Answer
a) 0.75;
b) 0.2375;
c) 0.487;
d) 0.464;
e) 0.63
Work Step by Step
We are given $f(x)=\frac{3}{16}(4-x^2), [0,2]$
a) The expected value is
$\mu=\int^2_0x(\frac{3}{16}(4-x^2))dx$
$=\frac{3}{16}\int^2_0(4x-x^3)dx$
$=\frac{3}{16}(2x^2-\frac{x^4}{4})|^2_0$
$=\frac{3}{4}=0.75$
b) The variance is
$V=\frac{3}{16}\int^2_0x^2(4-x^2)dx-\mu^2$
$=\frac{3}{16}\int^2_0 (4x^2-x^4)dx-(0.75)^2$
$=\frac{3}{16}(\frac{4x^3}{3}-\frac{x^5}{5})|^2_0-(0.75)^2$
$=0.2375$
c) The standard deviation is
$\sigma=\sqrt V \approx 0.487$
d) $P(x \geq \mu)=\int^2_{0.75}\frac{3}{16}(4-x^2)dx$
$=\frac{3}{16}(4x-\frac{x^3}{3})|^2_{0.75}$
$=0.464$
e) $P(\mu - \sigma \leq x \leq \mu + \sigma)=P(0.263 \leq x \leq1.237)$
$=\int^{1.237}_{0.263}\frac{3}{16}(4-x^2)dx$
$=\frac{3}{16}(4x-\frac{x^3}{3})|^{1.273}_{0.263}\approx 0.63$
