Answer
$p(x)$ is a probability density function on $[1, \infty)$.
and $\mu=\dfrac{3}{2} $
Work Step by Step
We have:
$\int_{1}^{\infty} 3x^{-4} \ dx=\lim\limits_{a \to \infty}\int_{1}^{a} 3x^{-4} \ dx\\=(-1) \lim\limits_{a \to \infty} \dfrac{1}{a^3}-1\\=(-1) (-1)\\=1 $
This implies that $p(x)$ is a probability density function on $[1, \infty)$.
The mean value is:
$\mu=\int_{1}^{\infty} (x) 3x^{-4} \ dx \\=\lim\limits_{a \to \infty}\int_{1}^{a} 3x^{-3} \ dx\\=\lim\limits_{a \to \infty} [\dfrac{-3}{2}x^{-2}]_1^a\\=\dfrac{-3}{2} \lim\limits_{a \to \infty} \dfrac{1-a^2}{a^2} \\=\dfrac{3}{2} $
