Answer
a. $\mu=\frac{37}{10}$
b. $\sigma=\frac{7\sqrt 3}{30}\approx0.404$
c. $1.289$
Work Step by Step
We are given $f(x)=\frac{5}{7}; [3;4.4]$
The mean of the distribution:
$\mu=\frac{1}{2}(4.4+3)=\frac{37}{10}$
The standard deviation of X is
$\sigma=\frac{1}{\sqrt 12}(4.4-3)=\frac{7\sqrt 3}{30} \approx0.404$
One standard deviation above the mean indicates the length of $4.4+0.404=4.804$
$P(T\geq4.804)=\int^{4.804}_{3}\frac{5}{7}dt=\frac{5}{7}t|^{4.804}_{3}\approx1.289$
