I am currently doing my homework and for one of the questions, it asks me to find the median of the random variable X with the probability density function: $f(x) = [1/x^2]$ with interval $[1, \infty]$. I keep getting 2 while the textbook says 1/2.
My interval for the integration is [1,m] and i'm solving for m?
Edit: My Work: $\int ^m _1 (1/x^2) dx = -x^-1 + c$ and then I plug in m and 1 which then becomes:$(-(1/m)) - (-1/1) = 1/2$ which then i solved to get m=2?
The median satisfies
$$\int_1^m f(x) dx = \frac{1}{2}$$
So then
$$\int_1^m f(x) dx = \int_1^m \frac{1}{x^2} dx = \frac{-1}{x} \Biggr \vert_{x=1}^{x=m} = \frac{-1}{m} - \frac{-1}{1} = 1 - \frac{1}{m} = \frac{1}{2}$$
$$\implies m = 2$$
If you check your result and calculate $\operatorname{P}(X \le 2)$, then you get
$$\int_1^2 f(x) dx = \int_1^2 \frac{1}{x^2} dx = \frac{-1}{x} \Biggr \vert_{x=1}^{x=2} = \frac{-1}{2} - \frac{-1}{1} = 1 - \frac{1}{2} = \frac{1}{2}$$
So most likely the answer on your textbook is wrong (if I did everything correctly).