Answer
a. $\mu=\frac{23}{8}$
b. $\sigma\approx0.072$
c. 1.288
Work Step by Step
We are given $f(x)=4; [2.75;3]$
The mean of the distribution:
$\mu=\frac{1}{2}(2.75+3)=\frac{23}{8}$
The standard deviation of X is
$\sigma=\frac{1}{\sqrt 12}(3-2.75)=\frac{\sqrt 3}{24} \approx0.072$
One standard deviation above the mean indicates the length of $3+0.072=3.072$
$P(T\geq3.072)=\int^{3.072}_{2.75}4dt=4t|^{3.072}_{2.75}\approx1.288$
