Answer
(a) The probability that the amount of REM sleep is between 30 and 60 minutes is $\frac{19}{32}$ which is 59.4%
(b) The mean amount of REM sleep is 40 minutes.
Work Step by Step
(a) We can find $P(30 \leq T \leq 60)$:
$P(30 \leq T \leq 60) = \int_{30}^{60}f(t)~dt$
$P(30 \leq T \leq 60) = \int_{30}^{40}f(t)~dt+\int_{40}^{60}f(t)~dt$
$P(30 \leq T \leq 60) = \int_{30}^{40}\frac{1}{1600}t~dt+\int_{40}^{60}\frac{1}{20}-\frac{1}{1600}t~dt$
$P(30 \leq T \leq 60) = \frac{1}{3200}t^2~\vert_{30}^{40}+(\frac{1}{20}t-\frac{1}{3200}t^2)~\vert_{40}^{60}$
$P(30 \leq T \leq 60) = \frac{(40)^2}{3200}-\frac{(30)^2}{3200}+(\frac{60}{20}-\frac{(60)^2}{3200})-(\frac{40}{20}-\frac{(40)^2}{3200})$
$P(30 \leq T \leq 60) = \frac{16}{32}-\frac{9}{32}+\frac{96}{32}-\frac{36}{32}-\frac{64}{32}+\frac{16}{32}$
$P(30 \leq T \leq 60) = \frac{19}{32}$
The probability that the amount of REM sleep is between 30 and 60 minutes is $\frac{19}{32}$ which is 59.4%
(b) We can find the mean amount of REM sleep:
$\mu = \int_{-\infty}^{\infty}t~f(t)~dt$
$\mu = \int_{0}^{40}t~f(t)~dt+ \int_{40}^{80}t~f(t)~dt$
$\mu = \int_{0}^{40}\frac{1}{1600}t^2~dt+ \int_{40}^{80}(\frac{1}{20}t-\frac{1}{1600}t^2)~dt$
$\mu = \frac{1}{4800}t^3~\vert_{0}^{40}+ (\frac{1}{40}t^2-\frac{1}{4800}t^3)~\vert_{40}^{80}$
$\mu = \frac{(40)^3}{4800}-0+ (\frac{(80)^2}{40}-\frac{(80)^3}{4800})-(\frac{(40)^2}{40}-\frac{(40)^3}{4800})$
$\mu = \frac{40}{3}+ \frac{480}{3}-\frac{320}{3}-\frac{120}{3}+\frac{40}{3}$
$\mu = 40$
The mean amount of REM sleep is 40 minutes.
