Answer
(a) The area under the graph is equal to 1. Therefore, this function is a probability density function.
(b) (i) $P(X \lt 3) = 0.15$
(ii) $P(3 \leq X \leq 8) = 0.75$
(c) $\mu = \frac{16}{3}$
Work Step by Step
(a) The area under the graph is equal to 1. Therefore, this function is a probability density function.
(b) (i) We can find the area under the graph on the interval $[0,3)$
The area is: $\frac{1}{2}(3)(0.1) = 0.15$
$P(X \lt 3) = 0.15$
(ii) We can find the area under the graph on the interval $[3,8]$
The area is: $\frac{3}{4}(5)(0.2) = 0.75$
$P(3 \leq X \leq 8) = 0.75$
(c) We can calculate the mean:
$\mu = \int_{0}^{10}xf(x)~dx$
$\mu = \int_{0}^{6}xf(x)~dx+ \int_{6}^{10}xf(x)~dx$
$\mu = \int_{0}^{6}(x)(\frac{x}{30})~dx+ \int_{6}^{10}(x)(-\frac{0.2x}{4}+0.5)~dx$
$\mu = \int_{0}^{6}\frac{x^2}{30}~dx+ \int_{6}^{10}(-\frac{0.2x^2}{4}+0.5x)~dx$
$\mu = \frac{x^3}{90}~\vert_{0}^{6}+ (-\frac{0.2x^3}{12}+\frac{x^2}{4})~\vert_{6}^{10}$
$\mu = \frac{216}{90}+(-\frac{200}{12}+\frac{100}{4})-(-\frac{432}{120}+\frac{36}{4})$
$\mu = \frac{216}{90}-\frac{200}{12}+\frac{100}{4}+\frac{432}{120}-\frac{36}{4}$
$\mu = \frac{144}{60}-\frac{1000}{60}+\frac{1500}{60}+\frac{216}{60}-\frac{540}{60}$
$\mu = \frac{320}{60}$
$\mu = \frac{16}{3}$
