Answer
a. $P(0\leq X \leq 1) \approx 0.098$
b. $P(1 \leq X \leq 3) \approx 0.096$
c. $P(X\geq 2)\approx 0.092$
Work Step by Step
We find:
a. $P(0\leq X \leq 1)$
$=\frac{1}{2}\int^{1}_{0}e^{-x/2}dx$
$=-\frac{1}{4}e^{-\frac{1}{2}x}|^{1}_{0}$
$\approx 0.098$
b. $P(1 \leq X \leq 3)$
$=\frac{1}{2}\int^{3}_{1}e^{-x/2}dx$
$=-\frac{1}{4}e^{-\frac{1}{2}x}|^{3}_{1}$
$\approx 0.096$
c. $P(X\geq 2)$
$ =\int^{\infty}_{2}\frac{1}{2}e^{-x/2}dx$
$=\frac{1}{2}\int^{\infty}_{2}e^{-x/2}dx$
$=-\frac{1}{4}e^{-\frac{1}{2}x}|^{\infty}_{2}$
$\approx 0.092$
