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

I was able to show the mean. Taking an approach in which I created a random variable, $X$ that was equal to $\sum_1^{100} X_i$ where $ X_i $ is a number 1 or 0, where the probability of it being 0 is $1-p_i$ and the probability of 1 is $ p_i $. I know that means the expected value is $(0)(1-p_i) + (1)(p_i) = p_i $, but I cannot figure out how to do the variance.

Wendell Blatt
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The total mark is $\sum_1^{100} X_i$, where the $X_i$ is the amount we get on Question $i$.

On the perhaps unreasonable assumption that the $X_i$ are independent, the variance of the sum $\sum X_i$ is the sum of the variances of the $X_i$.

Recall that the variance of a random variable $Y$ is $E(Y^2)-(E(Y))^2$.

You already found $E(X_i)$.

But let us repeat the calculation you made, in case you got the right answer for the wrong reason. We have $X_i=1$ with probability $p_i+\frac{1}{5}(1-p_i)$, and $X_i=-\frac{1}{4}$ with probability ${4}{5}(1-p_i)$. So $E(X_i)=p_i$.

Now we calculate $E(X_i^2)$. Note that $X_i^2=1$ with probability $p_i+\frac{1}{5}(1-p_i)$, and $X_i^2=\frac{1}{16}$ with probability $\frac{4}{5}(1-p_i)$.

So now we can find the variance of $X_i$. Add up for the variance of the total grade.

André Nicolas
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