5
$\begingroup$

A multiple choice exam has 100 questions, each with 5 possible answers. One mark is awarded for a correct answer and 1/4 mark is deducted for an incorrect answer. A particular student has probability $p_i$ of knowing the correct answer to the $i$th question, independently of other questions.

a) Suppose that if a student does not know the correct answer, he or she guesses randomly. Show that his or her total mark has mean $\sum p_i$ and variance $\sum p_i (1-p_i)+\frac{(100-\sum p_i)}{4}$.

b) Show that the total mark for a student who refrains from guessing also has mean $\sum p_i$ but with variance $\sum p_i (1-p_i)$.

b) is pretty easy, since it's a simple application of the binomial distribution. I can't really get a) though, mainly since I can't come up with a good-looking expression for the expected score of question number $i$.

$\endgroup$
  • 2
    $\begingroup$ is it correct you use the same variance twice? (the word 'but' suggests otherwis) $\endgroup$ – long tom Mar 23 '13 at 21:07
  • $\begingroup$ Guessing randomly has an expected gain of $0$, so if a student has probability $p_i$ of knowing an answer, she has an expected mark per question of $p_i$, for a total of $100p_i$. $\endgroup$ – Joe Z. Mar 24 '13 at 3:45
1
$\begingroup$

For part a $X_i$: score on question $i$.

$P(X_i=1)=p_i+\frac{1-p_i}{5}=\frac{1}{5}+\frac{4}{5}p_i$

$P(X_i=-\frac{1}{4})=(1-p_i)*\frac{4}{5}=\frac{4}{5}-\frac{4}{5}p_i$

So

$\mu=\Sigma(\frac{1}{5}+\frac{4}{5}p_i-{1\over4}(\frac{4}{5}-\frac{4}{5}p_i))=\Sigma p_i$

And

$\sigma=\Sigma((\frac{1}{5}+\frac{4}{5}p_i)(1)^2+(\frac{4}{5}-\frac{4}{5}p_i)(-\frac{1}{4})^2-p_i^2)$

$\sigma=\Sigma(1-p_i)(\frac{1}{4}+p_i)$

Which is not the result given in the question - is the question wrong?

$\endgroup$
  • $\begingroup$ Sorry, I copied the question wrongly. Check my edited version. $\endgroup$ – ithisa Mar 24 '13 at 13:13
-1
$\begingroup$

$X_i$: score on question $i$.
$P(X_i=1)=p_i+\frac{1-p_i}{5}=\frac{1}{5}+\frac{4}{5}p_i$.
$P(X_i=-\frac{1}{4})=(1-p_i)*\frac{4}{5}=\frac{4}{5}-\frac{4}{5}p_i$

$\endgroup$
  • 3
    $\begingroup$ This answer is quite unclear. $\endgroup$ – ithisa Mar 23 '13 at 22:43

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.