 3 '12 at 18:48
The is no need to put an upper bound, but you may put upper bound M such that the P(X>M) will be so small such that it won't make any difference –  checkmath May 3 '12 at 18:54

Hint: For a normal distribution, the probability can be converted to how many standard deviations you are away from the mean. For example, following the 68-95-99.7 rule the chance that you are over $2\sigma$ above the mean is about $0.025$. You should be able to find some $n$ such that the chance you are at least $n\sigma$ above the mean is $0.1212$. This says that $\mu + n\sigma$ is that value. If you have $n$ and $\sigma$, you can find $\mu$

-
Thanks for the push in the right direction, I will work through it from here. –  EulerChild May 3 '12 at 18:50

Let $\mu$ be the expected value of $X$. Normalizing gives, $$P\left(Z > \frac{16.34 - \mu}{10}\right) = .1212.$$ Using a normal distribution chart, the corresponding $z$-score is approximately 1.17. Thus, \begin{align*} 1.17 &= \frac{16.34 - \mu}{10}\\ 11.7 &= 16.34 - \mu\\ \mu &= 4.64 \end{align*}

-
got it thanks!!!! –  EulerChild May 3 '12 at 19:13

The random variable $X$ has no theoretical upper bound; $X$ is indeed unbounded, mathematically speaking. However, the right tail of its distribution, being a normal curve, decays extremely fast -- the exponential of a negative quadratic. That means that for all intents and purposes, you should not be worrying about $X>\mu+10\sigma$, since the chances of this are astronomically low.

As for the problem at hand, if you convert to a standard normal variate $$Z=\frac{X-\mu}{\sigma}$$ then $$0.1212=P(X > 16.34)=P\left(Z>\frac{16.34-\mu}{10}\right)$$ is the area of a right tail of the normal distribution. The corresponding cutoff value is $z\approx1.169$, which leads us to $$X=\mu+\sigma Z\qquad\implies$$ $$\mu=X-\sigma Z=16.34-11.69=4.65\,.$$

-
got it thanks!!!!! –  EulerChild May 3 '12 at 19:13