Answer
$0.86218$
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:
$P(\dfrac{30-50}{10} \leq Z\leq \dfrac{62-50}{10} )=\dfrac{1}{\sqrt 2 \pi}\int_{\frac{30-50}{10}}^{\frac{62-50}{10}} \exp(\dfrac{-t^2}{2}) \ dt$
or, $P(\dfrac{30-50}{10} \leq Z\leq \dfrac{62-50}{10} )=0.86218$