2
$\begingroup$

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?

$\endgroup$
12
  • $\begingroup$ 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... $\endgroup$
    – Ross Millikan
    Apr 6, 2018 at 14:45
  • $\begingroup$ 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. $\endgroup$
    – Jimmy R.
    Apr 6, 2018 at 14:45
  • $\begingroup$ Binomial probability:P(X = x)=0.10171948927 is this correct? $\endgroup$
    – David
    Apr 6, 2018 at 14:46
  • 1
    $\begingroup$ I also get $0.1017$ for exactly $13$ correct answers, so that's probably right. But that's not what the question asks about. $\endgroup$
    – Arthur
    Apr 6, 2018 at 14:57
  • 1
    $\begingroup$ I get $0.3325$. Although the exact answer seems to be $0.3414$. $\endgroup$
    – Arthur
    Apr 6, 2018 at 15:25

1 Answer 1

Reset to default
1
$\begingroup$

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:

$$P(X \le 13) = P(X \le 13.5) = P\left(\frac{X - \mu}{\sigma} \le \frac{13.5-15}{3.464}\right) \approx P(Z \le ??) = ??$$

enter image description here

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.