Answer
$0.38292$
Work Step by Step
Our aim is to find the probability of normal distribution given the mean and standard deviation.
Write the equation for standard normal curve.
$P(a \leq Z\leq b)=\int_a^b \exp(\dfrac{-t^2}{2}) \ dt$
where, $Z=\dfrac{x-\mu}{\sigma}$
Plug in the above equation the given values to obtain:
$P(-0.5 \leq Z \leq 0.5 )=\dfrac{1}{\sqrt 2 \pi}\int_{-0.5}^{0.5} \exp(\dfrac{-t^2}{2}) \ dt$
or, $P(-0.5 \leq Z \leq 0.5 )=0.38292$