Answer
(a) $\int_{-\infty}^{\infty} f(x)~dx = 1$
The function $f$ is a probability density function.
(b) $\mu = 5$
Work Step by Step
(a) We can verify that $f$ is a probability density function:
$\int_{-\infty}^{\infty} f(x)~dx = \int_{0}^{10}f(x)~dx$
$\int_{-\infty}^{\infty} f(x)~dx = \int_{0}^{10}0.1~dx$
$\int_{-\infty}^{\infty} f(x)~dx = 0.1x~\vert_{0}^{10}$
$\int_{-\infty}^{\infty} f(x)~dx = (0.1)(10)-(0.1)(0)$
$\int_{-\infty}^{\infty} f(x)~dx = 1$
The function $f$ is a probability density function.
(b) Since every real number between 0 and 10 has an equal probability of being selected, we would expect the mean to be the average number in this interval which is 5.
We can check our intuition by evaluating the following integral:
$\mu = \int_{-\infty}^{\infty} xf(x)~dx$
$\mu = \int_{0}^{10} xf(x)~dx$
$\mu = \int_{0}^{10} 0.1x~dx$
$\mu = \frac{0.1}{2}x^2~\vert_{0}^{10}$
$\mu = \frac{0.1}{2}(10)^2 - \frac{0.1}{2}(0)^2$
$\mu = 5-0$
$\mu = 5$
