#### Answer

$\mu=0.25$
$\sigma=0.25$
$P(\mu \leq t \leq \mu + \sigma) \approx 0.2325$

#### Work Step by Step

We are given $f(x)=4e^{-4t}; [0;\infty]$
The mean of the distribution:
$\mu=\frac{1}{4}=0.25$
The standard deviation of X is
$\sigma=\frac{1}{4}=0.25$
The probability of the random variable between 1 standard deviation above the mean:
$P(\mu \leq t \leq \mu + \sigma) = P(0.25 \leq t \leq 0.5) =\int^{0.5}_{0.25}4e^{-4t} =-e^{-4t}|^{0.5}_{0.25} \approx 0.233 \approx 0.2325$