Answer
$0.912737$
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:
$P(-1.71 \leq Z\leq 1.71 )=\dfrac{1}{\sqrt 2 \pi}\int_{-1.71}^{1.71} \exp(\dfrac{-t^2}{2}) \ dt$
or, $P(-1.71 \leq Z\leq 1.71 )=0.912737$
