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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 $i$.

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is it correct you use the same variance twice? (the word 'but' suggests otherwis) –  long tom Mar 23 '13 at 21:07
    
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 $100p_i$. –  Joe Z. Mar 24 '13 at 3:45

2 Answers 2

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?

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Sorry, I copied the question wrongly. Check my edited version. –  user54609 Mar 24 '13 at 13:13

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

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This answer is quite unclear. –  user54609 Mar 23 '13 at 22:43

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