Answer
a) 3.2;
b) 5.76;
c) 2.4;
d) 0.451;
e) 0.573.
Work Step by Step
We are given $f(x)=\frac{x^{-1/3}}{6}, [0,8]$
a) The expected value is
$\mu=\int^8_0 x(\frac{x^{-1/3}}{6})dx$
$=\int^8_0 \frac{x^{2/3}}{6}dx$
$=\frac{1}{10}x^{5/3}|^8_0 $
$=\frac{16}{5} = 3.2$
b) The variance is
$V=\int^8_0 x^2(\frac{x^{-1/3}}{6})dx-\mu^2$
$=\int^8_0x^{5/3}dx-(\frac{16}{5})^2$
$=\frac{1}{16}x^{8/3}|^8_0-\frac{256}{25}$
$\approx 5.76$
c) The standard deviation is
$\sigma=\sqrt V = 2.4$
d) The probability that the random variable has a value greater than the mean is
$P(x \geq \mu) = P(x \geq 3.2)=\int^8_{3.2}\frac{x^{-1/3}}{6}dx$
$=\frac{1}{4}x^{2/3}|^8_{3.2} \approx 0.457$
e) $P(\mu - \sigma \leq x \leq \mu + \sigma)= P(0.8 \leq x \leq 5.6)$
$=\int^{5.6}_{0.8}\frac{x^{-1/3}}{6}dx$
$=\frac{1}{4}x^{2/3}|^{5.6}_{0.8} \approx 0.573$
