A multiple-choice examination consists of $75$ questions- Probability A multiple-choice examination consists of $75$
 questions, each having possible choices a, b, c, d, and e. Approximate the probability that a student will get at most $13$
 answers correct if she randomly guesses at each answer. (Note that, if she randomly guesses at each answer, then the probability that she gets any one answer correct is $0.2$.) Use the normal approximation to the binomial with a correction for continuity.
I tried a bit 
Binomial Problem with n = 75 P(correct answer)=0.2
Binomial probability:P(X = x)=$0.10171948927$
I am I correct?
 A: You are correct that you need to look at $X \sim \mathsf{Binom}(n = 75, p = .2).$
Then the desired probability is $P(X \le 13).$ The normal approximation will be based on $\mu = np$ and $\sigma = \sqrt{np(1-p)}$ and the continuity correction
will use $P(X \le 13.5).$
The exact binomial answer from R statistical software is 0.3414, which would be tedious to find by using the formula for the binomial PDF and
adding probabilities $P(X=0) + P(X=1) + \cdots + P(X=13).$
pbinom(13, 75, .2)
## 0.3414021

A normal approximation using R gives 0.3325, but you will get a slightly
different approximation if you standardize and use printed normal tables.
n = 75; p=.2; mu=n*p; sg=sqrt(n*p*(1-p))
pnorm(13.5, mu, sg)
## 0.3325028

In this problem the normal approximation is rough, you can see from the graph below that the normal approximation is a little too small. I will leave details of computing the normal approximation from tables to you. Here is a start:
$$P(X \le 13) = P(X \le 13.5) = 
P\left(\frac{X - \mu}{\sigma} \le \frac{13.5-15}{3.464}\right)
\approx P(Z \le ??) = ??$$

Note: As @David comments, you correctly found $P(X = 13),$ but that is
not the answer to the question.
