Answer
(a) $\int_{-\infty}^{\infty} f(x)~dx = 1$
The function $f$ is a probability density function.
(b) $P(X \leq \frac{1}{3}) = \frac{17}{81}$
Work Step by Step
(a) We can verify that $f$ is a probability density function:
$\int_{-\infty}^{\infty} f(x)~dx = \int_{0}^{1}f(x)~dx$
$\int_{-\infty}^{\infty} f(x)~dx = \int_{0}^{1}30x^2~(1-x)^2~dx$
$\int_{-\infty}^{\infty} f(x)~dx = \int_{0}^{1}30x^2~(x^2-2x+1)~dx$
$\int_{-\infty}^{\infty} f(x)~dx = \int_{0}^{1}(30x^4-60x^3+30x^2)~dx$
$\int_{-\infty}^{\infty} f(x)~dx = (6x^5-15x^4+10x^3)\vert_{0}^{1}$
$\int_{-\infty}^{\infty} f(x)~dx = [6(1)^5-15(1)^4+10(1)^3]-0$
$\int_{-\infty}^{\infty} f(x)~dx = 1$
The function $f$ is a probability density function.
(b) We can find $P(X \leq \frac{1}{3})$:
$P(X \leq \frac{1}{3}) = \int_{0}^{1/3}f(x)~dx$
$P(X \leq \frac{1}{3}) = \int_{0}^{1/3}(30x^4-60x^3+30x^2)~dx$
$P(X \leq \frac{1}{3}) = (6x^5-15x^4+10x^3)\vert_{0}^{1/3}$
$P(X \leq \frac{1}{3}) = [6(\frac{1}{3})^5-15(\frac{1}{3})^4+10(\frac{1}{3})^3]-0$
$P(X \leq \frac{1}{3}) = \frac{2}{81} - \frac{15}{81}+\frac{10}{27}$
$P(X \leq \frac{1}{3}) = \frac{2}{81} - \frac{15}{81}+\frac{30}{81}$
$P(X \leq \frac{1}{3}) = \frac{17}{81}$
