Calculate probability using the central limit theorem A multiple choice test consists of 25 questions with 3 answers for every question. Only one of the three answers is correct for every question.
With which probability can someone, that is only guessing the test, get a positive grade, when more than half of the questions must be answered correctly to pass the test?
We have to use the central limit theorem.
As the questions are independant from each other, I think it should be possible to approximate it by: $$ P(a \le S_n \le b) \sim \phi\left( \frac{b+\frac{1}{2}-np}{\sqrt{np(1-p)}}  \right) - \phi\left( \frac{a-\frac{1}{2}-np}{\sqrt{np(1-p)}}  \right).$$
We've $25$ questions, and every answer can be correct with a probability of $ \frac{1}{3}$ as there are $3$ answers given for every question. So I got the values: $n=25$, $p = \frac{1}{3}$, $a=13$, and $b=25$.
When I calculate it with these values, I get $\phi(7.28)-\phi(1.77)$ which results in $3.84\%$ and doesn't seem to be correct. So how can I do it the right way?
 A: The number $X$ of correct questions by guessing has $X \sim \mathsf{Binom}(n=25, p=1/3).$ You seek $P(X \ge 13) = 1 - P(X \le 12).$ You seem to be roughly on
the right track using the normal approximation.
The exact binomial value for $P(X \ge 13) = 0.0415$ from R statistical software
1 - pbinom(12, 25, 1/3)            # 'pbinom' is CDF
## 0.04151368
x = 13:25; sum(dbinom(x,25,1/3))   # 'dbinom' is PDF
## 0.04151368

If you use the continuity
correction with the normal approximation [that is, start with $P(X \ge 12.5)$]
you should be able to match the exact binomial value to about two decimal
places.
[Note: BTW, yhe exact binomial value of $P(X=13)$ from software is  0.02513952.]
I will look at this page again in several hours to see your progress.

Later: I checked and it does seem that the normal approximation is
$1 - \Phi(1.767767)$ $=  0.03854994 \approx 0.0385,$ using software
and thus carrying more places of accuracy than you could get with
printed normal tables. 
Therefore, the exact binomial answer to two places is $.04$
and the normal approximation gives $.04$ to two places. So you are
about as close is can usually be expected with a normal approximation.
The figure below compares the exact binomial distribution (vertical bars)
with the normal distribution $\mathsf{Norm}(\mu=8.333333, \sigma=2.357023)$ that matches means and standard deviations. The binomial probability of interest is the
sum of the heights of the vertical bars to the right of the dotted red line,
and the normal approximation (with continuity correction) is the area
under the density curve to the right of that line. Within the resolution of
the graph, the two probabilities seem very close.

My only remaining question is: What 'answer' were you given that makes you
believe your normal approximation is incorrect? Everything seems to match
as closely as should be expected.
