A multiple choice exam has 20 questions, each with 4 answers, only one is correct. The students are guessing the answers in an independent way.

Question

Correct answer = +5, incorrect answer = -1, unaswered = 0. What are the expected value of the grades and their variance? What is the probabilty to get the same grade as the expected value?

The answers are: 10, 135, 0.2. I tried binomic distrubiton: p = 0.125 (0.5 if the students choose to answer the question and 0.25 if they're right) and the grades range from 0 to 100, but I don't get the correct answers. I'm assuming it's not Continuous uniform distribution, since the E(X) doesn't match as well.

Thank you.

Answer 1

Hints:

• based on the three desired official solutions, it seems that the students are answering every question, and picking answers at random

• so your binomial distribution for the number of correct answers should be $X\sim \text{Bin}(20, \frac14)$. All other answers, i.e. $20-X$, will be incorrect.

• You can find the expected value and variance of $X$

• You can find the marks $M$ as a function of $X$

• From this you can find the expected value and variance of $M$

• You can find which value of $X$ would make $M=\mathbb E[M]$ and the probability of this X

Answer 2

For $i=1,\dots,20$ let $Y_i$ denote the score for question $i$.

Preassuming that the student gives an answer to question $i$ we find:$$\mathbb EY_i=\frac14\times5+\frac34\times(-1)=\frac12$$

Note that this is positive, so a student must be out of his mind if he does not answer questions. Let us assume not with good reason that the student answers every question.

For finding variance be aware that $X_i=\frac16Y_i+\frac16$ has Bernoulli distribution with parameter $p=\frac14$ and consequently $X=\sum_{i=1}^{20}X_i=\frac16X+\frac{10}3$ has binomial distribution with parameters $p=\frac14$ and $n=20$.

(Observe that $X$ actually equals the number of questions that are answered correctly)

So: $$\mathsf{Var}X=np(1-p)=\frac{15}4$$ and from that we conclude for $Y=6X-20$ that: $$\mathsf{Var}(Y)=6^2\times\frac{15}4=135$$ Further we have $Y=10\iff X=5$ so that:$$P(Y=10)=P(X=5)$$where the RHS is easy to find.