Answer
a) 1.6;
b) 0.107;
c) 0.327;
d) 0.5904;
e) 0.698
Work Step by Step
We are given $f(x)=\frac{1}{4}x^3, [0,2]$
a) The expected value is
$\mu=\int^2_0x(\frac{1}{4}x^3)dx$
$=\frac{1}{4}\int^2_0x^4dx$
$=\frac{1}{4}(\frac{x^5}{5})|^2_0$
$=\frac{8}{5}=1.6$
b) The variance is
$V=\int^2_0x^2(\frac{1}{4}x^3)dx-\mu^2$
$=\frac{1}{4}\int^2_0x^5dx-(1.6)^2$
$=\frac{1}{4}(\frac{x^6}{6})|^2_0-(1.6)^2$
$=0.107$
c) The standard deviation is
$\sigma=\sqrt V \approx 0.327$
d) $P(x \geq \mu)=\int^2_{1.6}\frac{1}{4}x^3dx$
$=\frac{1}{16}x^4|^2_{1.6}$
$=0.5904$
e) $P(\mu - \sigma \leq x \leq \mu + \sigma)=P(1.273 \leq x \leq1.927)$
$=\int^{1.927}_{1.273}\frac{1}{4}x^3dx$
$=\frac{1}{4}(\frac{x^4}{4})|^{1.927}_{1.273}\approx 0.698$
