Answer
$\mu=3$
$\sigma=3$
$P(\mu \leq X \leq \mu + \sigma)\approx 0.2325$
Work Step by Step
We are given $f(x)=\frac{e^{-t/3}}{3}=\frac{1}{3}e^{-t/3}$
with $[0;\infty)$
The mean of the distribution:
$\mu=\frac{1}{\frac{1}{3}}=3$
The standard deviation of X is
$\sigma=\frac{1}{\frac{1}{3}}=3$
The probability that the random variable is between the mean and 1 standard deviation above the mean:
$P(\mu \leq X \leq \mu + \sigma)=\int^{3}_{6}(\frac{e^{-t/3}}{3})dt=\frac{1}{3}.(-3)e^{-t/3}|^{6}_{3}\approx 0.2325$
