Answer
a. 12.15;
b. 5.55;
c. 2.4;
d. 0.535;
e. 0.6
Work Step by Step
We are given $f(x)=\frac{\sqrt x}{18}=\frac{x^{1/2}}{18}, [0,9]$
a) The expected value is
$\mu=\int^9_0 x(\frac{x^{1/2}}{18})dx$
$=\int^9_0 \frac{x^{3/2}}{18}dx$
$=\frac{1}{45}x^{5/2}|^9_0 $
$=\frac{27}{5} = 5.4$
b) The variance is
$V=\int^9_0 x^2(\frac{x^{1/2}}{18})dx-\mu^2$
$=\frac{1}{45}\int^9_0x^{5/2}dx-(5.4)^2$
$\approx 5.55$
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
$P(x \geq \mu) = P(x \geq 5.4)=\int^9_{5.4}\frac{x^{1/2}}{18}dx$
$=\frac{1}{27}x^{3/2}|^9_{5.4} \approx 0.535$
e) $P(\mu - \sigma \leq x \leq \mu + \sigma)= P(3 \leq x \leq 7.8)$
$=\int^{7.8}_{3}\frac{x^{1/2}}{18}dx$
$=\frac{1}{27}x^{3/2}|^{7.8}_{3} \approx 0.6$
