#### Answer

$\mu=10$
$\sigma =10$
$P(\mu \leq X \leq \mu + \sigma)\approx 0.2325$

#### Work Step by Step

We are given $f(x)=0.1e^{-0.1x}$
with $[0;\infty)$
The mean of the distribution:
$\mu=\frac{1}{0.1}=10$
The standard deviation of X is
$\sigma=\frac{1}{0.1}=10$
The probability that the random variable is between the mean and 1 standard deviation above the mean.
$P(\mu \leq X \leq \mu + \sigma)=\int^{20}_{10}(0.1e^{-0.1t})dt=-e^{-0.1t}|^{20}_{10}\approx 0.2325$