1
$\begingroup$

In a test with 100 multiple choice questions, a student chooses to pick an option randomly. There are four options for each questions. Answering a question correctly will give the student 1 mark, while picking a wrong option will give the student a pentalty of - 0.25 marks. There is only one correct answer for each question. What is the probability that a student will get negative marks if they follow such a strategy?

I'm stumped at even trying to model this problem. I do observe a few things, but I can't make any substantial arguments from them:

  • We could have a random variable, X represent the marks a student gets while answering the 100 questions that way.
  • The range of values this variable can take is from -25 (all questions wrong) and 100 (all correct).
  • The expected value of this random variable is 6.25
  • How do I find the variance?
  • If we find the variance can we assume a normal distribution and see the area in the lower tail, below 0?
$\endgroup$
  • $\begingroup$ Find the distribution of random variable $X$ that counts the number of correct questions would be a good start. Then, if $X = k,$ what will the score be? What value of $X$ just barely gives a positive score. Etc. $\endgroup$ – BruceET May 26 at 14:16
  • 1
    $\begingroup$ Thanks. Can I take this random variable (that counts the no of correct questions) to be binomial? (with p, the probability of success as .25 and q, the probability of failure as .75). Then I can take the mean and variance of this variable and find the cumulative probabilities (or approximate with a normal). $\endgroup$ – WorldGov May 26 at 14:19
  • 1
    $\begingroup$ In R, pbinom(19, 100, .25) returns 0.09953041. What do you get using normal approx.? On this site it is permitted to answer your own question. $\endgroup$ – BruceET May 26 at 14:34
  • $\begingroup$ I've answered the question. Thank you! :) $\endgroup$ – WorldGov May 26 at 19:28
1
$\begingroup$

One way to model this is:

Instead of taking X to represent the score student scores, let X represent a random variable that represents how many answers a student gets right. Let this be a binomial distribution with $ p $, the probability of success $0.25$. Now, we should look for the number of questions one would need to get right to get a score that's close to being negative.

If the student gets 20 answers right, his score would be 0. So, we need to find the probability of X being lesser than 20. This probability distribution has mean and variance: $$ \text{Mean} = n * p = 100 * 0.25 = 25 $$

$$ \text{Variance} = n * p * (1 - p) = 18.75 $$

We can now use approximate this distribution with a normal distribution. We need to find the probability of X being lesser than 20. Since 20 has a $ z $-score of -1.16, we need to find the area under the standard normal curve in the tail to the left of $ - 1.16 $.

And that gives us the probability: $ .1230 $

$\endgroup$
  • 1
    $\begingroup$ Right idea (+1), but a bit of improvement is possible. For the best normal approximation, you should use a continuity correction: $P(X \le 19) = P(X < 19.5) = P(X < 20),$ so use $z = (19.5 - 25)/\sqrt{18.75} = -1.27.$ and $P(Z \le -1.27) = .1020,$ which is close to the exact binomial value .0995 in my previous comment. $\endgroup$ – BruceET May 26 at 22:12

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.