Answer
a. $m=2+\sqrt 2 \approx 3.414$
b. $\approx-1.792$
Work Step by Step
We are given $f(x)=\frac{x}{8}-\frac{1}{4}; [2;6]$
The expected value $\mu=\frac{14}{3}$
a. The median is
$\int^{m}_{2}f(x)dx = \frac{1}{2}$
$\int^{m}_{2}(\frac{x}{8}-\frac{1}{4}) dx= \frac{1}{2}$
$\frac{1}{16}x^{2} - \frac{1}{4}x|^{m}_{2}= \frac{1}{2}$
$\frac{1}{16}m^{2} - \frac{1}{4}m + \frac{1}{4}= \frac{1}{2}$
$\frac{1}{16}m^{2} - \frac{1}{4}m - \frac{1}{4}= 0$
$m=2+\sqrt 2$
b. The probability between mean and median
$P(4 \leq X \leq 2+\sqrt 2)=\int^{2+\sqrt 2}_{4}(\frac{x}{8}-\frac{1}{4})dx =\frac{1}{16}x^{2}-\frac{1}{4}x|^{2+\sqrt 2}_{4}=\frac{-43}{24}\approx-1.792$
