Answer
$0.19146$
Work Step by Step
Our aim is to find the probability under the standard normal curve.
Write the equation for standard normal curve.
$P(a \leq Z\leq b)=\int_a^b \exp(\dfrac{-t^2}{2}) \ dt$
Plug in the above equation the given values to obtain:
$P(0 \leq Z\leq 0.5)=\dfrac{1}{\sqrt 2 \pi}\int_0^{0.5} \exp(\dfrac{-t^2}{2}) \ dt$
or, $P(0 \leq Z\leq 0.5)=0.19146$