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!

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"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 $\mu$

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 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*}

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

There are several ways to approach this. One is: if $X$ is normally distributed with mean $\mu$ and variance $\sigma^2 = 100$, then $Z = \frac{X - \mu}{\sigma}$ is standard normally distributed with mean $0$ and variance $1$. So then

$$P(X > 16.34) = P\left(\frac{X - \mu}{\sigma} > \frac{16.34 - \mu}{\sigma}\right) = P\left(Z > \frac{16.34 - \mu}{10}\right) = 1 - P\left(Z \leq \frac{16.34 - \mu}{10}\right) = 1 - \Phi\left(\frac{16.34 - \mu}{10}\right) = 0.1212.$$

Doing a little algebra, we get

$$\Phi\left(\frac{16.34 - \mu}{10}\right) = 0.8788.$$

We can now do e.g. a table lookup on the value $0.8788$ to find that

$$\Phi(1.169) \approx 0.8788.$$

This tells us that

$$\frac{16.34 - \mu}{10} = 1.169 \quad \Rightarrow \quad 16.34 - \mu = 11.69 \quad \Rightarrow \quad \color{blue}{\mu = 4.65}.$$

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 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 $$X=\mu+\sigma Z\qquad\implies$$ $$\mu=X-\sigma Z=16.34-11.69=4.65\,.$$

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