Central Limit Theorem (Normal Approximation to Binomial) A fair six-sided die is rolled 100 times. Let $X$ be the number of rolls that the face showing a six turns up. Find $P(15 \le X \le 20).$
(a) 0.4871
(b) Approximately 0
(c) 0.0054
(d) 0.2310
Ok I did the computation and got 0.561. Where did I go wrong?
 A: Perhaps, by trusting that the right answer is one of the four provided.
Let $X$ be the number of 6's in 100 rolls. Then $X \sim Binom(n=100, p=1/6).$
I got $\mu = E(X) = 16.667$ and $\sigma = SD(X) = 3.727.$
Then 
$$P(15 \le X \le 20) = P(14.5 < X < 20.5) \approx 
P((14.5-\mu)/\sigma < Z < (20.5 - \mu)/\sigma) = 0.5677,$$
where $Z$ is standard normal. 
Computations in R: Because I used software instead of printed tables,
there may be differences due to rounding. (Or maybe there's a typo I didn't
see.) The normal approximation
may not be at its best for $p = 1/6$ even with $n = 100.$ When I did
the exact binomial computation, I got 0.5607.
mu = 100/6
sg = sqrt(100*5/36)
mu;  sg
## 16.66667
## 3.72678
diff(pnorm(c(14.5,20.5), mu, sg))
## 0.5676711
diff(pbinom(c(14,20), 100, 1/6))
## 0.5606912

But my best guess is that you did a continuity correction, and the
author of the problem didn't.
diff(pnorm(c(15,20), mu, sg))
## 0.4870929

Below is a relevant figure showing exact binomial probabilities and
the approximating normal curve.

