Answer
$0.13983$
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: $\sigma=\sqrt{npq}=\sqrt {200 \times \dfrac{1}{6} \times \dfrac{5}{6}}=5.2705$ and $\mu=np=200 \times \dfrac{1}{6}=33.333$
Plug in the above equation the given values to obtain:
$P(a \leq X \leq b )\implies P(a-0.5 \leq Y\leq b+0.5)$
Now, we have: $ P(\dfrac{0-33.333}{5.2705}-0.5 \leq Z \leq \dfrac{0-33.333}{5.2705}+0.5 )=0.13983$
