normal approximation with continuity correction a fair die is rolled 100 times. What is the probability that "6" appears more than 15 times? Use the normal approximation with continuity correction.
I've found the mean to be $100/6$ or $50/3$ and $\sigma$ as $3.727$ but I'm unsure of how to use continuity correction? Do I do a $Z$-Test ? 
 A: Let random variable $X$ be the number of $6$'s. Then $X$ has mean $\mu=100\cdot \frac{1}{6}$ and variance $\sigma^2=100\cdot \frac{1}{6}\cdot\frac{5}{6}$.
The random variable $X$ is a sum of a fairly large number of reasonably nice independent identically distributed random variables, so the cdf of  $X$ is reasonably well approximated by the cdf of a normal $W$ with the same mean and variance.
The probability that $X\le 15$ is approximately the probability that $W\le 15$. One can expect a better approximation by the probability that $W\le 15.5$. (This is the continuity correction.) We have
$$\Pr(W\le 15.5)=\Pr\left(Z\le \frac{15.5-\mu}{\sigma}\right),$$
where $Z$ is standard normal.
Compute. If the resulting probability is $p$, then our required probability is approximately $1-p$.
Remark: It is unfortunately quite possible to make an "off by $1$" error when using the continuity correction. To avoid that, it is useful to reduce our problem always to one of the shape $\Pr(X\le n)$, where $n$ is an integer, and to remember that this is approximately the probability that the approximating normal is $\le n+\frac{1}{2}$. 
Alternately, if we are using the normal $W$ to approximate a random variable $X$ that only takes on integer values $k$, use the fact that $\Pr(X=k)\approx \Pr(k-\frac{1}{2}\le W\le k+\frac{1}{2}$.
A: You have $\mu = \dfrac{100}{6}$ and $\sigma = \sqrt{\dfrac{500}{36}}$, and that is okay. $\checkmark$
You are thus looking for $\mathsf P\left(Z\gt \dfrac{15.5-\mu}{\sigma}\right)$, where $Z\sim\mathcal N(0,1^2)$, as your Normal Approximation value.
Plug in your values and use your Z-tail tableau, online calculator, mathematical package, or whatever.
