Answer
$0.068461$
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:
$P(\dfrac{70-100}{15} \leq Z\leq \dfrac{80-100}{15} )=\dfrac{1}{\sqrt 2 \pi}\int_{\frac{70-100}{15}}^{\frac{80-100}{15}} \exp(\dfrac{-t^2}{2}) \ dt$
or, $P(\dfrac{70-100}{15} \leq Z\leq \dfrac{80-100}{15})=0.068461$
