# 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?

• Have you looked up the normal approximation to the binomial? It will give you values for the expected mean and standard deviation. Then compute the Z score... Apr 6, 2018 at 14:45
• The Binomial Distribution with $n=75$ and $p=0.2$ is correct, but the exercise asks you to use the Normal approximation (of the Binomial Distribution) instead. Apr 6, 2018 at 14:45
• Binomial probability:P(X = x)=0.10171948927 is this correct? Apr 6, 2018 at 14:46
• I also get $0.1017$ for exactly $13$ correct answers, so that's probably right. But that's not what the question asks about. Apr 6, 2018 at 14:57
• I get $0.3325$. Although the exact answer seems to be $0.3414$. Apr 6, 2018 at 15:25

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.