Answer
$0.51$
Work Step by Step
Write the equation for standard normal curve.
$P(a \leq Z\leq b)=\int_a^b \exp(\dfrac{-t^2}{2}) \ dt$
where, $Z=\dfrac{x-\mu}{\sigma}$
For binomial distribution, we have: $\sigma=\sqrt{npq}=\sqrt {100 \times \dfrac{1}{6} \times \dfrac{5}{6}}=3.7268$ and $\mu=np=100 \times \dfrac{1}{6}=16.667$
Plug in the above equation the given values to obtain:
$P(\dfrac{10-16.667}{3.7268}-0.5 \leq Z \leq \dfrac{15-16.667}{3.7268}+0.5 )=0.51$
