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$$f(x) = \begin{cases} 0 &\text{if} \,\, x < -1 \\ 1/3 &\text{if}\,\, -1 \leq x \leq 0 \\ 2/3 &\text{if}\,\, 0 < x \leq 1 \\ 0 &\text{if}\,\, x > 1 \end{cases}$$

So I had to graph the density which I did, and than illustrate using the cumulative distribution function which I did as well. What's troubling me is that the question wants to know what the median of X is and whether the median appears to be greater than or less than the mean of X. Know, my professor doesn't want us using integrals since the course only requires basic prob and stat so I need another method besides integration to find or intuitively deduce that median and mean.

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up vote 1 down vote accepted

From the post, it looks as if you know the cdf. From that, you can find the median. We will go through the details without using the term cdf.

We want to find the point $m$ such that half the "total area" under the density curve is in the part with $x\le m$, and half the area is in the part with $x\ge m$. (I assume you have graphed the density function.) The total area is $1$, so we want the area up to $m$ to be $\frac{1}{2}$.

It is intuitively clear that $m\gt 0$. The area up to $0$ is $\frac{1}{3}$. The area from $0$ to $m$ is $m\cdot \frac{2}{3}$ (base times height). So $$\frac{1}{3}+\frac{2m}{3}=\frac{1}{2}.$$ Solve. We get $m=\frac{1}{4}$.

Now for the mean. Here I am not sure that what I write will make sense to you. We can think of the mass $\frac{1}{3}$ of the part before $0$ as being concentrated at its midpoint $-\frac{1}{2}$, and the mass $\frac{2}{3}$ of the right-hand part as being concentrated at the point $\frac{1}{2}$.

Think of it as a teeter-totter (seesaw) in a playground, with a small kid of mass $\frac{1}{3}$ sitting at $-\frac{1}{2}$, and a big kid of mass $\frac{2}{3}$ sitting at $\frac{1}{2}$. We want to know where to put the fulcrum so the kids balance. Say we put it at distance $x$ from $-\frac{1}{2}$. Then $$\frac{1}{3}x=\frac{2}{3}(1-x).$$ Solve. We get $x=\frac{2}{3}$. So the mean is at $-\frac{1}{2}+\frac{2}{3}$, which is $\frac{1}{6}$.

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I actually understood that last part completely. That was super helpful and I appreciate it. Thanks! – USC Feb 5 '13 at 7:09
Good. For someone who knows basic Physics, thinking in terms of mass rather than area can be helpful. We could hav done the median in that language, finding the point $m$ such that half the total weight is to the left of $m$. The calculation is the same. – André Nicolas Feb 5 '13 at 7:15

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